LOCUS
1
Tips on how to use the study-material effectively
Read the following lines carefully to understand how you can derive the maximum benefit out of this study-material. (Don’t just read them! Follow them sincerely, and you should have an excellent chance of succeeding at the JEE). 1) Give a general general reading to the study-material study-material before coming to the class for this topic. This will ensure ensure that you have at least an overview of what is being taught in the class and you can follow the classroom discussion in a more effective effe ctive manner. 2) After the class, class, go through through this material again, this time time in detail, observing observing all the the fine points, working out the examples on your own, completing completi ng the exercises sincerely, and pondering over the subject matter. This will complement what you have studied in the class. Particularly Parti cularly about the exercises, if you cannot solve any particular problem, don’t just give up and ask your teacher for the solution! Keep trying tr ying and you are sure to hit upon the solution sol ution sooner or later. 3) After going through through the study-mate study-material rial in the detailed fashion fashion described described above, above, reread your your classroom classroom notes and solve on your own all the problems discussed in the class. This will ensure your complete mastery over the topic. 4) From an examination examination point point of view, view, take self-tests self-tests at home on this topic so that you actually get the feel on how to solve questions on this particular topic in i n an exam-like situation. This will additionally addit ionally help in building your examination temperament. 5) Prepare for for the monthly monthly tests seriously seriously,, as you would prepare prepare for an an actual exam. 6) Every few days, days, pick up up this study-material study-material and go go through it again, again, so that that the topic remains remains strong and and fresh in your memory, and you are always exam-ready in relation to this topic. 7) Lastly, Lastly, and very importantly importantly,, try to frame new questions questions on your own (and (and also solve them!), them!), by modifying modifying the questions discussed in this study-material. Have discussions with your friends on these ‘new’ questions. This will really help in expanding expandi ng your mind’s horizons and give you hours of intellectual pleasure.
KINEMATICS
LOCUS
2
Kinematics REFERENCE FRAME: A reference Frame is the collection of reference axes (i.e., coordinate axes) and a reference point (i.e. origin). Every direction and distance would be defined with respect to reference directions direct ions and reference point respectively. Observations made in one frame may or may not be true tr ue for a different frame. fr ame. A 2-dimensional reference frame : y O → reference point x, y → reference directions
x
O (origin)
REST AND MOTION: The state and character of motion of any particle are not absolute quantities and actually depend on the observer. observer. In other words we can say that the state of rest or motion of any particle are relative entities and can not be defined absolutely. To To put precisely, there exists no universal, fixed or absolute frame of reference. ref erence.
POSITION: The position of any particle P in a given reference frame is characterized by a length (distance from origin) and a direction (angle with x or y axis, generally with x-axis). Therefore, position is a vector quantity and the vector
defining the position of a particle is known as position as position vector r of that particle. y y
P
P
r
θ
O
≡
r
x
x O
Now, you can note that to get the position vector of any point you y ou have to join that point to the origin. or igin. DISPLACEMENT :
Suppose that a particle P moves from A to B on the path shown in the figure below, then the displacement the displacement for this duration is defined as the change in the position of the particle. Therefore, KINEMATICS
LOCUS
3
displacement = change in position = ∆r
= rf − r i (a vector quantity)
y
A ∆ s
∆ →
r = → r f – → r i
B
r i r f
x
O
Therefore, the displacement vector is obtained by joining the final position to the initial position.
DISTANCE: Distance, Distance,
∆s, is defined as the actual actual length of the path traversed. traversed. (a scalar quantity).
Here you should note that from point A to B there ther e are many (in fact, infinite) infinit e) possible paths but for all paths displacement is same y ∆ s3
A
∆ s1 ∆
r
∆ s2
B x
O Note:
*
| ∆ r | ≤ ∆ s i.e. magnitude of displacement < distance.
*
Equality holds Equality holds if particle particle move movess from from A to B on a strai straight ght line line and it moves moves cont continuo inuously usly alon along g same direction.
AVERAGE VELOCITY & AVERAGE SPEED: Average velocity is defined as the average rate rat e of change in position w.r.t. w.r.t. time, i.e. i. e.
average velocity veloci ty,, v =
KINEMATICS
∆r rf − r i = = (vector) ∆t ∆t
LOCUS
4
Note:
* Aver verag agee veloc velocity ity is alwa always ys alon along g
∆r (displacement).
Average speed spe ed is defined as the average rate of covering distance i.e. y distance covered average speed, v = time taken
=
∆s ∆t
)
A ( t 0 v
∆ s
∆
r
(scalar
∆r ∆s | ∆r | ≤ ∆s ⇒ ≤ ⇒ ∆t ∆t
B (t +∆t ) 0
Note:
*
| v |≤ v
x
i.e. (magnitude of average velocity) < (average speed)
O
INSTANTANEOUS VELOCITY : Suppose that a particle is moving in such a way that its average velocities velociti es measured for a number of different time intervals do not turn out to be constant. This particle is said to be moving with variable vari able velocity. Then, we must seek to determine a velocity of the t he particle at any given instant of time, t ime, called, instantaneous velocity. veloci ty. It is defined as an instantaneous rate of change in position. posit ion. While defining average velocity if we decrease ∆t , i.e., letting B approach A, we find that although the displacement
becomes extremely small, the ratio of ∆r &
∆t is not necessarily a small quantity. Similarly, Similarly, while growing
approaches a limiting direction of the tangent to the path of the ∆t → 0), the displacement vector approaches particle at A. This limiting value of ∆r ∆t is called the smaller (i.e.
y
A
∆
r
B x
O instantaneous velocity at the point A, or the velocity of the particle at the instant t o.
∴ velocity, v = lim ∆ r = d r ∆t → 0 ∆t dt
Note:
y
→
dr
→
v
*
d r is displacement for the time interval dt
*
d r is always along tangent to the path
*
v is always along d r i.e. along tangent to the path.
*
Derivative of position w.r.t. w.r.t. time gives gi ves velocity.
KINEMATICS
→
v O
x
LOCUS
5
INSTANTANEOUS SPEED: We can define it exactly in the previous manner. It is defined as instantaneous rate of covering distance. speed (also called speed), Therefore, instantaneous speed (also
∆s ds = ∆t → 0 ∆t dt
v = lim
Note:
*
i n the time interval dt . ds is the distance travelled along the path in
*
Displacement d r is also along the path, therefore, | d r |= ds
*
Deri De riva vati tive ve of dis dista tanc ncee w.r w.r.t .t tim timee give givess spe speed ed..
v
*
= dr ⇒ dt
|v |
=
| dr | | dt |
= ds = v dt
i.e. magnitude of velocity = speed If v be the speed and t tˆ be the unit vector along the tangent to the path at some moment of time,
*
then at that t hat moment velocity, v
= v . t ˆ
ACCELERATION: For a given time interval ∆t , average acceleration is defined as average rate of change in velocity velocit y, i.e., average acceleration,
< a >=
change in velocity time taken
∆v = ∆t
=
v f
t f
− vi − t i
At any given instant of time t o, instantaneous acceleration (or acceleration) is defined as instantaneous instant aneous rate of change in velocity, i.e.
∆v dv lim = ∆t →0 ∆t dt
acceleration, a =
Example : 1 A point traversed half a circle cir cle of radius R with constant speed u (as shown in figure). Find : (a) Duratiton of motion, t o ; (b)) displacement for the time interval [0, t 0], assuming that at t = 0 particle is at origin; (b (c) average velocity for the time intervals [0, t 0] and [0, t 0 /2];
y
(d)) magnitude of average velocity for the time interval [0, t 0 /2]; (d
j^
speed = u (constant)
(e) average acceleration for the time interval [0, t 0].
O KINEMATICS
x B i →^
LOCUS
6
Solu So luti tion on:: (a (a)) As speed is constant, its average value for any interval would be same,
∴
⇒
av.speed =
u=
π R
⇒
t0
t 0
=
π R u
(b)) at t = 0 particle was at origin and at t = t 0 particle is at point B, so the vector from O to B (b would give the required r equired displacement. displacement.
∆r = 2 Riˆ
y
Therefore, for [0, t 0] :
j^
(c) for [0, t 0] : v
∆r 2Riˆ = ∆t t 0
=
O
2 R ˆ 2u ˆ i i = π R = π u
∆r
B ^ i →
∆r Riˆ + Rˆj = ∆t t 0 2
for [0, t 0 /2] : v =
y t /2 • 0
=
2 R ˆ ˆ (i + j ) π R
u
r
∆
O
2u ˆ ˆ (i + j )
=
π
(d)) for [0, t 0 /2]: Magnitude of average velocity = | v | (d
=
2u ˆ ˆ 2 2u |i + j | =
π
π
∆v = ∆t
(e) for [0, t 0]: a =
v (t0 ) − v (0)
=
KINEMATICS
t 0
−0
v f t f
− vi − t i
R^ i
R j^
x
x
LOCUS
7
(−u ˆj ) − (u ˆj ) π R = u
y
vi 2 2u ˆ j = − π R
O
x
B v f
∴ Magnitude = (2u 2 ) /(π R ) Example : 2 For a moving particle, if the position posit ion vector as a function of time is given as, r (t ) = α tiˆ − β t 2 ˆj , where α and β
are positive constants, then find :
(a) velocity as a function of time, v (t ) ; (b)) acceleration as a function of time , a (t ) (b
(c) v & a for the time interval [t 1, t 2]
(d)) position of the particle when its speed is minimum. (d
Solu So luti tion on:: (a (a)) v (t ) =
dr
=
dt
dt = α iˆ dt
(b)) a (t ) = (b
dv dt
2 d (α ti t iˆ − β t ˆj )
dt
(c) v =
dt
=d
(α iˆ − 2β tˆj )
r (t2 ) − r (t1 )
dt
t1 − t 1
=
=d
(α iˆ) dt
−d
KINEMATICS
−2β ˆj
( 2β tˆj ) dt
α (t2 − t1 )iˆ + β (t12 − t 22 ) ˆj
a = a (∵ a is constant)
=
dt
2
(t2 − t 1 )
= α iˆ − β (t2 + t1 ) ˆj
−
2 d (β t jˆ )
− β ˆj dt = α iˆ − 2β t ˆj dt
=
d (αti t iˆ)
= −2 β jˆ
LOCUS
8
(d)) speed, v = | v | (d
α 2 + 4 β 2 t 2
=
∴ vmin
= | (α iˆ − 2 β t ˆj ) |
= α (at t = 0)
∴when speed is minimum, the particle is at the origin TRY YOURSELF - 1 1.
For the situation given in example 1, find the magnitudes of average average velocity and average acceleration for the time interval [t 0 /2, t 0] .
2.
In 1.0 s, a particle goes from point A to point B, moving in a semicircle . The magnitude of the average velocity is: (a) 3.14 m/s (b) 2.0 m/s (c) 1.0 m/s (d) zero
3.
1 .0 m
A particle moves such that its position vector r at time t is given by:
r
= (3t 2 + 2t )iˆ + (2t 2 + 4t ) ˆj ; determine the magnitude and direction of its velocity and acceleration
at t = 5 sec. ***************
ACCELERATED MOTION:
Till now we have seen that if we have the knowledge of r then we can find v by differentiating it and by
differentiating v we can find the acceleration also. Now, Now, we will proceed in the reverse order. order. Using a we will try to find v and using v we will try to find r .
We have have,, dv
=a
dt
⇒ dv
= a.dt = change in velocity in time interval dt .
Therefore, the change in velocity for the time interval 0 to t is t
t
∆v = ∫ d v = ∫ a.dt
0
0
t
⇒
∫
v (t ) − v (0) = a.dt ⇒
0
t
∫
t
∫
v (t ) = v (0) + a.dt = u + a.dt
0
KINEMATICS
0
LOCUS
9
t ⇒ = u + ∫ a.dt ⇒ d r = u .dt + ∫ a.dt .dt dt 0 0 dr
t
= change in position duri during ng time interval dt .
⇒ ∆ r (displacement) for time interval [0, t]
t
=
t
∫
∫
t
d r = u.dt +
0
0
t
∫0 ∫ 0 a.dt .dtdt
t ⇒ r (t ) − r (0) = u.t + ∫ ∫ a.dt.dt 00 t t u t + ∫ ∫ a.dt .dt ⇒ r (t ) = ri + ut 00
t ⇒ r (t ) = r (0) + ut + ∫ ∫ a .dt .d t 0 0
t
t
t
Note:
*
∫
t t
v (t ) = u + a.dt & r (t ) = ri + ut +
0
*
∫ ∫
a.dt.dt
0 0
If mo moti tio on sta tart rtss fr fro om or oriigi gin n, th theen t t
r (t ) = u .t +
∫ ∫
a.dt.dt
0 0
*
If a is constant (i.e. uniform), then
v
= u + at
& r (t ) = ri + ut +
1 2
at 2
and, if motion starts from origin with uniform acceleration, acceleration, then
r (t ) = ut +
*
ds dt
=| v | ⇒
1 2
2
at
ds = | v | .dt
∫
t
t
∫ v .dt = ∫ (speed).dt
⇒
distance, ∆s = ds =
*
For uniform a , we have r (t ) = ri
0
0
Here we see that ut is along u and
1 2
ut + + ut
1 2
y
2
at
u
at 2 is along a , therefore,
0 t =
r i
we can assume that in the time duration durati on ‘t ’, ’, u contributes ut an and da
contributes KINEMATICS
1 2
2 at to the displacement of the particle.
a
O
u t
r
) ( t )
1 2
2
a t
t x
LOCUS
10
Example : 3 Two bodies were thrown simultaneously simult aneously from the same point: one, straight strai ght up, and the other, at an angle of θ = 60° to the horizontal. The initial velocity of each body is equal to u = 25 m/s. Neglecting the air drag, find the distance between the bodies t = 1.7 s later. Solution: At some time ‘t ’, ’, positions of particles parti cles are shown in figure. We have,
r1 (t ) = u1t +
1 2
gt 2
1 2 = 25t ˆj − gt ˆj 2
and an d r2 (t ) = u2t +
1 2
gt 2
1 2 cos θ iˆ + u sin θ ˆj )t + gt = (u co 2
g
∴
required distance
j^
y
(t)
r 1
= r2 − r 1
r 2 – r 1
u
u
2
1
25 ˆ 25 i+ 2 − 3 ˆj .t = 2 2
(
=
25 2
× (1)2 + ( 2 −
θ
)
3
O
) ( t )
r 2
i^ x
2
) ×1.7 = 22 m.
***************
MOTION IN ONE DIMENSION : Suppose a particle is moving along a particular straight line. In this case we need only one reference direction (parallel to line of motion) because particle will be moving either along this direct ion or opposite to the this direction, therefore, theref ore, this single direction can define all possible motions along the given line. In this case we will drop the t he arrow sign from the head of the vector symbols (for example. a will be denoted as a) because any vector can be either along reference direction or opposite to reference direction. If vector is along reference direction then we define it as +ve and if it is in the opposite direction of the reference direction, then we define it as –ve. Therefore, in this t his case direction can be defined with the help hel p of sign (+ve or –ve) only.
Reference frame : x
+ve x
direction
–ve x
direction
O
Reference Point (Origin)
KINEMATICS
Reference Direction
LOCUS
11
Notations: Position → x, velocity → v, speed = | v | , acceleration → a Definitions:
=
Av.vel. v
change in position time taken
=
∆ x ; ∆t ∆x = ∆ t → 0 ∆t
Instantaneous velocity or velocity vel ocity,, v = lim
Avv. acceleration, a A
=
dx ; dt
∆v ; ∆t ∆v dv = ; ∆t → 0 ∆t dt
Instantaneous acceleration or acceleration, a = lim SOME CASES:
v<0 a < 0
v > 0
a<0
O
x > 0
X
P
O
x > 0
X
P
particle v<0
P
v<0
x < 0
X O
P
x < 0
a>0 a<0
Note:
*
when x is increasing: v =
dx
*
when x is decreasing: v =
dx
*
when v is increasing: a > 0; when v is decreasing, a < 0
dt
(rate of change of x x) > 0
dt
(rate of change of x x) < 0
ACCELERATED MOTION : According to the analysis done in the t he previous section, t
t t
∫
v = u + adt , x(t ) = xi + ut + 0
0 0
t t
x( t) = ut +
∫ ∫ a. dt. dt (if x x = 0). i
0 0
KINEMATICS
∫ ∫ adt.dt,
O
X
LOCUS
12
If acceleration is uniform:
Example : 4 2 If the position of a moving particle, with respect to time be, x( t) = 1 + t − t , then, find:
(a) v(t ), ), a(t ) ; (b)) (b
v & a for the time interval [3, 2];
(c) displacement and distance travelled for the time interval [0, 2].
Solu So luti tion on:: (a (a)) v(t ) =
(b)) v (b
=
dx dt
= 1 − 2t ,
∆x = ∆t
a(t ) =
x(3) − x( 2)
3− 2
=
dv dt
= −2
(− 5) − (− 1) 1
= −4
a = a (∵ a is constant) = –2 (c) displace displacement ment = ∆ x= x(2) − x(0) = ( −1) − (1)
= −2 1
distance
=
2
2
0
0
∆s = ∫ | v | .dt = ∫ | 1 − 2t | .dt = 12
=
∫
2
(1 − 2t ).dt
0
+
∫
( 2t − 1).dt =
12
2
2
∫ | 1 − 2t | .dt + ∫ | 1− 2t | .dt 0
1
2
5 2
ALTERNATE METHOD: (for ∆s & ∆ x) : At t = 0: x = + 1 v=+1 a = –2
Here, we see that at t = 0 particle is at x = 1 and is moving along +ve x direction, but its accelera acceleration tion is along –ve x-direction, therefore, particle will move in + ve x direction with decreasing velocity and at some time (t = ½ sec) its velocity becomes zero. At the t = ½ sec the particle still has the negative acceleration, therefore, now it will start star t moving in the –ve x direction or we can can say that it reverses its direction of motion at the t = ½ sec. KINEMATICS
LOCUS
13
∴
∆ x = –2
for [0, 2] :
∆s = 1/4 + 1/4 + 2 =
5 2
a
–1 t =
O
t =
0
v
1
5/4
2
X
=0 t = 1/2 x = 5/4 v
path of motion
2
1/4
∆ x= –2
Example : 5 A particle moves in a straight line with constant acceleration. If it covers 10 m in the first second and 20 m in the next second, find its initial velocity and acceleration. accelerati on. Solution: Let v0 be the initial velocity and a be the acceleration.
Applying x = v0 t + 10 = v0 ×1 +
1 2
1 2
2
at for the first second of motion, we get,
× a × 12 ⇒
10 = v0 +
a
2
...(i)
and applying the same for the first two seconds of motion, we get, 30 = v0 × 2 +
1 2
× a × 22 ⇒
solving (i) & (ii), we get, v0
30 = 2v0 + 2 a
...(ii)
= 5 m/s and a = 10 m/s².
Example : 6 A ball is dropped from a tower of height h. Find its speed and duration of motion when it reaches bottom of the tower. Solution: Let the ball be dropped at the instant t = 0. In this case the point from where the ball is dropped is being considered as origin and the downward direction is being considered as the +ve x direction. If bott om of the tower, then applying v be the velocity of the ball at the bottom
⇒ KINEMATICS
v
2
= u 2 + 2ax, we get,
v
2
= 02 + 2 gh
v
=
2 gh
LOCUS
14
If the ball reaches the bottom at the instant t = t 0, then applying x= ut ut+ h = 0+
⇒
t0
=
1 2 1 2
O
t = 0 • u=0
2
at , we get + ve ve
X
g
h
gt 02
2h g
Example : 7 A stone is dropped from a balloon ascending with wit h v0 = 2 m/s, from a height h = 4.8 m. How much time will the t he ball take to reach the ground ? Solution: This case may be compared with the situation given in i n example 4.
Here the initial velocity velocit y is in the upward direction but acceleration is in the downward direction, therefore, firstly firstl y particle will move in the upward direction with decreasing speed. After some time its velocity will become zero and then it will start moving in the downward direction. Here, I have assumed the point of projection as the origin and upward direction dir ection as the +ve x direction (as shown in the figure). Applying x= ut ut+
1 2
+ve x
2
at , we get, v0
1
−4.8 = 2t0 − × 9.8t 02
t =0
2
⇒
9.8t
2 0
⇒
t 0 =
∴
t 0 =
g h
− 4t 0 − 9.6 = 0 − ( − 4) ±
O
16 − 4 × 9.8× (− 9.6)
t = t0
2 × 9.8 4 + 16 + 4 × 9.8 × 9.6 2 × 9.8
[ − ve time is not acceptable]
= 1.22 sec.
Example : 8 A particle moves on a straight line with deceleration whose magnitude magnitude varies as | a |= α v , where α is a +ve constant. The initial velocity of the particle is v0. What distance will it traverse before it stops? What time will it take to cover that distance ? KINEMATICS
LOCUS
15
Solution: We have, a = −α v
dv
⇒
dt
= −α
v
dv
⇒
v
= −α dt
⇒ ⇒
dx dt
= v0 − α
v0 t +
α 2t 2 4 a= α v0
⇒
∫
x
∫
t
∫
t
2
dx = v0 dt − α v0 t.dt +
0
0
α
0
4
t
∫
v
a=0 0
v=0
2
t .dt t = 0
0
t = t 0
x= 0
⇒
x = v0 t −
α v0 t 2 2
t 0
=
2 v0
α
x = v0 × 3 2 0
= v
+ α
t
... (ii)
12
Let particle stops at t
⇒
2 3
= t 0 , then v (t 0 ) = 0 ⇒
v0
−
α t 0 2
=0
, substituting t 0 in (ii), we get
2 v0
α
−
α v0 2
α 2 8v03 2 × 2+ . 3 12 α α 4v0
2 − 2 + 2 = 2v03 2 α α 3α 3α
As particle is continuously moving in same direction, required distance = x (t 0 ) =
3 2
2v0
3α
TRY YOURSELF - 2 1.
A body starts from rest, moves in a straight line with constant acceleration accele ration and covers a distance of 64 m in 4 sec. (a)
What is the final velocity?
(b)) (b
How much time is required to cover half the total distance?
(c)
What is the distance covered in one-half the total time?
(d)) (d
What is the velocity when half the total distance dist ance has been covered?
(e)
What is the velocity after one half the total time?
KINEMATICS
LOCUS
16
2.
A particle with an initial velocity of 10m/s moves along a straight line with a constant acceleration. When the velocity of the particle is 50 m/s, the direction of its acceleration accelerati on is reversed. Find the velocity of the particle when it reaches the starting point again.
3.
An object is thrown vertically verticall y upward. It has a speed of 10 m/s when it has reached one half its maximum height. (a)
How high does it rise?
(b)) (b
What is its velocity and acceleration accelerat ion 1 sec after it is thrown?
(c)
What is its velocity and acceleration 3 s after it i t is thrown?
(d)) (d
What is the average velocity during the t he first half second?
4.
From an elevated point A, a stone is projected vertically upwards. When the stone reaches a distance h below A, its velocity is double of what it i t was at a height h above A. Show that the greatest height attained att ained by the stone is equal to 5/3 h.
5.
A point mass starts moving in a straight line with a constant acceleration α . At a time t 1 after the beginning of motion the acceleration changes sign, remaining the same in magnitude. Determine Determi ne the time t from the beginning of motion in which the point mass returns to the initial position.
6.
A particle of mass m moves on the x-axis as follows : it starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times ( 0 < t < 1). If α denotes the instantaneous acceleration of the particle, then: (a)
α cannot remain positive for all t in the interval 0 ≤ t ≤ 1
(b)) (b
α cannot exceed 2 at any point in its path
(c)
α must be
(d)) (d
α must change sign during the motion, but no other assertion can be made with the information given
≥ 4 at some point or points in its path
7.
A particle moves along a straight line such that its displacement at any time t is given by x = t3 − 6 t2 + 3 t + 4 . What is the velocity of the particle when its it s acceleration is zero?
8.
A motor boat starts from rest with an acceleration given by the law α =
c
( x + 4)
2
2
m / s , where c is a
positive constant. Given that the velocity of the t he boat when its displacement is 8 m is i s 4 m/sf find:
9.
(a)
the magnitude of c
(b)) (b
the position of the boat when its speed was 4.5 m/s
(c)
the maximum velocity of the boat.
An object moves along the x-axis such that is acceleration is given as a = 3 – 2 t . Find the initial speed of the object such that the particle will have the same x-coordinate at t = 5.0 s as it had at t = 0. Also find the object’s velocity at t = 5.0 s.
KINEMATICS
***************
LOCUS
17
GRAPHS RATE OF CHANGE : Position-time graph for a straight line motion: xi = 0, then ∆ x = x f – xi = x f . ⇒ Position at time t ≡displaceme (Note that if x displacement nt for the time interval [0, t ]) ])
x
x= ( f) t
x
B
2
x –x 2
1
θ
x
1
t –t
A
O
2
P
1
t
t
1
t
2
From the given position-time graph it is clear that at t = t 1 the positon is x1 and at t = t 2 the position is x2. If we connect A and B then the slope of the chord AB is tan θ =
⇒
− x1 ∆x = = av.velocity for the time interval [t , t ] 1 2 t2 − t1 ∆t
x2
slope of the chord on x − t graph = av. velocity
and an d slope of chord on v − t graph =
∆v ∆ t
= av. acceleration
If ∆t → 0, then B approaches towards A and finally the chord joining from A to B becomes tangent to the graph at point A. We say that the chord AB becomes a tangent to the curve at
x
A as B → A.
∴
∆t → 0 (slope of chord) = ∆t → 0
⇒
slope of the tangent = instantaneous rate of change
lim
B
lim (av. rate of change)
* Slope of the tangent on x - t graph = velocity
A
* Slope of the the tangent on s - t graph = speed * Slope of the tangent on v - t graph = acceleration.
KINEMATICS
O
t
LOCUS
18
Example : 9 BASIC APPROACH: Go through the analysis of following graphs carefully to learn the application of concepts described on previous page. x
x is increasing
slope is increasing O
t
⇒ ⇒ ⇒
v is +ve;
⇒ ⇒ ⇒
v is +ve;
⇒ ⇒ ⇒
v is +ve;
⇒ ⇒ ⇒
v is –ve
⇒ ⇒ ⇒
v is –ve;
⇒ ⇒ ⇒
v is –ve;
v is increasing a is +ve.
x
x is increasing
slope is constant O
v is constant a is zero.
t
x
x is increasing
slope is decreasing
O
a is –ve
t
x
x is decreasing
slope is increasing
O
v is decreasing
v is increasing a is +ve.
t
x
x is decreasing
slope is constant constant
O
v is constant a is zero.
t
x
x is decreasing
slope is decreasing O
KINEMATICS
t
v is decreasing a is –ve
LOCUS
19
Example : 10 x
I.
Analyzing a graph: The graph given below describes the motion of a particle moving along the x-direction. The entire motion is i s being analysed in a very basic way way.. Carefully go through the complete analysis and see the amount of information informatio n that can be gathered from a single graph. *
At t = t 2, slope of the tangent tangent = 0
⇒
at t = t 2,
t 2
t 1
O
x ↑, v >0 slope ↑ , v↑ a>0
v=0
Note :
t 3
t
t 4
↑x v, >0 ↓x v, <0 ↓x v, <0 slope↑ , slope ↓ , slope ↓ , v ↓, v ↓, v ↑, a<0
a<0
a>0
↑ denotes increasing sense and ↓denotes decreasing sense.
v
II.
For the given v– t graph, plot the x – t graph, if x xin = 0. For the interval (0, t1): v>0
⇒
o
x is increasing;
v is increasing
t 1
t 2
t
x
⇒ slope of ‘ x x – t ’ is increasing
for [t 1, t 2]: v is constant and + ve
⇒
o
x is increasing and slope of x x – t is constant.
t 1
t 2
t
Here you should notice that if you draw tangents to x – t graph just before t 1 and just after t 1 then these tangents coincide with each other. I have drawn the x – t graph in this way only because velocities just before t 1 and just after t 1 are equal (in calculus you will study this property as continuity of a function), therefore, slopes of tangents on x – t graph just before t 1 and just after t 1 must be equal (In calculus you will study this property as differentiability differenti ability of a function).
III.
For the given ‘a – v’ graph,
a
plot ‘v – t ’ graph, if v = 0 at t = 0 O
KINEMATICS
v
LOCUS
20
Solution: At t = 0. v = 0 and a > 0
⇒
v
v should increase
But as v increases a decreases
⇒
with increasing v, slope of ‘v – t ’ must decrease. ***************
t
O
AREA UNDER GRAPHS : Area =
∫
dA =
v
∫
v.dt
dA
= v (t).dt dt
= ∆ x = displacement
area = d=A .
v dt
v (t )
t
t t+d t t 1
t 2
*
area under v– t grap graph h = displacement
*
area under | v | −t graph = ∫ | v | .dt = ∫ speed.dt = distance
*
area under a – t graph = ∫ a.dt = ∫ acceleration.dt = ∆v = change in velocity. veloci ty.
> 0, A 1
*
< 0, A 2
>0 A 3
v
Distance = | A1 | + | A2 | + | A3 |
A3
A1
Displacement = A1 + A2 + A3 O
t
A2
Example : 11 For the given velocity-time graph, select the correct alternatives out of the following : (a) parti particle cle turns turns back exactly exactly once once during during its motion motion (b) acc acceler eleratio ation n is always always –ve
V
(c)) av (c aver erag agee acceleration for [0, t o] is +ve (d) avera average ge velocity velocity for [0, [0, t 0] is +ve (e)) aver (e average age spee speeds ds for [0, [0, t 0] and [t 0, 2t 0] are equal KINEMATICS
O
t
0
2t
0
3t
0
t
LOCUS
21
Solu So luti tio on: (a (a)) Co Corre rrect ct : For t < t 0, v > 0, at t = t 0, v = 0 and
for t > t 0, v < 0
∴
Particle’s velocity changes sign at t = t 0. Before t = t 0, particle is moving along the +ve x direction and after t = t 0 it is moving along –ve x direction.
∴
Particle turns back at t = t 0.
(b) Co Corr rrect ect:: Velocity is always alway s decreasing, therefore, therefo re, a < 0.
Alternatively: Slope of v – t graph is always –ve, therefore acceleration is always –ve. (c)) In (c Inco corr rrec ect: t: Slope of ‘v – t ’ is constant
∴ a = constant ⇒ < a > = a ∴ for any time interval av acceleration is –ve. (d)) Co (d Corr rrect ect:: For [0, t 0]: Displacement is +ve
⇒ average velocity is +ve. (e) Co Corr rrec ect: t: For [0, t 0] and [t 0, 2t 0]: M agnitudes of areas (i.e. distances) are equal, therefore
average speeds are equal. ( ∵ lengths of intervals are also equal).
Example : 12 For a particle if motion starts from the origin and its velocity varies with time according to given velocity-ti me graph, find: (a) the nature of acceleration (– ve or +ve); (b)) the sign of average velocity for time intervals (0, t 0), (b
(t 0, 2t 0) and (0, 3t 0);
v v0
(c) compare average speeds for time intervals
(0, t 0) and (t 0, 2t 0);
O
t 0
2t 0
(d)) plot acceleration-time, position-time (d
and distance-time graphs, for the time interval (0, 2t 0)
Solu So luti tion on:: (a (a)) Acceleration is always negative, because velocity is continuously decreasing
Alternate:∵ slope of v - t graph is –ve
∴
a is –ve.
(b)) time interval(0, t 0): area is +ve (b
⇒ ⇒ KINEMATICS
displacements +ve average velocity is +ve
3t 0
t
LOCUS
22
(t 0, 2t 0) : area is –ve ⇒ displacement is negative
⇒
av.. velocity is –ve av
(0, 3t 0) : area is –ve ⇒ displacement displacement is i s negative
⇒
av.. velocity is –ve. av
(c) Magnitudes of areas for (0, t 0) and (t 0, 2t 0) are same. Therefore, distances travelled in these two intervals (with equal lengths) are ar e same, and hence, average speeds for these two intervals are equal to each other. othe r. (d) From the given v–t graph, we have tan α = v0 / t 0 .
∴ slope of v–t graph = tan θ = tan (π − α )
= − tan α = −v0 / t 0 . For the time interval(0, t 0), velocity is +ve and is decreasing, therefore, x should increase with decreasing slope. At t = t 0, v becomes zero, therefore, at this moment slope of x – t grap graph h also becomes zero. After t = t o, velocity becomes negative and still decreases, therefore, x should decrease with decreasing slope. Now, Now, I will encourage you to think about the path of this motion.. Till motion Till the instant t 0, the particle is moving along +ve x–axis; at t = t 0, the particle momentary comes to rest and after that it starts moving in –ve x-direction. After t = t 0, whatever distance it covers along the –ve x-direction must must also be include in cluded d in the net distance distance travelled. travelled.
v v0
θ = π−α α O
t 0
2t
t 0
2t
t
0
a
O
t
0
–v0
t 0
x
s(t ) x(t )
O
KINEMATICS
t 0
t
2t 0
LOCUS
23
Example : 13 For the given position time graph, find the time t 0 such that the velocity at t 0 is the same as average velocity for the time interval [0, t 0].
KINEMATICS
LOCUS
24
Solution:
It is given that av.. velocity = instantaneous velocity av vel ocity
⇒
slope of chord = slope of tangent
Therefore we should look for an instant t 0 such that the chord t = 0 to t = t 0 is also a tangent to the curve at t = t 0. Therefore, t 0 = 50 sec.
The slope of the chord OB gives the average velocity of the particle upto t = t 0. This chord is also a tangent to the cureve at B(t = t 0) and hence the slope of this chord also gives the instantaneous velocity at B.
KINEMATICS
LOCUS
25
TRY YOURSELF - 3 1.
The figure below shows displacement-time graph of a particle. Find the time interval (starting from t = 0) during motion such that the average velocity of the particle parti cle during that period is zero. x(m) 20 10 0
2.
5 10 15 20 25 t(s)
Plot the velocity-time and speed-time graphs corresponding to the position-time graph given below. x x
0
t O
3.
t
3t
2t
0
0
0
x – t ) graph find the interval in which For the given position-time ( x x
O
t
1
t
2
t
3
(a) ve velo locit city y is ze zero ro (b) both velocity velocity and accele acceleration ration are negativ negativee (c) velocity is positive positive but but acceleration acceleration is negative negative (d) velocity is negative negative but but acceleration acceleration is positive.
KINEMATICS
t
4
t t 5
6
t
LOCUS
4.
26
Plot the approximate acceleration-time, position-time and distance-time graphs for the velocity-time graph given below below. v
O
5.
t 0
2t
t
0
The velocity of a particle that moves in the positive X-direction varies with its position, positi on, as shown in figure. Find its acceleration in m / s 2 when x = 6m. v
4 m/s 2 m/s x 4m 8m 6.
A particle starts moving from rest. Its acceleration (a) versus time (t ) varies as shown in the figure below. The maximum speed of the particle will wil l be: a 10m/s
2
(a) 110 m/ (a) m/ss (b) 55 m/s (c) 550 m/s (d) 660 m/s 11 7.
t(s)
Repeat example 12, if the origin is chosen on ground surface (the point where the ball will hit the ground) and the upward direction is chosen as +ve X direction directi on ************
KINEMATICS
LOCUS
27
MOTION IN A PLANE : If a particle is moving in a plane, say the X –Y plane, any two convenient mutually perpendicular directions in the Now, we will resolve all vectors along x and y axes. x– y plane are chosen as the X and Y axes. Now,
r = xˆi+ yˆj and v
We have
= v xiˆ + v y ˆj
Y v y
But,
⇒ ⇒
dr
v
v = dt v x iˆ + v y ˆj = v x iˆ + v y ˆj =
⇒
v x
=
dx
and an d
v y
=
dy
dt
dt
P( x, y)
d ( xiˆ + yˆj ) dt
v x
y r
dx ˆ dy ˆ i+ j dt dt
(velocity component along x )
O
x
X
(velocity component along y )
Similarly, a x =
dv x dt
and a y
=
dv y dt
Now, x, v x and a x are related by equations of straight line motion along the t he x-axis (the same argument holds for components along y-axis). The problem of motion in a plane is thus, broken up into two independent problems of straight line motion, one along the x-axis and the other along the y-axis. If a x and a y are constant, then
and an d
v x = u x + a xt ,
x = u xt +
1
v y = u y + a yt ,
y = u yt +
1
2
2
a xt 2 , v 2x = u 2x + 2 a xx ; a yt , v y = u y + 2 a y y . 2
2
2
At this juncture, I would like to encourage you to think about the sense in which these two independent sets of equations are related to each other.
Example : 14 A particle moves in the X –Y plane, making an angle of 37° with the x-axis. At t = 0 the particle particl e is at the origin and and its velocity is 8.0 m/s along x-axis. Find the velocity and the position of the particle at t = 4.0 s. Solution:
As discussed above, here we will first find f ind the x and y components of initial velocity and acceleration. Using these components we can completely analyze the motions along these two axes. KINEMATICS
LOCUS
Solution:
28
u x = 8.0 m/s, a x = a cos 37° = (1.5 m/s²) ×
u y = 0, a y = a sin 37° = (1.5 m/s²) ×
3 5
4
y
5 a =1.5 m/s²
= 0.9 m/s².
37°
t = 0
At
∴
t = 4.0 s
u = 8.0 m/s
x = u xt +
1
y = u y t +
1
2
axt 2
2
a yt
2
1 1 6 m = 41.6m. = 8 × 4 + × 1.2× 16 2
= 7.2 m
Y
v x = u x
+ axt = (8 + 1.2 × 4) m s = 12.8 m
v y = u y
+ a y t = 3.6 m
v=
2 x
v
=
12.8
v
v y
s
α
+ v = 13.3 m 3 .6
a =1.5 m/s²
s
2 y
and tan α = v y v x
x
=
v x y
s
O
9
X
x
32
Hence, at t = 4 s, the particle is at (41.6 m, 7.2 m) and its velocity is 13.3 m/s at an angle of tan −1 (9 32) with the +ve ve x x-direction.
Example : 15 Projectile motion This is the two-dimensional motion of a particle thrown obliquely into the air. It is an example of curved motion with constant acceleration. acceler ation. We assume that the effect of air can be neglected.
Y (v y >0) g u
v
KINEMATICS
=
v x
= u co cosθ (v = 0 y
v x
v x
α
O
v
H
v y
(v y < 0) v
X Range R
LOCUS
29
The particle is projected with speed u (called as projection speed) at an angle α with x-axis (called as angle of projection). The motion of this particle is i s an accelerated one with constant acceleration ‘g’ in the downward ( –ve y) direction. The particle moves on the path shown in the figure. H is called maximum height and R is the horizontal range. r ange. Total Total time of motion mot ion is known as the time of flight, fli ght, T. Here, I will suggest you to analyse the above the figure thoroughly and notice not ice all the given informations infor mations carefully. We have have,,
=0
u x = u cos α , a x
∴ and
v x = u x + axt = u cos α x = u x t +
1
(constant)
....(i)
2
axt
2
= u cos α .t Again, u y
∴
= u sin α ,
v y = u y
a y = –g;
+ a y t = u sin α − gt
y = u y t +
and an d
....(ii)
1 2
ayt
2
....(iii)
= (u sin α ) t −
1 2
2
gt
...(iv)
Using the four equations above we can find the velocity and position of particle at any time during its flight. Time of flight, T T :: At t
⇒
T
= T,
0 = (u sin α ).T −
1 2
gT
2
g
R =
= T,
x=R
H =
⇒
R = u cos α .T
= u cos α .
2u sin α g
2
sin 2α u sin g
Maximum height, H: At y = H, v y= 0
⇒
⇒
= 2u sin α
Horizontal Range, R: At t
⇒
y=0
⇒
2
v
= u2 y+ 2 a yy ⇒
y
0 = ( usin α )2
u 2 sin 2 α
2g
Equation of trajectory: Time independent independent relation between x and y coordinates of any point on the path is the equation of the path.
We have have,, KINEMATICS
x = u cos α .t
⇒
t =
x u cos α
LOCUS
30
y = u sin α .t −
again,
⇒
y = u. sin α .
Note:
2
gt gt 2 2
x u cos α
y = x tan α −
⇒
1
−
gx
g x . 2 u cos α
2
2 2 2u cos α
*
Thee same Th same app approa roach ch co could uld be fol follow lowed ed for any typ typee of of motio motion n in a plan plane. e.
*
Each Ea ch po point int on the pa path th mu must st sa satis tisfy fy the eq equa uatio tion n of of the the traj trajec ector tory y
*
R(α) = R(π /2 – α).
*
For α ∈[0, π /2], R is symmetrical about θ = 45°. For α ∈ (0, π /4), R increases with angle and for α∈(π /4, π /2), R decreases with angle
*
R is ma maximum at θ = 45°
Example : 16 A ball is projected horizontally with speed u from height H above ground surface (horizontal). Find the time t ime of motion and horizontal distance travelled, before this ball hits the ground. Solution: If the ball hits the ground in T seconds, then H =
1 2
2
gT (Since ball is projected horizontally its motion in
vertical direction is due to g only) T =
⇒
2 H g
Again , R = u.T
(As there is no horizontal acceleration, only u is responsible for horizontal movement of the particle.)
= u.
2 H g
ALTERNATE METHOD: u x = u, a
and an d KINEMATICS
x = u x t +
=0 ⇒
x
1 2
a xt
2
v
= u.t
= u x+ a tx = u
x
LOCUS
31
= +g
Again,
u y = 0,
a
and an d
y = u y t +
1
1
=
2
⇒
y
2
v
= u y+ a ty = gt
y
2
a yt
O
2
gt
H
at t = T , y = H ⇒ H= T
=
u
x
g
If T be the time of flight, then
⇒
1 2
Y
2
gT R
2H g
Again, at t = T , x = R
∴
R = x(T ) = u.T = u 2 H g
Example : 17 A body projected with the same velocity at two different angles covers the same horizontal distance R. If T 1 and T 2 are the two time of flight, prove that R =
1 2
gT1.T 2
Solution: Since it is given given that the ranges ranges of two projectiles projectiles are are same, if one one is projected projected at an angle angle θ, the other
must be projected at
T 1 =
∴ ∴
2u sin θ
T 1.T 2 = 1 2
g
2
− θ . Thus,
and T 2 =
2u sin(90 − θ )
2.2u 2 sin θ . cosθ g
g T .T = 1 2
=
2
2.2u 2 sin θ . cosθ g u 2 sin2θ
= R
KINEMATICS
π
g
g
=
2u cos θ g
LOCUS
32
Example : 18 A particle is thrown thr own over a triangle from one end of a horizontal horizont al base and grazing the vertex falls fall s on the other end of the base. If α an d β be the base angles and θ be the angle of projection, prove that tanθ = tan α + tan β. Solution: Let u be the projection speed and P( x x, y) be the vertex of then triangle then from the figure, fi gure, we have
tan α =
y
... (i)
x
and tan β =
y
... (ii)
R − x
y
P x ( , y) u
Again, equation of trajectory is gx
y = x tan θ –
⇒ ⇒ ⇒
2
2u cos θ
x
= 1 −
θ 2
y = x tan θ 1 − y
2
2u
cos θ . sin θ
⇒ R tan θ + tan β
yR = tan θ x( R − x)
= y+ x
β
α x
gx
2
x
tan θ = tan α
O
y
x
R – x
y R− x
[using (i (i ) & (ii)]
Example : 19 Projectile motion on Inclined plane: The figure along side shows an inclined plane at an angle angle α with the horizontal, with initial initi al speed u.
β with the horizontal and a particle is projected at an
In such a case we take our reference x and y axes in the directions along and perpendicular to the incline plane plane as shown. We have, u x = u cos(α a x =
− β ) ;
Y
− g sin β ;
X
∴
v x = u x + axt = u cos(α − β ) − g sin β t
and
x = u xt +
KINEMATICS
1 2
ax t
2
= u cos(α − β ).t −
1 2
u
i n β g s
2
g sin β .t
g O
β g cosβ A
LOCUS
33
Again u y = u sin(α
− β ) ;
− g cos β ;
a y =
+ a yt = u sin(α − β ) − g cos β .t
∴
v y = u y
and an d
y = u y t +
1 2
ayt
2
= u sin(α − β ). t −
1 2
2
g cos β .t
Now, using above equations we can analyze the motion of the particle part icle completely. Time of flight, T : If the particle is projected at t = 0, then at t = T , y again becomes zero, i.e., at t = T , y = 0
⇒ ⇒
u sin (α − β).t – T =
1 2
2
g cos β .t
=0
2.u.s . sin(α − β ) g cos β
Maximum height, H : If H H be the maximum height with respect to plane, then at y = H , v y = 0
+ 2.a y .H
⇒
02 = u y
⇒
0 = [u sin(α
⇒
2
H =
− β )]2 – 2 ( g cos β ) H
u 2 sin 2 (α − β )
2.g cos β
Note that the maximum height with respect to inclined plane under consideration is different from maximum height with respect to horizontal plane OA. In this case, maximum height is in fact maximum distance of particle from inclined plane. Range, R : If R R be the range on inclined plane, then at t = T , x = R
⇒ R = ⇒
u cos(α
R =
− β ).T −
u 2 sin 2(α − β ) g cos β
1 2
−
g sin β .T 2
2u 2 sin β . sin 2 (α
−β)
g cos2 β
Maximum Range, R max :To find maximum range on inclined plane corresponding to particular u, we use
and we get,
⇒
KINEMATICS
Rmax
=
u2 g (1 + sin β )
dR d α α
=0
LOCUS
34
Example : 20 A child throws a ball so as to clear a wall of height h at a distance x from it. Find the minimum speed required for clearing the wall. Solution: If we join the top most point of the wall with the point of projection and consider this line as an inclined i nclined
plane with inclination, β = tan −1 u
∴
Rmax=
⇒
u > g h+ h
⇒
x
, then distance x 2 + h 2 must be maximum range on the assumed inclined plane.
2
g (1 + sin β )
(
umin =
h
(
2
+x
2
≥
h
)
∵
g h + h2 + x 2
2
+ x2 sin β = 2 2 x + h h
u 2
x h + 2
h
β
)
x
TRY YOURSELF - 4 1.
An airplane flying horizontally at 100m/s drops a box at an elevation of 2000 m. (a) How much time is required for the box to reach earth? (b)) How far does it travel horizontally while falling? (b (c) Find the horizontal & vertical components of its velocity when it strikes the earth.
2.
A ball is thrwon from the ground into air at an angle with the horizontal. At a height of 9.1 m the velocity is observed to be v = 7.6 i + 6.1 j m/s. Find: (a) the maximum height to which it will rise. (b)) horizontal distance traveled by the ball. (b (c) velocity of the ball at the instant of striking the ground.
3.
A stone is thrown from the top of a tower of a tower of height 50 m with a velocity of 30 m per second at an angle of 300 above the horizontal. Find. (a) the time during which the stone will be in air (b) the distance from the tower base to where the stone will hit the ground (c) the speed with which the stone will hit the ground, (d) the angle formed by the trajectory of the stone with the horizontal hori zontal at the point of hit.
4.
A stone is thrown horizontally with a speed of 10 m/s from a balloon ascending witha velocity of 5 m/s and it reaches the ground in 8 sec.
KINEMATICS
LOCUS
35
(a) How high is the balloon at the moment the stone is released? (b)) How far the stone will move horizontally before striking the ground. (b (c) What is th impact velocity with the ground ? ( take g
= 10m / s 2 )
5.
The trajectory of a projectile in a vertical vert ical plane is y = ax− bx2 , where a,b are constants, and x and y are ar e respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is .........and the angle of projection from the horizontal is........... is.... ....... .
6.
The maximum range of a particle with wit h a certain speed on a horizontal plane is R. Find its maximum range when projected on an inclined plane with inclination 300 .
7.
A gardener shower jet is placed at a distance d from the t he wall of a building. If R is i s the maximum range of the jet that is produced when the bowl is connected to the nose of a fire engine, show that the por tion of the wall that is hit by the jet of water is bounded by a parabola whose height is
2 2 ( R − d )
2 R
and breadth is
2 R 2 − d 2 . 8.
The coordinates of a bird flying in the xy-plane are x = 2 − α t and y = β t 2 , where α = 3.6 m / s and
β = 1.8m / s 2 . Calculate the velocity and acceleration vectors and their magnitude as a functions of time. Also find the magnitude and direction of bird’s velocity and acceleration at t = 3.0 s. From the given data can you find whether at this instant, bird bir d is speeding up, speeding down or it is taking a turn. If so in which direction. ***********
RELATIVE MOTION : For analysing the motion of a moving body, the frame of observation can be chosen according to the convenience
of the problem. The position r , velocity v and acceleration a of a particle depend upon the frame chosen. Now,, we have to relate these quantities Now quantiti es for a particle measured in i n two different frames.
Consider two frames of reference S and S´ and suppose a particle P is observed from both the frames. The frames may be moving with wi th respect to each other. Let,
OO '
= posititon of S' w.r.t. S = r S ' S
OP
= posititon of P w.r.t. S = r PS
KINEMATICS
LOCUS
36
O ' P
= posititon of Pw.r.t. S ' = r PS '
Y
Y′ P
we have,
⇒
rPS
OP
=
OO '
+ O ' P
= rS ' S + r PS '
S′
Differentiating Differenti ating both sides, we get`
O′
S
O
v PS
X′ X
= vS ' S + vPS '
Differentiating both sides one more time, we get
aPS Note:
= aPS ' + aS ' S
d
*
It is as assu sume med d tha thatt the the me mean anin ing g of of tim timee and and
*
Rear Re arra rang ngin ing g th thee ab abov ovee eq equa uati tion onss ab abov ove, e, we ge get. t.
r PS ' = rPS v P S ' = v PS
dt
are same in both the frames fr ames correspondingly.
− r S ' S − vS 'S
aPS' = aPS ' − aS ' S
Thus , if the velocities of two bodies are known with respect to a common frame (here S), we can find velocities of bodies with respect to each other. other. The same holds true for position and acceleration. *
Posi Po siti tion on,, vel veloc ocit ity y and and ac acce cele lera rati tion on of ob obse serv rver er ar aree alw alway ayss zer zero o wit with h res respe pect ct to it itse self lf..
Example : 21 I.
Positions of points A and B are shown in some frame:
y A
r
A
•B
r
B
O fr ame of A, then But if we analyze B from the frame
r w.r.t. A = r A − r A AA = position of A w.r.t.
=0
(i.e. A would be treated as origin now)
KINEMATICS
x
LOCUS
37
A
− r B
=Bposi p tion n of B w.r.t.A w.r.t. A = r A ositio
r
r
B
A
A
r
A
•B
r A
O
≡
r
B –
r
r B
•
A
B
Note:
*
Vecto torr jo join inin ing g B fr fro om A is posit itio ion n ve vecto torr of of B w.r .r..t. A.
*
Here,, you Here you sh shou ould ld no note te th that at wh when en I mov moved ed to th thee fra frame me of A, I sub subtr trac acte ted d r A from both positions of A and B. If there would be a third t hird particle C, then I would have done the same with its position vector. vector.
II.
Velocities of points A and B are shown in some frame:
Y v A v B
A
B B X
O
But, if we analyze B from frame of A, then w.r.t. A v AA = velocity A of w.r.t.
= v A
v
− vA =
0
= Bvel vAeloc ocit ity y of B w.r.t w.r.t.. A = v
− vB
A
Note:
*
Have Ha ve al also so yo you u sh shou ould ld no note te th that at wh when en I mov moved ed to th thee fr fram amee of A, I sub subtr trac acte ted d v A frome the velocity of each
particle. In this way, we get velocities of all particles in the new frame. fr ame. *
We will fo foll llo ow the same approach for accelera rati tio on.
KINEMATICS
LOCUS
III.
38
At t = 0 cars A and B are moving in the same direction with constant velocities velociti es of 10 m/s and 5 m/s respectively and A behind B, find time at which A catches B if initially the distance between them is 50 m.
Solution:
at t = 0
at t = 0
v A=10m/s
v B=5m/s
A
Ground frame:
B at t = t 0
50 m
Lett t = t 0, be the instant at which A catches B then the distance Le dist ance travelled by B + 50 = distance travelled by A.
⇒
5t0 + 50 50 = 10t0
⇒
t0
=
50 5
= 10s.
Let us solve for the same from the frame of B. As discussed earlier, we will subtract velocity of B from f rom velocities of both A and B to get their thei r velocities in the new frame. v A
v B
Frame of B:
v B
v A v B
v B
B
A
∴
t0
=
50 v A − vB
=
B
A
50 m
Rest
50 m
50 10 − 5
= 10 s.
It is trivial to see that B is at rest in this frame. IV.. IV
A is chasing B on a straight str aight road. A is moving with constant speed v and B starts moving away from A with constant acceleration a when A is at a distance d away away.. Find the minimum value val ue of v for B to be caught.
Solution: At t = 0: v
Ground frame:
a
A
B d
a
v
a
Frame of B:
A
a
a B
≡
REST
v A
B d
In frame of B, if speed of A becomes zero before covering a distance d , it can never catch B. For minimum required value of v, speed of A becomes zero when it just covers the distance d , i.e., it just catches B.
KINEMATICS
LOCUS
39
∴
02 = vm2 in
− 2ad
⇒
vmin
Therefore, for A to be caught v ≥
V.
=
2ad
2 ad
Two particles, 1 and 2, move with constant velocities v1 and v 2 . At the initial moment their position positi on
vectors are r 1 and r 2 . How must these four vectors be interrelated for the particles to collide ? Solution:
Ground frame:
Frame of Particl Particlee 1:
Analyzing motion of 2, from frame of 1, we see that 2 and 1 will collide if and only if v1 – v 2 is
antiparallel to r 2 – r 1 , i.e., 1
v2 − v1
v2
– v
r
2
–
1
r
1
⇒ |v −v | 2 1
=−
r2 − r1 unit vector along v2 − v1
| r2 − r1 | = −(unit vector along r2 − r 1 )
2
VI.
Raindrops are falling on a horizontal straight road at an angle of 30° with vertical. verti cal. A man is running at 10 km/h and he finds that raindrops are hitting hit ting his head vertically. verticall y. Find the speed of raindrops with respect to (a) the road, (b) the moving man.
Solution:
Ground frame:
Man’s frame:
Analyzing from the moving man’s frame, we see that raindrops are falling vertically (i.e. vrm, velocity of raindrop w.r.t. w.r.t. moving man, is in vertically downward direction). Therefore, horizontal component of vr must cancel out vm, i.e., vr sin30 ° = vm
⇒
vr
= 2vm = 20 km/h
[Ans (a)]
Again, vrm (vel (veloc ocit ity y of of rai raind ndro rop p w.r w.r.t .t.. mo movin ving man man KINEMATICS
= v)r cos cos 30° =
km/h [Ans(b)]
LOCUS
40
Example : 22 SWIMMER & RIVER CASE : Let us analyze different cases for a man swimming in flowing flowi ng water. water w.r. w.r.t. t. ground ground vr : velocity of water vm : velocity of man w.r.t. w.r.t. to still water
w.r.t. to ground. vmG : velocity of man w.r.t. *
v
v
r
m
:
We see that, the swimmer wishes to reach point A but will reach point B. Distance AB is called drifted distance. A
Time of crossing, t c
=
vm
vmG vm
d
drifted distance, x = tc × vr
*
B x
d
=
d vm
×
vr
v r v r
vm at an angle θ with vr : t c
=
d vm sin θ A
B vm
d
vmG vr
≡
vm sin θ
vmG
θ v r
vr+vm cos θ x = tc × ( vr
*
+ vm cos θ )
Therefore, we have t c
=
width velocity along width
x = t c × velocity along flow.
*
Minimum time of crossing :
t c
=
d vm sin θ
KINEMATICS
,
LOCUS
∴
41
t cmin
=
d vm
at θ
= π 2
Therefore, time of crossing is minimum when vm is along width of the river. ri ver. *
Zero drifting :
vnet (i.e. vmG ) must be along width
∴
vr
and an d
vnet
From (i), we have
Therefore,
= vm sin θ
...(i)
= vm cos θ
...(ii)
sin θ = vr vm . Please note that θ = 0° and θ = 90° are not possible
0 < sinθ < 1
⇒
vr vm
<1 ⇒
vr
< vm
Therefore, if river flow velocity is equal to or greater than velocity of swimmer (w.r.t. (w.r.t. still water), it is not possible to have zero drifting at any angle.
TRY YOURSELF - 5 1.
An aeroplane takes off from Mumbai to Delhi with velocity 50 kph in north-east direction. Wind is blowing at 25 kph from north to south. What is the resultant displacement of aeroplane in 2 hrs.
2.
Two trains, one travelling tr avelling at 54 kph and the t he other at 72 kph, are ar e headed towards each other on a level track. When they are two kilometers apart, both drivers drivers simultaneously apply apply their brakes. If their brakes 2
produces equal retardation in both the trains at a rate of 0.15m / s , determine whether there is a collision or not. 3.
A motorcycle and a car start from rest at the same place at the same time and they travel in the same direction. The cycle accelerates uniformly at 1 m / s 2 up to a speed of 36 kph and the car at 0.5 m / s 2 up to a speed of 54 kph. Calculate the time and the distance at which the car overtakes the cycle.
4.
A river 400 m wide is flowing at a rate rat e of 2.0 m/s. A boat is sailing at a velocity of 10.0 m/s m/ s with respect to the water, in a direction perpendicular perpendicul ar to the river. (a) Find the time taken by the boat to reach the opposite bank. (b) How far from the point directly direct ly opposite to the starting starti ng point does the boat reach the opposite bank.
5.
A swimmer wishes to cross a 500 m wide wi de river flowing at rate 5km/hr. His speed with respect to water is 3 km/hr. (a) If he heads in a direction making maki ng an angle θ with the flow, find the time he takes to cross the t he river. (b) Find the shortest possible time ti me to cross the river.
KINEMATICS
LOCUS
6 .
42
A man running on a horizontal road at 8 km/hr finds the rain falling fal ling vertically vertical ly .He increase the speed to 12 km/hr and finds that the drops make an angle 300 with the vertical. Find the speed and the direction of the rain with respect to the road.
7.
On a frictionless horizontal surface, assumed to be the x–y plane, a small trolley troll ey A is moving along a straight line parallel to the y-axis with a constant velocity of ( 3 − 1)m / s . At a particular instant when the t he line OA makes an angle of 450 with the x-axis , a ball is thrown long the surface from the t he origin O. Its velocity makes an angle φ with the x-axis and it hits the trolley. y A
45
o
0
x
(a) The motion of the ball is observed from the frame fr ame of the trolley. Calculate the angle θ mad madee by the velocity vector of the ball with the x-axis in this frame. (b)) Find the speed of the ball with (b wit h respect to the surface, if φ
= 4θ / 3 .
************
Example : 23 A person travelling on a straight line moves with an uniform velocity v1 for some time and with uniform velocity v2 for the next equal interval of time. Find the average velocity for the entire duration. Solution: If the particle moved with velocity v1 for time t 0, then total time of motion = 2t 0.
∴
av. velocity =
displacement time
=
v1t0
+ v2 t 0 2t 0
=
v1 + v2
2
Example : 24 A particle moves along the x-axis as x = u (t − 2s ) + a (t − 2s )2 . Find : (a) position of particle at t = 0 s. (b)) initial velocity of particle (b (c) acceleration of particle. Solu So luti tio on: (a) x (a) x (0) = 4 (0 –2) + a (0 –2) 2
= (–2u + 4a) m. KINEMATICS
LOCUS
43
(b)) v = (b
dx dt
= u + 2 a (t − 2)
∴ initial velocity = v(0) = u – 4a dv
(c) acceleration =
dt
= 2a
Example : 25 Show that time taken by a particle to slide from any point on a vertical circle along a smooth chord terminating at the lowest point of the circle is constant. Solution: Consider any chord QB making an angle θ with the vertical diameter AB. For motion of particle along QB, u = 0, a = g cos θ, s = QB
using
s = ut + 1
QB =
i.e.
t =
2
1 2
at², we get
A
. g cos θ. t ²
2.QB
... (i)
g .cos θ
Q
In right angle triangle BQA, we have cosθ =
QB d
θ
where d is diameter of the given circle.
substituting for cos θ in (i), we get t =
2d g
B = constant.
Example : 26
If position vector of a moving particle varies with time t as r is a positive factor. Find:
= ωt (1 − α t ), , where ω is a constant vector and α
(a) velocity v and acceleration a of the particle as functions of time;
(b)) the time interval ∆t taken by particle to return to the initial point, (b point , and distance s covered during that time.
Solu So luti tion on:: Note: Initially (at t = 0) particle is at origin and r is always along constant vector ω , therefore,
particle is moving along a straight line parallel to vector ω .
KINEMATICS
LOCUS
44
=
dr
=
dv
(a) v
dt
= ω (1 − 2α t )
a
dt
= −2αω
(b)) At t = 0 particle is at origin. Let at t = t 0, it comes back to origin. (b
∴ ωt0 (1 − α t 0 ) = 0
⇒
t0 (1 − α t 0 ) = 0
⇒
t0
=0
t0 =
or
1
α
Therefore, particle comes back to origin at t =
α
. Again, distance travelled during this time is
1 α
t 0
s=
1
∫ speed . dt = ∫ | ω (1 − 2α t ) | .dt
0
0
1 α
= |ω |
∫ | (1 − 2α t) |.dt 0
1 α 1 2α − + − ω α α t d t t d t | | ( 1 2 ) . ( 2 1 ) . ∫ = ∫ 0 1 2α
| ω |
=
2α
Example : 27 The velocity of a particle moving in the positive direction of the x axis varies as v = α x , where α is a positive constant. Assuming that at the moment t = 0 the particle was located at the point x = 0, find : (a)
the time dependence of the velocity and the acceleration of the particle part icle ;
(b)) (b
the mean velocity of the particle particl e averaged over the time that the particle parti cle takes to cover the first s meters of the path.
Solu So luti tion on:: (a (a)) We have, v = α x
KINEMATICS
LOCUS
45
⇒ α 2t 2 d 4 α 2 dx = ∴ v= = dt
dt
Again, a =
dv dt
=
dt
2
α 2t
=
dt
4
2 d (α t 2)
=
dt
2
α 2 dt α 2 . = 2 dt 2
(b)) As velocity is always positive, displacement is same as distance travelled. Therefore, (b s (t ) = x (t ) =
α 2t 2 4
.
Again, required average velocity =
=
s
4s
=
αs 2 s
=
displacement time
=
s t
α s 2
α 2
Example : 28 The acceleration of a point started at t = 0, varies with time as shown in given graph. Find the t he distance travelled in 30 seconds and draw the velocity-time and position-time graphs if initial velocity and position are zero. a
(m/s²)
5.0
O
20
10
30
t (second
5.0 – 5.0
Solution: As given in the acceleration -time graph,
for time interval (0, 10) : a is constant and +ve
⇒
velocity increases linearly linearly..
⇒
slope of x x - t graph must be increasing
for time interval (10, 20) : a is zero
⇒ ⇒ KINEMATICS
velocity is constant x – t graph is constant slope of x
LOCUS
46
for time interval (20, 30) : a is constant and – ve
⇒
velocity decreases linearly li nearly..
⇒
x - t graph is decreasing slope of x
At t = 10 s, v = u + at = 0 + 5 × 10 = 50 m/s.
a +5
0
10
20
10
20
30
t
– 5 V 50 0
t
30
x
t 10
O
20
30
As we see from v - t graph, velocity is always nonnegative for time interval (0, 30), therefore distance travelled, s = displacement = area under v – t graph
=
1 × 10 × 50 + 50 × 10 1 2 ( 1 0 ) + 2 × 1 0× 5 0
= 250 + 500 + 250 = 1000 m. Note:
*
At t = 30 sec, v = 0
∴ *
x - t graph also becomes zero. at t = 30 sec, slope of x
For time interval (0, 30) area under a - t graph is zero
⇒
KINEMATICS
change in velocity = 0
⇒
v f in = vin = 0 m/s.
LOCUS
47
Example : 29 A ball is released from rest from a height h above the ground surface. Assuming the point of release as origin, the downward direction as positive and gravity to be uniform, plot the following fol lowing graphs for one complete trip (It is given that collision between ball and ground surface is purely elastic and duration of collision is very short): (a)
velocity-time
(b)
speed-time
(c)
acceleration-time
(d)
displacement-time
(e)
position-time
(f) distance-time. Solution: Let the ball be released at t = 0 so that it collides with ground groun d at t = t 0. Downward and upward motions of the ball are shown in following figures. When t < t At t=0 sec (downward motion) • u= u= 0 O O 0
When t > t (upward motion) 0
O
x
g h
x
h
h g
x
v g
v
For t < t 0, we have v = u + at = 0 + gt = gt gt ;
x= u+ t
at t = t 0
1 2
a2t = 0 +
x = h
1 2
⇒
1
g2t =
2
h=
1 2
g2t; gt 02
just before collision, speed = gt0
⇒ =
t 0
=
2h g
2 gh
just after collision, speed = 2 gh For motion in upward direction (let t ´ = t – t 0, then upward motion starts at t ´ = 0) initial velocity is
−
´) = u + at ´, ´, we get 2 gh and acceleration is +g, therefore, using, we get v(t ´) v (t ') = − 2 gh
⇒ KINEMATICS
+ gt '
v(t − t0 ) = − 2 gh + g(t − t0 )
LOCUS
48
and using x(t ´) ´) = xi + ut ´ + we get x( t') = h− 2 gh t'+
⇒
x( t − t0 ) = h − 2 gh( t − t0 ) +
1 2 1 2 1 2
at '2 , 2
gt' g( t − t0 ) 2
Note that at t ´= ´= t 0 (i.e. at t = 2t 0) v = 0 and x = 0. x(t – t 0) and v(t – t 0) can be obtained by first plotting graphs of x(t ) and v(t ) and then shifting Graphs of x rightwards by t 0 units. a
g O
t
2t
0
t
0
v + 2 gh
t
O – 2 gh
speed, |v| 2 gh
t
O
x h t
O
s 2h h
O
t t 0
2t
0
Note: * Since motion starts from origin, displacement-time and position-time graphs are identical. * For the instant t = t 0 analysis will be done in later parts of mechanics. (This involves the concept of collisions).
Example : 30 A car accelerates from rest at a constant rate α for sometime after which it decelerates at constant rate β to come to rest. If the total time lapse l apse is t 0 seconds, evaluate (i)) (i the maximum velocity reached and (ii) total distance travelled.
KINEMATICS
LOCUS
49
Solution: Let velocity is maximum at t = t 1. For t ∈ (0, t 1) : a = slope of graph
v
= tan θ1
⇒
α = vm t 1
vm
....(i)
For t ∈ (t 1, t 0) : a = slope of of graph = tan tan (π – θ2)
⇒
O
− β = − vm
(t0 − t 1 )
⇒
θ2
θ1
= –tanθ2
β =
vm t0
− t 1
t 1
t 0
t
.... (ii)
Eliminating t 1 from (i) and (ii), we get. vm =
αβ t 0 α + β
From the graph, we see that the velocity is i s never negative, hence distance
= di displacement = area under velocity- time graph =
1 2
.vm .t 0
αβ t 02 = 2(α + β )
Example : 31 For the given acceleration-time graph, plot the velocity time graph. It is given that initial velocity is v0 and for t > t 0, magnitude of acceleration is inversely proportional to time. a t 0
O
KINEMATICS
t
LOCUS
50
Solution: From a – t graph, we see that for t ∈ (0, t 0) velocity must be decreasing (∵ a is – ve) and its slope
should also decrease simultaneously (∵ a is decreasing) in a linear fashion. a t 0
t
O
V V 0
t O
t 0
When t ∈ (t 0, ∞), velocity should decrease decrease but the magnitude of its slope should be increasing. increasing.
Example : 32 Position vector of a point relative to origin varies with time t as r
= atiˆ − bt 2 ˆj,
where a and b are positive
constants. Findthe equation of the point’s trajectory. Solution:
It is given taht r
= at iˆ − bt 2 ˆj,
but we know r
= x iˆ + y ˆj,
therefore, x= at and y= − bt2 Eliminating t , we get 2
y =
x −b a
⇒
y = −
bx a
2
2
Example : 33 For a particle moving in the x - y plane, if a
= (3iˆ − 2 ˆj ) m/s² and
particle when its y-coordinate is maximum. Solution: we have, a x = 3 m/s², a y = –2 m/s² u x = 2 m/s, u y = 4 m/s.
therefore, KINEMATICS
v y = u y
+ a y t = (4 − 2t ),),
vin
= ( 2 iˆ + 4 ˆj )
m/s, then find position of
LOCUS
51
y = u y t + v x = u x
1 2
= 4t − t 2 ,
+ axt = 2 + 3t ,
x = u x t +
1
axt 2
2
When y is maximum, v y
⇒
a yt 2
= 2t + dy
= 0 or
dt
3 2
t2
=0
4 – 2t = 0 ⇒ t = 2 s.
At t = 2 s, x = (2 × 2) 2) +
3 × 2 2 2
= 10 m. and y = (4 × 32) – (2) 2 = 4 m. Hence, particle is at point (10, 4), and position vector of particle is yˆj= (10 ˆi + 4 ˆj) m r = xˆi + yj
Example : 34 A particle moves in the plane xy with constant acceleration a directed along the negative y-axis. The equation of motion of the particle has the form y= α x− β x2 , where α and β are positive constants. Find speed of particle at origin of coordinates. Solution: We have, a x = 0
and
a y = – a
Equation of path is y= α x − β x
⇒ ⇒
v y = α v x
dv y dt
⇒
− 2β x.vx
= α.
dv x dt
⇒
ay =
α .a
⇒
ay =
− 2 β v x2
⇒
–a =
− 2 β v x2
KINEMATICS
2
− 2 β x.
dy dt
= α .
....(i) dv x dt
− 2 β .
dx
.v x dt
− 2β x.a x− 2 β .V 2x
x
[∴ a x = 0]
dx
−2β x.
dt
dx dt
LOCUS
52
⇒
a
v x =
At origin:
a 2β
v x =
v y = α .v x
[using (i) and x = 0] 0] a
= α .
∴
(constant)
2β
2 β
Speed = v x2
+ v 2y =
a
2 β
(1 + α 2 )
Example : 35 If R be the range of a projectile on horizontal plane and H 1, H 2 be maximum heights for its two t wo possible trajectories, find the relation between the given parameters. Solution: We have, have , 2
H 1 =
H 2 =
2
u sin θ
2g
u 2 sin 2 (9 (90 − θ )
2g
∴ H 1 H 2 = ⇒ R
;
=
u 2 cos 2 θ
2g
u 2 sin 2 θ .u 2 cos 2 θ
4g 2
=
;
(u 2 sin 2θ ) 2 16 g 2
=
R2
16
= 4 H 2 H 2
Example : 36 A ball is projected with wit h some speed at an angle of 45° with the horizontal. If it just clears a wall at a distance ‘ a’ from the point of projection and falls at a distance ‘b’ from the wall, then find the height of the wall. Solution: Let the projection speed be u and the projection angle be θ , then equation of path is y = x tan θ −
KINEMATICS
gx
2
2 2 2u cos θ
LOCUS
53
= x.tan 45° –
⇒
gx
2u
gx
y = x −
u
2
2
× cos
2
Y
45°
2
θ
g (a + b) u
2
g
⇒
2
u
2
=
h (a+b, 0)
Putting (a + b, 0) in the path equation, we get 0 = (a + b) −
(a, h)
u
g
2
a
1
X
b
.....(i)
a+b
Again, putting (a, h) in the path equation, we get, h = a−
⇒
h=
ga u
2
2
=a−
a
2
[using (i)]
a +b
ab a+b
Example : 37 A particle is projected up an inclined plane of inclination incli nation β at an elevation α to the horizontal. Show that (a)
tan α
= cot β
(b)) (b
tan α
= 2 tan β , if the particle strikes the plane horizontally.
ri ght angles + 2 tan β , if the particle strikes the plane at right
Solu So luti tion on:: (a (a)) Let t be the time of flight from O to A. Then X Y A
u
–
α
( o s u c
β )
–
( α i n
β )
α
s u
β O
t =
2u si s in(α − β )
... (i)
g cos β
Now,, we shall consider the motion of the particle Now particl e along x-axis. Here u x = u cos (α – β);
v x = 0
(∵ Particle strikes the inclined plane at right angles)
a x = – g sin β;
Applying v x = u x + a xt space along x-direction, we get 0 = u cos(α
KINEMATICS
− β ) − g sin β .t
LOCUS
54
⇒
t =
u cos(α − β )
... (ii)
g sin β
Dividing (i) by (ii), we get 2 tan(α
− β )
cot β
⇒
=1
2 tan(α
− β ) = cot β
tan α − tan β ⇒ 2 1 + tan α . tan β = cot β ⇒ 2 tan α − 2 tan β =cot β + tan α ⇒ tan α = cot β + 2 tan β Hence proved. (b)) When the particle strikes the plane horizontally, (b horizontally, along the vertical direction, we have,
0 = v sin α – gt ⇒ t =
v sin si n α
... (iii)
g
substituting t from (iii) in (i), we get v sin α g
=
2u si s in(α − β ) g cos β
⇒
sin α . cos β = 2 [sin α . cos β
⇒
2 co cos α . sin β = sin α . cos β
⇒
2tan β = tan α. Hence proved.
− cos α . sin β ]
Example : 38 A particle is fired at an angle θ = 60° along the direction of the breadth of a rectangular building of dimension 7m × 8m × 4m so as to sweep the edges. Find the range of the projectile. Solution: Since the projectile projecti le touches A & B, both of these
y
points lie on the path of the projectile. After putting the coordinates of A in the trajectory equation of projectile, we obtain
u
θ
x
y = h = x tan θ –
gx
... (i)
2 2 2 v0 cos θ 2
v0 sin sin 2θ g
2
KINEMATICS
x + h) = 2( x
x(
v0 sin 2θ g
... (ii)
h +2
A
B
C
D 2h R
2
As we know R = x + x + 2 h =
⇒ R =
( x, h)
,h )
x x
LOCUS
55
Using (i) & (ii) 2
R g − h R 2 h = − h tan θ − 2 gR 2 cos 2 θ sin2θ
1R R ⇒ h = − h tan θ − − h R 2 2
⇒ ⇒
2
. ta n θ
2 R − 4 Rhcot θ − 4 2h = 0
R=
4h co c ot θ
±
16h 2 cot 2 θ + 16h 2 2
1 + cosθ + = θ h c o s e c ) 2 ⇒ sin θ ⇒ R = 2h cot (θ /2) Substituting θ = 60° & h = 4 m R = 2h(cot θ
R = 2 × 4 cot 30° = 8 3 m.
Example : 39 Particles are projected from the same point in a vertical plane with velocity
2 gk ; prove that the locus of the
vertices (points with maximum height) of their paths is ellipse : x2 + 4 y( y − k) = 0. Solution: If α be the angle of projection and ( x1 , y1 ) be the coordinates of the vertex of one of the trajectories, then
x1 =
u 2 sin α . cos α g
=2
gk sin α . cos α
g
[∵ u =
= 2k sin α. cos α and an d y1 =
2 2 u sin α
2g
=
2 gk sin 2 α 2g
= k sin 2 α
Eliminating α from these two expressions, we get
y1 1 − y1 = 4 y (k − y ) 1 1 k k
x12 = 4k 2 sin 2 α .c . cos2 α = 4k 2
KINEMATICS
2 gk ]
LOCUS
56
or
x12
+ 4 y1 ( y1 − k) = 0
Generalising, we get 2
x
+ 4 y( y − k) = 0 Hence proved.
Example : 40 A steamer going downstream overcame a raft at a point P. P. 1 hr later it turned back and after some time passed the raft at a distance 6 km from fr om the point P. P. Find the speed of river if speed of steamer steam er relative to water remains r emains constant. Solution: Let u be the speed of steamer relative to water and v be the speed of river flow. f low. During first the 1 hr. distances travelled by the steamer and raft are ( u + v) km and u km respectively. respectively. Now when the steamer turns back (say from point M), its velocity becomes ( u – v) km/hr and say it passes raft at point Q after time t from the time of crossing M.
Therefore, we have
(u + v)×1 = (u – v)t + v (1 + t ) = ut + v
⇒
t = 1 hr.
Again, we have, v(1 + t ) = 6
⇒ v=
6 1 + t
6
= = 3 km/hr 2
Alternatively: If we analyze the motion of the steamer from the raft frame, then in this frame raft and river water have zero speed and therefore time taken by steamer to come back to raft is same as time for which it i t had moved away.. Hence, total time of motion is 2 hrs. But we know that during this interval away i nterval raft has moved a distance of 6 km relative to ground.
∴
speed of raft (i.e. speed of river flow) =
KINEMATICS
6 2
= 3 km/hr.
LOCUS
57
Example : 41 A car and a truck start moving movi ng simultaneously along the same straight line l ine and from the same point. The car moves with constant velocity of 40 m/s m/ s and the truck starts from rest with wi th constant acceleration of 4 m/s². m/s² . Find the time t 0 that elapss before the truck catches the car. Find also the greatest distance between them prior to it and the time at which this occurs. Solution: If motion starts at t = 0, then at some time t , position of car is x1= 40 t
and position of truck is x2 =
1 2
at t = t 1
At t = 0 C •
× 4 × t 2 = 2t 2
T
40 m/ s 4 m/s²
⇒
2t02
= 40t0
T
C
•
•
x1
When truck catches the car x2 = x1
at t = t 1
⇒
t 0
x2
= 20 sec
S
Before this distance between them is s = x1 − x2
= 40 t – 2t 2 when s is maximum,
ds dt
=0 ⇒
40 – 4t = 0
⇒ t = 10 sec and, smax = s (10) = 40 × 10 − 2 × (10) 2 = 400 − 200 = 200m.
Alternatively: v
C
c
aT
Ground frame: T aT
Frame of truck:
C
at t = 0
aT
vc aT
T
aT
at t=0
vc
C •
T
(Rest )
In this frame, as initial velocity and acceleration of car are in i n opposite directions and acceleration is constant, the car will first fir st go away from the truck with decreasing speed. After some time tim e its speed becomes zero and then it will start coming back towards the truck with wit h increasing speed. Obviously the truck is at rest rest in this frame. KINEMATICS
LOCUS
58
Let the car come back to the truck (i.e. truck catches it in ground frame) at some time t , then x = ut +
applying
vc t −
0=
⇒
1 2
1 2
2
at , we get
⇒
aT t 2
0 = 40t −
1 2
× 4× t 2
t = 20 sec
At smax, velocity of car must be zero, therefore, applying applyi ng v 2 0 2 = ( 40)
2
− 2 × 4 × smax
⇒
smax
=
1600 8
= u 2 + 2 ax , we get
= 200m.
If car is at maximum distance from truck at time t 0, then applying v = u + at, we get 0 = 40 – 4 × t 0
⇒
t 0 = 10 sec.
Example : 42 Two ships, 1 and 2, move with constant velocities veloci ties 3 m / s and 4 m / s along two mutually perpendicular straight tracks toward the intersection point O. At the moment t = 0 the ships were located at the distances 120 m and 200 m from the point O. How soon will the distancebetween distancebetween the ships become the shortest and what is it equal to to ? 3 m/s
120 m 4 m/s 2
Solution: Frame of ship 2:
O
2 00 m
Let v12 be the velocity of ship 1 w.r.t. w.r.t. ship 2. We see that in this frame ship2 is at rest and ship 1 moves along the straight line coinciding with the direction fo v12. The distance between the ships is minimum when ship shi p 1 reaches the point P. P. Therefore, we have to find out the distance BP BP.. 4ms
A
1
θ
θ v12 4 ms
We have, v12 (velocity of 1 relative to 2) =
4 2 + 32
tanθ = 4/3 KINEMATICS
3ms
= 5 m/s
v12
≡
4ms
2
1
O
2 (Rest) B
θ
Q
P
O
LOCUS
59
∴
4
sinθ =
5
and cosθ =
3 5
In triangle AQO, QO
tanθ =
∴
⇒ QO = AO. tanθ = 120 ×
AO
4 3
= 160
BQ = 200 – 160 = 40 40 m.
In the same triangle, cosθ =
AO
3
⇒
AQ
5
=
120
⇒
= 200 .m AQ
AQ
In triangle BPQ, BP
⇒
cosθ = BQ
∴
BP= BQ.cos θ
Shortest distance = BP = BQ. cosθ = 40 ×
Again,
QP
∴
5
⇒
sinθ = BQ = 40 ×
3
4 5
= 24m. QP = BQ.sin θ
= 32m.
Time taken by ship 1 to reach point P is AP v12
=
+ QP (200 + 32) m 232 AQ v12
=
5m s
=
5
sec.
= 46.4 sec. ALTERNATE AL TERNATE METHOD: METH OD:
suppose that that ship 1 starts moving moving from point B (at t = 0) and ship 2 from point A (at t = 0 only), then after time ‘t ’, ’, 1 must have travelled 4t units along +ve x-axis and 2 must have travelled 3 t units along–ve y axis. Therefore, at some ‘t ’ ship 1 is at point (–(200–4t),0) and ship 2 is at point (0,120–3t ). ). Therefore, distance between 1 and 2 is d
= ( x1 − x2 )2 + ( y1 − y2 )2
A•
=
When d is minimum, d² is also minimum. Therefore, derivative derivat ive of d² w.r.t. w.r.t. time must be zero. KINEMATICS
2
B 4t
3t (0, 120 120 – 3t )
1 (–(200–4 (–(2 00–4t ), 0)
O
LOCUS
60
∴
2( 200 − 4t )( −4) + .(120 − 3t )( ) ( −3)
⇒ ⇒
−1600 + 32t − 720 + 18t = 0 50t = 2320 ⇒ t = 2320/50 = 46.4 sec.
=0
Substituting t = 46.4 sec in equation (i), we will get the minimum distance, which comes out to be 24 m.
Example : 43 An elevator is descending with uniform acceleration. To measure measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor fl oor in 1 second. Calculate from these data the acceleration accelerat ion of the elevator. (g = 32 ft/s²) Solu So luti tion on:: Met Metho hod d 1:
Ground frame : Initial velocities of both elevator and coin are zero.
a
Suppose that the coin falls on floor in t seconds, then the distance travelled by coin = distance travelled by
6 ft
g
floor + initial distance between coin and floor 1 2
gt 2 = 6 +
1 2
at 2
gt ²
at ²
2
2
Substituting t = 1, we get g = 12 + a
⇒
/ s². a = 32 – 12 = 20 ft
Method 2:
Elevator frame:
≡
In this frame, elevator is at rest and coin starts falling f alling with zero initial speed and constant acceleration (g – a). If coin falls on the floor after t seconds, then
KINEMATICS
LOCUS
61
6=
1 2
( g − a )t 2
substituting t = 1, we get a = g − 12 = 32 − 12 = 20 ft / s²
Example : 44 Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d at t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. CA. At what time will the particles meet each other ? Solution: The paths traced out by the particles A, B and C are roughly shown in the figure. figur e.
A
By symmetry they always maintain an equilateral triangle (of reducing size) and will finally meet at the centroid of the t he triangle. If we focus on only two particles and find time of collision then at that instant third must be colliding with these two. Hence, we focus on motion of A
B
C
from the frame of B. Frame of B:
We see that relative velocity of A along AB is
3v 2
. Therefore, time of collision =
d 2 d = 3v 3v 2
Now, I would like to suggest you y ou to compare this case with wit h illustration illustrat ion V of example 21. Why am I asking you to do this? Because in this case velocity veloci ty of A is not along B (in the frame fram e of B) but collision still st ill occurs. You have to figure out the reason behind this.
Example : 45 An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The speed of plane with respect to t o air is 150 m/s. (a) Find the direction di rection in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B. KINEMATICS
LOCUS
62
Solution: Suppose that the pilot heads the plane at an angle θ with the line joining A an and d B (as shown in figure). As the net velocity is along the AB, we must have vw sin 30° = vP sinθ
⇒
20 ×
1
2 sin θ = 150
∴ θ
= sin
−1
=
North
1 15
vnet
1 15
[Ans (a)] vw
Again, vnet = vw cos 30° + vP cos θ
= 20 ×
B
+ 150 ×
3 2
30°
− 12 15
15
vP
θ
2
East A
= 10 3 + 10 224 =
∴
time taken by plane to reach from A to B is
∆t =
AB
= 500 ×103 ×
vnet
1 60
min
= 50 min [Ans (b)].
Example : 46 Two boats, A and B, move away from a buoy anchored at the middle of a river along mutually perpendicular straight lines : The boat A along the river, and the boat B across the river. Having moved off an equal distance from the buoy the boats returned. Find the ratio of the times of motion of boats, T A / T t he each T B , if the velocity of the boat with respect to water is n = 1.2 times greater than the stream velocity velocity.. Solution: Let u be the stream velocity, velocity, then velocity of each boat(with respect to water) is v = 1.2 u. Moving away from buoy : If A moves away a distance d in time t A and B in t B, then t B =
d v B
=
d v cos θ
d
= v.
v2
− u2 v
vB
v
θ
d
=
and an d t A =
KINEMATICS
v2
− u2
d v+u
[∵ v sin θ = u i.e. sin θ = u / v]
u
M
v A =v
+u
LOCUS
63
Coming back to buoy: If A takes a time of t A' seconds and B takes t B' seconds in coming back to
buoy, then d
t B' =
and an d t A' =
v2
− u2
d
u
v−u
θ
Therefore, total time of motion of B is
v2
vB
M
2d
B+ t 'B = 2t B =
T B = t
v
v A=v –u
− u2
and that for A is T A = t A
=
+ t A' =
d v+u
+
d v−u
2 vd v
Hence
2
− u2 T B T A
= =
2vd v 2 − u 2 v 2 − u 2 = 2 2 v v 2 − u 2 v − u 2d
v
2
− u2 v
(1.2u )2 − u 2
=
1.2u
1.2
Hence T A T B =
(1.2)
2
−1
=
(1.2) 2 − 1 1.2
= 1.8
Example : 47 A spider has fastened one end of a ‘super-elastic ‘super-elastic’’ silk thread of length 1 m to a vertical wall. A small caterpillar is sitting somewhere on the thread. The hungry spider, whilst not moving from its original position starts start s pulling the other end of the thread with uniform speed, v0 = 1 cm/s. Meanwhile the caterpillar starts fleeing towards the wall with a uniform speed of 1 mm/s with respect r espect to the moving thread will the caterpillar reach the wall ?
v
KINEMATICS
0
LOCUS
64
Solution: The velocity of the thread at a distance of x x meters from the wall is obviously proportionately smaller than the velocity of the end of the thread, i.e. it is xv0.
If this value is greater than t han the speed of the caterpillar, then the latter will move away from the wall. Its I ts situation will become more and more hopeless, and it will never reach the wall. v0
x m.
1 m.
On the other hand, if vcaterpillar > xv0, the net velocity of the caterpillar is towards the wall and increases as time passes, with the consequence that the caterpillar will certainly reach the wall. The limiting case corresponds to x = vcaterpillar / v0 = 0.1 m. Starting at this point, the caterpillar does not move in either direction.
Example : 48 A boatman sets off from one bank of a straight, uniform canal for a mark directly opposite the starting point. The speed of the water flowing in the canal is v everywhere. The boatman rows steadily at such a rate that, were there no current, the boat’s speed would also be v. He always sets the boat’s boat’s course in the direction of the mark, but the water carries him downstream. How far downstream does the water carry the boat ? What trajectory does it follow with respect to the bank? Solution: Denote the width of the canal by d and draw a straight line perpendicular to its bank at a distance d downstream from the boat’s starting point A (see figure). d /2 /2 F
v
vnet d
v
v A
d
The boat is initially at distance d both from F on the opposite bank and from this straight line. As both the speed of the flow and that of the t he boat with respect to water are v, the water flow takes the boat downstream by the same distance as is covered by the boat in the direction of F . KINEMATICS
LOCUS
65
This means that the boat is always equally far from the point F and the straight line. The path of the boat is, therefore, a parabola with F as its it s focus and the straight line as its directrix. directr ix. After a very long time, the boat approaches the opposite bank at a point d /2 from F .
Example : 49 Smugglers set off in a ship in a direction perpendicular to straight shore and move at constant speed v. The costguard’s cutter is a distance a from the smugglers’ ship and leaves the shore at the same time. The cutter always move at a constant speed in the direction of the smugglers’ ship and catches up with the criminals when at a distance ‘a’ from the shore. How many times greater is the speed of costguard’s cutter than that of the smugglers’ ship ? Solution: Let kv denote the speed of coastguard’s cutter, i.e., k is the required ratio of the speeds of the two vessels.
At a general time t , as shown in the figure, the t he distance x between the ships (initially a ) decreases by dx = kvdt – v sin α.dt in time dt. Meanwhile, the distance for the cutter from the shore increases by dy = kv sin α.dt ,
where α is the angle the instantaneous velocity of the cutter makes with the shore. We know, t 0
∫ v.dt = a, 0
a
and an d
∫ dx = a
⇒
0
t0
t 0
0
0
∫ kvdt − ∫ v sin α .dt = a
t 0
⇒
∫
k . a – v sin α .dt
=a
0
t 0
⇒
∫
v sin α .dt = ka − a 0
Again,
a
t 0
0
0
∫ dy = a ⇒ ∫ kv sin α .dt = a t 0
⇒
∫
k .v sin α .d t 0
KINEMATICS
=a ⇒
k .( .(ka – a) = a
LOCUS
66
⇒
k2
− k − 1 = 0 ⇒
k =
1+ 5 2
≈ 1.68
[This value (k ≈ 1.68) is the famous ‘golden mean’ associated with the fibonacci f ibonacci series.]
KINEMATICS
LOCUS
67
EXERCISE OBJECTIVE LEVEL – 1 1.
Which statements can be possible cases in one/two dimensional motion : (a) A body has zero zero velocity and still be accelerating (b) The velocity of an object reverses its direction when acceleration is constant constant (c) An object be increasing in speed speed as its acceleration decreases decreases (d) None of of these. these.
2.
A car is moving eastwards with velocity 10 m/s. In 20 sec, the velocity changes to 10 m/s northwards. The average acceleration acceleration in this time is : 2 2 m / s towards N–W
(a) 1
(b) 1
(c) 1 2 m/ s 2 towards N – W 3.
2 2 m / s towards N – E
(d) 1 2 m/ s 2 towards N.
Choose the wrong statement : (a) Zero velocity of a particle does not necessarily mean that its acceleration is zero (b) Zero acceleration of a particle does not necessarily mean that its velocity is zero. (c) If speed of a particle is constant, constant, its acceleration must be zero. (d) None of of these. these.
4.
Choose correct statements : (a) average speed is never less than magnitude of average velocity (b) average velocity is always along velocity velocity (i.e. instantaneous velocity) velocity) (c) rate of change of speed is acceleration acceleration (d) magnitude of acceleration is rate of change of speed.
5.
A small block slides without friction fricti on down an inclined plane stating from rest. Let sn be the distance travelled from t = n-1 to t = n. Then (a)
6.
2n − 1 2n
sn sn +1
(b)
is :
2n + 1 2n − 1
(c)
2n − 1 2n + 1
(d)
2n 2n + 1
Pick up the correct statements : (a) Area under a – t graph gives velocity (b) Area under a – t graph gives change in velocity (c) Path of projectile as seen by by another projectile is parabola parabola (d) A body, body, whatever be its motion, is always at rest in a frame of reference fixed to body itself.
KINEMATICS
LOCUS
7.
68
A particle has initial velocity 2iˆ + 4 ˆj and retardation 4iˆ + 8 ˆj . The distance traveled by particle from t = 0 to t =1s is (a) 5 units
8.
6 units
(c)
7 units
(d) data is insufficient.
A stone projected from the ground level falls fal ls on the ground after 4 second. Then the height of the stone 1 sec after the projection is : (a) 5 m
9.
(b)
(b) 10 m
(c) 15 m
(d) 20 m.
A stone is dropped into a well in which level of water is h meters below the to top of the well. If v be the velocity of sound, the time T after which the t he splash is heard is : (a)
2h v
2h
(b)
g
+
h
2h
(c)
v
g
+
h
h
(d)
2v
2g
+ 2h . v
10.
A body thrown vertically up from the ground passes the height 0.2 m twice at an interval of 10 sec. What was its initial velocity. [g = 10 m/s²] (a) 52 m/s (b) 26 m/s (c) 35 m/s (d) None of these. these.
11.
A lift starts moving from ground from rest. Its acceleration is plotted against time in the following graph. When it comes to rest its height above its starting point is : (a) 20 m
a(m/s ²)
(b) 64 m (c) 32 m
2
(d) 36 m. 12.
8
0
12
4
t(s)
–2
The velocity - time graph of a moving particle is shown in the figure. The maximum acceleration is (a) 1 m/s² (b) 2 m/s² (c) 3 m/s² (d) 4 m/s².
V(m/s) 60 40 20 0
13.
t(s )→ 1 0 2 0 3 0 4 0 5 0 60 7 0 8 0
For given velocity time graph, average acceleration is zero for (a) [ 0, t 1 ] (b) [t1 , t 2 ] (c)
[t1 , t 3 ]
(d) [t 2 , t 4 ] .
KINEMATICS
V
15 10 5 O
t 1
t 2
t 3
t 4
t
LOCUS
14.
69
The co-ordinates of a moving particle at any time t are given by x= ct2 and y= bt2 . The speed of the particle is given by (a) 2t ( c + b)
15.
2u 2 3g
2g
.
(b) 1.2 s
(c) 1.4 s
(d) 1.6 s.
(b) tan
−1
1 2
(c) tan
−1
1 5
(d) sin
−1
1 5
.
(b) u 2
(c) u
(d) u
2.
(b) v0 sin θ 1
(c) v0 cos θ 1
(d) v0 sin θ 1 .
(b) 5 / R 2
(c) 2 R
(d) 3 R.
(b) 3
( c) 4
(d)
41 .
Rain is falling vertically downwards. To a man running eastwards, the rain will appear to be coming from (a) east
23.
2
A boat, which has a speed of 5 km/hr in i n still water, crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in Km/hr is : (a) 1
22.
(d)
3g
u
A particle is thrown from a platform with velocity v at an angle of 60° with the horizontal. The range obtained is R. If the platform moves horizontally in the direction of target with velocity v /2, the range will increase to : (a) 3/ R 2
21.
2
Two particles are projected project ed from same point on a horizontal plane with same initial speed v0 in two different directions such that their horizontal ranges are same. Ratio of their maximum heights will be : (a) tan 2 θ 1
20.
(c)
2g
u
A body is projected with a speed u at an angle to the horizontal to have maximum range. At the highest point the velocity is : (a) u/2
19.
(b)
3u 2
A bomber plane is moving horizontally with a speed of 500 m / s. If a bomb released from it strikes the ground in 10 sec. The angle at which it strikes the ground is ( g = 10 m / s²) (a) tan −1 5
18.
(d) 2t c 2 + b 2 .
A particle is thrown with a speed of 12 m / s at an angle 60° with the horizontal. The time interval between the moments when its speed is 10 m / s is (g 10 m / s²) : (a) 1.0 s
17.
(c) t c 2 + b 2
The speed at maximum height of a projectile is half of its initial velocity u. Its range on the horizontal plane is. (a)
16.
(b) 2t c 2 − b 2
(b) west
(c) north east
Displacement time graphs for two particles P1 & P2 are shown in figure. Their relative velocity :
KINEMATICS
(d) south west.
LOCUS
70 x
(a) is zero
P1
(b) is non zero zero and constant constant (c) continuously decreases (d) continuously increases. 24.
The velocity of a particle is zero at t = 0.
P2
O
t
(a) The acceleration at t = 0 must be zero (b) The acceleration at t = 0 may be zero (c) If the acceleration acceleration is zero zero from t = 0 to t = 10 s, the speed is zero in this interval. (d) If the speed is zero from t = 0 to t = 10 s, the acceleration is zero in this interval.
KINEMATICS
LOCUS
71
OBJECTIVE LEVEL – II 1.
A point moves rectilinearly. rectilin early. Its displacement x at time t is given by x 2 (a)
2.
1 x
(b)
3
x
2
(c)
−t x
(d) None of these. these.
2
(b) decreases
(c) stays constant
(d) becomes zero.
(b)
1 K
ln( Kut )
(c)
1 K
ln(1 + Kut )
(d) K ln(Kut ). ).
Velocity of a particle part icle moving along x-axis is given as v = x 2 − 5 x + 4 (in m/s) where x denotes the position of the particle in meters. The magnitude of acceleration of the particle when the velocity of the particle is zero is : (a) 0 m/s²
5.
1
A point moves in a straight line under the retardation Kv². If the initial velocity is u, then distance covered in ‘t ’ seconds is : (a) Kut
4.
x
−
The velocity of a car moving on a straight road increases linearly according to equation V = a + bx, where acceler ation in the course of such motion : ( x is the distance travelled) a and b are positive constants. The acceleration (a) increases
3.
1
= t 2 + 1 . Its acceleration at time t is :
(b) 2 m/s²
(c) 3 m/s²
(d) None of these.
The position of a particle moving along x-axis is given by x =3t ² – t ³ where x is in meters and t is in seconds. Consider the following statements : (i) displacement of particle after 4 s is 16 m (ii) distance traveled traveled by particle upto 4s is 24 m (iii) displacement of the particle after 4s is –16 m (iv) distance covered by the particle upto 4s is 22 m
6.
(a)) St (a Stat atem emen entt (i (i)) an and d (i (ii) i) on only ly ar aree co corr rrec ectt
(b)) St (b Stat atem emen ents ts (i (ii) i) an and d (i (iii ii)) on only ly ar aree co corr rrec ectt
(c)) Sta (c tate tem ments (i (i)) and (i (iii ii)) on only ly are corr rreect
(d)) no (d none ne of th thes esee.
The velocity of a particle defined by the relation v = 8 – 0.02 x, where v is expressed in m / s and x in meter knowing that at t = 0, x = 0 then at t = 0, acceleration is (a) – 0.16 m/s²
7.
(b) – 0.02 m/s²
(c) 7.08 m/s²
(d) 8 m/s².
A particle moves according to the equation t = ax² + bx, then the acceleration of the particle when x =
(a)
−
a b
KINEMATICS
3
(b)
−
2 a gb
3
(c)
−
2 a 27 b
3
(d) None of these. these.
b a
is
LOCUS
8.
72
A particle is moving with velocity v x
x
(a)
(c)
t
T
O
t
T
(d) None of these. these.
t
T
O
Which of the following graph correctly represents velocity-time velocity- time relationship for a particle released from rest to fall freely under gravity ? V
V
(a)
(c) t
O
t
V
V
(b) O
10.
x
(b) O
9.
t = v0 1 − . Which one is the correct graph for x - t T
(d) t
O
. t
O
A ball is dropped from a height d above the ground. It hits the ground and bounces vertically up to a height /2. Neglecting air resistance, its velocity varies with height above the ground, as. d V
V
V
V
d d
(a)
O
d h
(b)
(c)
O
h
O
(d)
h
.
h
O
11.
d
The acceleration time graph of a particle moving on a straight line is as shown in figure. At what time ti me the particle acquires its initial velocity ? a(m/s²) (a) 12 sec
10
(b) 5 sec (c) 8 sec (d) 16 sec. 12.
O
t ( sec)
The trajectory of a particle in a vertical plane is y= − x2 + 6 x− 4 where x and y are respectively the horizontal and vertical distance of the projectile from the point of projection on ground. The maximum height attained is (a) 4 m
13.
4
(b) 5 m
(c) 7 m
(d) None of these.
A particle moves in x - y plane with a velocity v x = 8t – 2 and v y = 2. If it passes through the point x = 14 and y = 4 at t = 2 s, the equation of the path is (a) x = y2
KINEMATICS
− y+ 2
(b) x = y + 2
(c) x = y 2 + 2
(d) x = y2 + y + 2 .
LOCUS
14.
73
A particle projected from O and moving freely under gravity strikes the horizontal plane through O at a distance R from it as shown in the figure. Then : (a) There will be two angles of projection projection if Rg < u² (b) the two possible angles of projection are complementary (c) The product of possible times of flight from O to A is 2 R/g Rg = u². (d) There will be more than two angles of projection projection if Rg u
A
α
O R
15.
A ball is projected horizontally from an inclined plane with a velocity v0 as shown in the figure. It will strike the plane after a time (a)
v0
3g
(b)
2v0 3g
(c)
v0 g
(d)
2 3v0 g
.
v
0
60°
16.
Two particles start from rest simultaneously and are equally accelerated. Throughout the motion, the relative velocity of one with respect to other is is : (a) zero (b) none zero and is directed directed parallel to acceleration acceleration (c) non zero and is directed opposite opposite to acceleration (d) non zero and is directed directed perpendicular perpendicular to acceleration.
17.
18.
Two particles part icles P1 and P2 are moving with constant velocities v1 and v2 respectively. Which of the statement about their relative velocity is true ? (a) it can not be greater than v1 + v2
(b) it can not not be greater greater than v1 – v2
(c) it is always always greater than v1 + v2
(d) it is always smaller than v1 – v2.
What are the speeds of two objects if they move uniformly towards each other, they get 4 m closer in each second and if they move uniformly uniforml y in the same direction with wit h the original speeds they get 4 m closer in each 10 sec ? (a) 2.8 m/s and 1.2 m/s
(b) 5.2 m/s and 4.6 m/s
(c) 2.8 m/s and 2.1 m/s
(d) 2.2 m/s and 1.8 m/s.
KINEMATICS
LOCUS
19.
74
Two particles positioned at A(5, A( 5, 3) and B(7, 3) are moving with constant velocity veloci ty 2 iˆ + 3 ˆj and xˆi+ yˆj respectively. After 2s they collide, then the values of x x and y are respectively (a) 2, 2
20.
(b) 1, 3
(c) 3, 2
(d) 1, 1.
Two particles start moving simultaneously from points (0, 0) and (1, 0) respectively in the OXY plane with uniform velocities v1 and v2 as shown in the figure. It is found that they collide. Then (a) v1
= v2
(b) v1
= v2
(c) v2
=
(d) v2
2
y
2v 1
= v1
3 2
v
v .
45°
30°
O 21.
2
1
(1, 0)
x
Pick the correct statements : (a) Average speed of a particle in given time is never less than the magnitude of the average average velocity.
(b) It is possible to have have a situation in which which ;
dv
dt
≠ 0 but
d dt
v
= 0.
(c) The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval. (d) None of of these. these. 22.
Mark the correct statements for a particle moving in a straight str aight line : (a) If the velocity & acceleration have opposite sign, the object is slowing down (b) If the position & velocity have opposite sign, the particle is moving towards the origin. (c) If the velocity is zero at an instant, the acceleration should also be zero at that instant (d) If the velocity is zero for a time interval the acceleration is zero at any instant within the time interval.
KINEMATICS
LOCUS
75
SUBJECTIVE
LEVEL – I
1.
A person travelling on a straight line moves with a velocity velocit y v1 for some time and with uniform velocity v2 for the next equal time. Find average velocity. velocity.
2.
A point traversed half the distance with a velocity v0. The remaining part of the distance was covered with velocity, v1 for half the time, and with velocity v2 for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
3.
Show that time taken by a particle to slide from any point on a vertical circle along a smooth chord terminating at the lowest point of the circle is constant.
4.
A car accelerates from rest at a constant rate a for sometime after which it decelerates at constant rate b to come to rest. If the total time lapse is t 0 seconds, evaluate (i) the maximum velocity velocity reached reached and (ii) total distance distance travelled. travelled.
5.
A body is released from a height and falls freely towards the earth surface. Exactly 1 sec later another body is released from the same point. What is the distance between the bodies bodies 2 sec after the release of the second body if g = 9.8 m/s² ?
6.
A ball projected vertically vert ically upwards from f rom A, the top of a tower, reaches the ground in t 1 seconds. If it is projected vertically downwards from A with the same velocity, it reaches the ground in t 2 seconds. If it falls freely from A, show that it would reach the ground in
t1t 2 seconds.
7.
A ball is allowed to slip from rest down a smooth incline plane and the distances dist ances are marked every 2.0 s. If the second mark is made 1.6 m from the starting point, where are the first and fourth marks ?
8.
For a particle moving in a straight line with a constant acceleration if t particle when its velocity is zero.
9.
Instantaneous velocity veloci ty of a particle moving in +ve ve x x direction is given as v
= =
x + 3 , find the position of the
3
. At t = 0, particle starts x + 2 x = 2) and Q ( x x = 4) of its from origin. Find the average velocity of the particle between the two points P( x motion path.
KINEMATICS
2
LOCUS
76
10.
For given velocity-time velocity-ti me graph find time interval of maximum length for which av. acceleration is zero
11.
For given path find point P such that av. av. velocity of motion moti on from A to P is along velocity at point P. y
A x
O
12.
Plot x – t and a – t graphs for given v – t graphs. It is given that initially particle is at x = 10. v
O
13.
t 0
t
A particle starts from the origin at t = 0. It moves in a plane with velocity given by v
= uiˆ + ( aω cos ω t ) ˆj .
Find the equation of trajectory of the particle. 14.
Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed u at an angle θ with the horizontal
15.
A bomb is dropped from a plane flying horizontally with uniform speed. show that the bomb will explode vertically below the plane. Is the statement true if the plane flies with uniform speed but not horizontally ?
16.
If R be the range of a projectile on horizontal plane and H 1, H 2 be the maximum heights for its two possible trajectories, find the relation between the given parameters.
KINEMATICS
LOCUS
77
SUBJECTIVE
LEVEL – II
1.
2.
A street car moves rectilinearly from fr om station A to the next stop B with an acceleration varying according to the law a = α – β x where α and β are positive constants and x is its distance from A. Find the distance between these two stations and maximum velocity of the car. The graphs A, B and C in figure shows, the position s, velocity v
=
ds dt
, and acceleration a
=
dv dt
=
2
d s 2
dt
of a body moving along y-axis as function of time t. Which graph is for displacement, velocity and acceleration ? x/v/a
t O
A
B
C
3.
For given velocity-time graph, plot approximate acceleration, speed, displacement and distance variations with respect to time V v0 2t 0 O
t 0
t
–v0
4.
For given acceleration-time graph plot approximate velocity-time velocity-ti me and position-time graphs. It is given that motion starts from rest and initially particle is at origin. a
O
KINEMATICS
t 0
2t 0
t
LOCUS
5.
78
For given ‘a – t ’ graphs plot approximate ‘v – t ’ graphs. a
a
a
a
0
0
t O
t 0
3t
2t
0
0
0
(A)
6.
t
t
O (B)
For given ‘v – x’ graph plot ‘ x – t ’ graph. v v
0
O
7.
x
x
0
xin = +5. For given ‘v – x’ graph plot ‘ x – t ’ and ‘a – t ’ graphs, if x v v=α x
O
8.
x
2 A radius vector of a point A relative to the origin (i.e. position vector) varies vari es with time t as r = α t iˆ − β t ˆj , where α and β are positive constants, and iˆ and jˆ are the unit vectors of the x and y axes. Find :
(a) The equation of the point’s trajectory y( x x); plot this function ; (b) velocity, speed, speed, acceleration and magnitude of acceleration as functions of time; (c) the time dependence of the angle θ between velocity and acceleration ; (d) average velocity and its magnitude for first t seconds of motion. 9.
A body is projected up such that its position vector varies with time as r
(a) in initial velocity 10.
(b)initial speed
= 6tiˆ + (8t − 5t 2 ) ˆj . Find the
(c) ti time of flight.
y plane from the ground level with an initial velocity A cricketer hits a ball in a vertical x- y vel ocity u
= (10
3iˆ + 30 ˆj )
m/s. Find the time in which velocity vector makes an angle of 30° with horizontal. 11.
A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal
KINEMATICS
LOCUS
79
road with an acceleration of 1 m/s² & the projection velocity in the vertical verti cal direction is 9.8 m/s. How far behind the boy will the ball fall fal l on the car ? 12.
A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed & the angle of projection (a) as seen from the truck, (b) as seen from the road.
13.
A boat moves relative to water with a velocity which is 2.0 times tim es less than the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting ?
14.
Two particles move in an uniform gravitational field with an acceleration (downward) g. At the initial moment the particles were located at one point and moved with wit h velocities v1 = 3.0 m/s and v2 = 4.0 m/s horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.
15.
Two particles are projected simultaneously simult aneously with same speed u in same vertical plane with angles of elevation θ and 2θ, where θ < 45° : At what time will their velocities be parallel ?
16.
A balloon moves up vertically such that if a stone is thrown from it with wit h a horizontal velocity v0 relative to it, the stone always hits the ground at a fixed point 2 v02 g horizontally away from it. Find the height of the balloon as a function of time.
17.
A gun is mounted on a trolley which can move uniformly unif ormly with speed v m/s along the x-axis. Two shots are fired from the origin with the gun making an angle 30° with the horizontal such that in the first case the trolley is moving along the x-axis and in the second case moving along the –ve x-axis. The respective range of the projectile is 250 m and 200 m, along the x-axis. Find the velocity of the trolley troll ey.. (Assume height of the trolley to be negligible)
18.
A sailor in a boat, which is going due east with a speed of 8 m/s, observes that a submarine is heading towards north at a speed of 12 m/s and sinking at a rate r ate of 2 m/s. The commander of submarine observes a helicopter ascending at a rate of 5 m/s and heading towards t owards west with 4 m/s. Find the actual speed of the t he helicopter and its speed with respect to boat.
19.
Two swimmers at point A on one bank of the t he river have to reach point B lying l ying right across on the other oth er bank. One of them crosses the river along the straight line l ine AB while the other swims at right angles to the stream and the walks the distance that he has been carried away by the stream to get t o point B. What was the velocity u of his walking if both swimmers swimmer s reached the destination simultaneously? The stream velocity velocit y v0 = 2.0 km/hr and the velocity v of each swimmer with respect to water equals 2.5 km/hr.
20.
What is the maximum angle to the horizontal at which a stone can be thrown and always be moving away from the thrower ?
21.
An object A is kept fixed at the point x =3 m and y =1.25 m on a plank P raised above the ground. gr ound. At time t = 0 the plank starts moving along the + x-direction with an acceleratio acceleration n 1.5 m / s 2 . At the same instant a stone is projected from the origin with a velocity velocit y u as shown .
KINEMATICS
LOCUS
80
y A P
1.25m
u
O
x 3.0 m
A stationary person on the ground observes the stone hitting hit ting the object during its it s downward motion at an angle of 450 to the horizontal. All the motions are in x- y y plane. Find u and the time after which the stone hits the object. Take g
KINEMATICS
= 10m / s 2
