Brilliants Kinematics

KINEMATICS

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2 In the previous lesson, we examined cause of motion  the the kinetics. In this lesson we shall study the geometry of motion i.e., kinematics. Kinematics is used to relate displacement, velocity, acceleration and time without reference to the cause of motion. We shall discuss about kinematics of the particle here. Use of the word ‘particles’ does not mean that our study will be restricted to small

corpuscles; rather, it indicates that in this lesson the motion of bodies possibly as large as men, cars, rockets, or aeroplanes will be considered without regard to their shape and size. The entire lesson is divided into two sections. In the first section we shall study about the motion in a straight line. In the second section motion of particle in a plane specially projectile motion and the concept of the relative motion of one particle with respect to another will be di scussed.

IIT JEE Syllabus: Kinematics in one and two dimension (Cartesian coordinates

only), projectiles; Relative velocity. Study More with www.puucho.com Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396

KINEMATICS

SECTION

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I

Study More with www.puucho.com Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396

KINEMATICS

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1. KINEM KINEMATIC ATIC VARIA VARIABLES BLES

Kinematics is study of geometry of motion without considering its cause. It deals with position, distance travelled by a body and its displacement, speed and velocity and acceleration of a body. Before going in details about different kinds of motion separately, let us know about these kinematics variable first. 1.1

DISTANCE AND DISPLACEMENT DISPLACEMEN T

When a particle is moving its successive position in general may lie on a curve, say, ABC shown in figure. The curve is then called the path of the particle. The total length of the path followed by the body is called the distance traversed by the body . body . Its unit in S.I. system is ‘metre ‘ metre’. ’. It is a scalar quantity. Sometimes we may not be interested in the actual path of the particle but only in its final position C relative to the initial position A. A. The directional distance between initial and final positions of the particle AC in the figure is called displacement . It is a vector quantity.

C

A

B

Fig. (1) 1.2

SPEED AND VELOCITY

Speed is the rate at which a moving body describes its paths . The path may be a curve or a straight line and its shape need not be considered to decide the speed. If a particle traverses a distance s during a time t, Average speed, speed,

v 

s t

…(1 …(1)

If the interval of time t is infinitesimally small approaching zero, this ratio is called of instantaneous speed or sometimes referred as speed of particle. i.e.,

instantaneous speed v  Lt

t 0

s ds  t dt

…(2 …(2)

Speed is a scalar quantity and in S.I. system it is measured in metre/second metre/second (m/s). Velocity is defined as the rate of change of position . If during a time internal t, a particle changes its position by r . Average velocity, velocity,



v av 

r t

…(3 …(3)


KINEMATICS

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v  Lt

Also, instantaneous instantaneous velocity,

t 0

r d r  t dt

…(4 …(4)

Velocity is a vector quantity having direction same as that of displacement and is measured in metre per second (ms1).

1.3

ACCELERATION

The rate of change of velocity is called as acceleration. The change in velocity can be either in the magnitude of the velocity or in the direction of velocity or in both of them simultaneously. If v is change in velocity which takes place in time interval 

Average acceleration, acceleration, aav 

v t

a  Lt

Also, instantaneous instantaneous acceleration, acceleration,

t, then during this interval

t  0

…(5 …(5)

v d v  t dt

…(6 …(6)

Acceleration is a vector quantity quantity and is measured measured in metre/second2 (ms2). I llustra llustratti on 1

Question:

A particle moves along a semicircular path of radius R = constant speed. For the particle calculate (take

(i)

distance travelled

(ii)

average speed

(iii)

average acceleration

m in time t = 1s with

2

= 10)

R

A

B

S olution: oluti on: (i)

Distance = length of path of particle = AB = R = 10 m

(ii)

Average speed, v = =

(iii)

Average acceleration =

distance time

=

R t

= 10 m/s

Change in velocity time taken

=

2v

t



2R 2

t

2

= 20 m/s

Can a body have (a) zero instantaneous instantaneous velocity and yet be accelerating; (b) zero average speed but nonzero average velocity; (c) negative acceleration and yet be speeding up?


KINEMATICS

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2. MOT MO T ION IN ONE DIMENSION As discussed earlier for the analysis of motion, we need a reference frame made of origin and a set of co-ordinate axis depending on the type of motion. If a motion is taking place on a straight line, we need one axis for the analysis of motion. So we call this motion as motion in one dimension. Similarly Similarly if motion is taking t aking place in a plane, we need two t wo co-ordinate axes and we call motion as motion in two dimension. Three co-ordinate axes are required if motion is random motion in space and we call such motion as motion in three dimension. Here we are going to discuss motion on a straight line i.e., motion in one dimension.

Consider a particle moving on a straight line AB. AB. For the analysis of motion we take origin, O at any point on the line and x -axis -axis along the line. Generally we take origin at the point from where particle starts its motion and rightward direction as positive x -direction. -direction. At any moment if particle is at P then its position is given by OP = x . A

O

P

B

X

x

Fig. (2)

v 

Velocity is defined as,

Acceleration is is defined as, a 

2.1

dx dt dv d 2 x v dv   dt dt 2 dx

MOTION IN A STRAIGHT LINE WITH UNIFORM VELOCITY If motion takes place with a uniform velocity v on on a straight line, then displacement in time t , s = v .t

…(7 …(7)

acceleration of particle is zero. 2.2

MOTION IN A STRAIGHT LINE WITH UNIFORM ACCELERATION-EQUATIONS ACCELERATION-EQUATIONS OF MOTION

Let a particle move in a straight line with initial velocity u (velocity at time t = 0) and with uniform acceleration a. Let its velocity be v at the end of the interval of time t (final velocity at time t ). ). Let S be its displacement at the instant t . Now, acceleration a = or,

v u t

v = = u + at

… (8 (8)

If ‘u’ and ‘a ‘a’ are in the same direction, ‘a ‘ a’ is positive and hence the velocity increases with time. If ‘a ‘a’ is opposite to the direction of ‘u ‘ u’, ‘a ‘a’ is negative and the velocity decreases with time.


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Displacement during the time interval t = = average velocity x velocity x t

S

u  v  t 2

... (9)

Eliminating v from from equations (8) and (9), we get

S

or

u  u  at 2

S  ut 

1 2

 t

at 2

... (10) 10)

Another equation equation is obtained obtained by eliminating eliminating t from from equations (8) and (9) v = = u + at or,

a =

S =

or,

aS =

= or,

v u t

v  u  t 2 v  u v  u   t 2 t v 2 u 2 2

v 2 – u2 = 2aS 2aS v 2 = u2 + 2aS 2aS

... (11) 11)

Distance traversed by the particle in the nth second of its motion The velocity at the beginning of the nth second = u + a (n – 1) – 1) The velocity at the end of the nth second = u + an Average velocity velocity during the nth second =

u  a (n  1)  u  an 2

= u

1 2

a (2n  1)

Distance traversed during this one second

 

Sn = average velocity  time = u 

1 2

 

a (2n  1)  1


KINEMATICS

i.e., Sn

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1 2

 u  a (2n  1)

... (12) 12)

The five equations derived above are very important and are to be memorized. They are very useful in solving problems in straight-line motion.

Calculus method of deriving equations of motion The acceleration of a body is defined as

a i.e.,

dv dt

dv = = a dt

Integrating, we get, v = = at + A + A Where A Where A is is constant of integration. integration. By the initial condition when t = = 0, v = = u (initial velocity), we get A get A = = u



v = = u + at

we know that the instantaneous velocity v  

dS dt

.

Displacement of body for duration dt from from time t to to t + + dt is is given by dS = v dt = (u + at ) dt On integration we get,

S At,



1

constant.  ut  at 2  B , where B is integration constant. 2

t = 0, S = 0 yields B = 0. S

= ut + +

Acceleration =



a

1 2

at 2

dv dv dS dv  .  v . dt dS dt dS

=v

dv dS

a.dS = v.dv Integrating we get,


KINEMATICS

aS = aS =

v 2 2

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is integration constant. constant.  C , where C is

Applying initial initial condition, condition, when S = 0, v = = u we get 0

or



u2 2

 C

u2 C = – = – 2 v 2



aS 



v 2 – u2 = 2aS 2aS



v 2 = u2 + 2aS 2aS

2



u2 2

If S1 and S2 are the displacements during n seconds and (n (n – 1) – 1) seconds S1 = u n + n +

1 2

an2

S2 = u (n – 1) – 1) +



1 2

a (n – 1) – 1)2

Displacement Displacement in nth second Sn = S1 – S2 = un +

Sn = u +

1 2

1 2

(n – 1) an2 – u (n – 1) – –

1 2

(n – 1) a (n – 1)2

a (2n (2n – 1) – 1)


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Question:

A certain automobile manufacturer claims that its super-delux sport’s car will –1 accelerate from rest to a speed of 32.0 ms in 8.0 s. Under the important assumption that the acceleration is constant. –2

(a)

determine the acceleration of car in ms .

(b)

find the distance the car travels in 8.0 s.

(c)

find the distance the car travels in 8 second.

th

Solution: (a)

We are given that u = 0 and velocity after 8 s is 32 m/s, so we can use v = u + at to to find acceleration

a (b)

v  u t



32.0  0 8.0

distance travelled in 8.0 s, We can use, S = ut + +

(c)

–2

= 4.0 ms

1 2

2

at = 0 +

2

 4.0  8 2  128 m

th

distance travelled in 8 second, We have, Sn = u + (2n (2n –1) –1)

= (2 × 8 – 8 – 1) 1) ×

2.3

1

a 2

4.0 2

= 30 m

GRAPHICAL REPRESENTATION OF MOTION

(i) Displacement-time Displacement-time graph: If displacement of a body is plotted on Y -axis -axis and time on X -axis, -axis, the curve obtained is called displacement-time graph. The instantaneous velocity at any given instant can be obtained from the graph by finding the slope of the tangent at the point corresponding corresponding to the time.


KINEMATICS

If the graph obtained is parallel to time axis, the velocity is zero (ab in figure). If the graph is an oblique straight line, the velocity is constant (oc (oc and ef in figure). If the graph obtained is a curve (od ( od in figure) whose slope increases with time, the velocity goes on increasing i.e., the motion is accelerated. If the graph obtained is a curve of type og in in figure whose slope decreases with time, the velocity goes on decelerated.

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y ) 0 y y = s ( t n e m e c y 0 a l p s i DO

v=0

a

b

g

decreasing velocity

c

e

d

v = constant

v = constant increasing velocity

f x

t 0 Time t

Fig. (3) Velocity-time graph: graph: Similarly a graph can be drawn between velocity and time of a moving body. The curve obtained will be similar to the one shown in figure. If the graph is a straight line parallel to time axis, the velocity is constant and acceleration is zero ( AB AB in figure). If the graph is an oblique straight line, the motion is uniformly accelerated (positive slope) or uniformly decelerated (negative slope). The velocity-time curve may be used to determine displacement, velocity and acceleration i.e., it is used to specify the entire motion. To obtain the velocity at any time t we draw a perpendicular from given instant on the curve and noting the corresponding point and dropping a perpendicular from this point on the velocity axis. The slope of the tangent at the point corresponding to the particular time on the curve gives instantaneous acceleration. The area enclosed by velocity-time graph and time axis for a time interval gives the displacement during this time interval. In figure the shaded area pqt 2t 1 represents the net displacement during the time interval between t 1 and t 2.

y

A Velocity constant ( a = 0) B g

) v ( y t i c o l e V

c

a decreasing

d f

Retardation Retardation (a) constant

a = constant a = increasing e x

Time (t )

Fig. (4)

y ) v ( y t i c o l e V

q P

t 1

Work t t + t

t 2 2

x

Time (t (t ) Fig. (5)

Acceleration-time Acceleration-time graph: It is a graph plotted between time and acceleration. If the graph is a line parallel to time axis, the acceleration is constant. If it is a straight line with positive slope, the


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acceleration is uniformly increasing. The co-relation of the graph explained above follows directly dv dS from the differential expressions expressions v  and a  .  dt dt The area enclosed between acceleration-time graph and time axis gives during this time interval change in velocity.

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Question:

–2

At t = 0 a particle is at rest at origin. Its acceleration is 2ms for the first 3s 2

and –2 ms for the next 3s. Plot the acceleration versus time, velocity versus time and position versus time graph.

S olution: oluti on:

We are given that for first 3s acceleration –2 –2 is 2 ms and for next 3s acceleration is – –2 –2 2ms . Hence acceleration time graph is as shown in the figure.

Y -2

a(ms )

X

O

The area enclosed between a-t curve and t-axis gives change in velocity for the corresponding interval. Also at t = = 0, v = = 0, hence final velocity at t = 3s will increase 1 to 6 ms . In next 3s the velocity will decrease to zero. Hence the velocity time graph is as shown in figure.

3

6

t (s) (s)

Y (ms ) v (ms -1

6

O

3

6

t (s)

Note that v – – t curves curves are taken as straight line as acceleration is constant. Now for displacement time curve, we will Y use the fact that area enclosed between x ( (m) v t curve and time axis gives 18 displacement for the corresponding interval. Hence displacement in first three second is 9 m and in next three second is 9m. Also the x the x -t curve curve will be of parabolic 9 nature as motion is with constant acceleration. Therefore x -t curve is as shown in figure.

O

X

3

6

(s) t (s)


X

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If acceleration-time graph is a straight line inclined with time axis, then what will be the nature of velocity-time graph.

2.4

VERTICAL MOTION MOTIO N UNDER GRAVITY

When a body is thrown vertically upward or dropped from a height, it moves in a vertical straight line. If the air resistance offered by air to the motion of the body is neglected, all bodies moving freely under gravity will be acted upon by its weight only. This causes a constant vertical acceleration g having value 9.8 m/s 2, so the equation for motion in a straight line with constant acceleration can be used. In some problems it is convenient to take the downward direction as positive, in such case all the measurement in downward direction are considered as positive i.e., acceleration will be + +g g . But sometimes we may need to take upward as positive and in such case acceleration will be g . Projection of a body vertically upwards Suppose a body is projected vertically upward from a point A point A with with velocity u. If we take upward direction as positive

B

(i) At time t , its velocity v = = u  gt (ii) At time t , its displacement displacement from A from A is is given by S = ut 

1 2

C

gt 2 u

(iii) Its velocity when it has a displacement S is given by A 2

2

v = u

 2gS 2gS

(iv) When it reaches the maximum height from A, A, its velocity v = 0. This u happens when t  . The body is instantaneously at rest at the highest point B. g (v) The maximum height reached u2 H  2g (vi) Total time to go up and return to t he point of projection =

2u

g

Since, S = 0 at the point of projection, S = ut 

1 2

gt 2


KINEMATICS

0 = ut 

1 2

= gt 2 or t =

Since the time of ascent =

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2u

g u

2u

u

u

, the time of descent =   g g g g

(vii) At any point C between A A and B, where AC = S, the velocity v is given by

v   u 2 2gS The velocity of body while crossing C upwards upwards and while crossing C downwards downwards is

  u 2 2gS

 u 2 2gS . The magnitudes of the velocities are the

same. A body body is fired fired upward upward with speed speed v 0. It takes time T to reach its maximum height H . True or false. H T v H (a) It reaches in (b) It has speed 0 at (c) it has speed v 0 at 2T . 2 2 2 2

I llustra llustratti on 4

Question:

A body is projected upwards with a velocity 100 m/s. Find (a) the maximum height reached, (b) the time taken to reach the maximum height, (c) its velocity at a height of 320 m from the point of projection, (d) speed with which it will cross down the point of projection and (e) the time taken to reach back the point of projection. Take g = 10 m/s

2

S olution: oluti on: (a)

The maximum height reached Initial upward velocity u = 100 m/s Acceleration a = ( g ) = – = – 10 10 m/s

2

Maximum height reached H is is given by 2

2

v = u + 2aS 2aS 2

0 = 100 + 2(  10)H 10)H H = =

(b)

100 100 2 2  10

= 500 m

The time taken to reach the maximum height


KINEMATICS

t  (c)

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100 u 100 = 10 s  g 10

Velocity at a height of 320 m from the point of projection 2

2

v = u + 2aS 2aS 2

v = 10000 + 2( 10)320 v = = 60 m/s + 60 m/s while crossing the height upward and 60 m/s while crossing crossing it downward. (d)

Velocity with which it will cross down the point of projection 2

2

v = u + 2gS 2gS At the point of projection S = 0



v = =  u

While crossing the point of projection downwards, speed = u = 100 m/s The velocity has the same magnitude as the initial velocity but reversed in direction. (e)

The time taken to reach back the point of projection

t 

2.5

100 2u 2 100

g



10

= 20 s

GENERAL MOTION IN A STRAIGHT LINE

Saying general motion means motion it is other then motion with uniform velocity or motion with uniform acceleration. In such motion we will be given the relation between two variables among position, velocity, acceleration and time and other are to be calculated. For this we need to use, Velocity,

v = =

Acceleration, a 

dx dt dv vdv  dt dx

and then using calculus we find the unknown variables.



KINEMATICS

Question:

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–2

A particle moving in a straight line has an acceleration of (3 t – 4) ms at time t seconds. The particle is initially 1 m from O , a fixed point on the line, with a velocity 1 of 2 ms . Find the time (> 4/3s) when the velocity is zero. Find also the displacement of the particle from O when t = = 3.


KINEMATICS


a

Using

dv dt



dv dt

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gives

 3t  4

v

t

2

0

 dv   (3t  4) dt



v – 2 – 2 =



v=

3t 2 2

3t 2 2

 4t

 4t  2

The velocity will be zero when

3t 2 2

 4t  2  0

i.e., when (3t (3t – 2) – 2) (t (t – 2) – 2) = 0 2



t = =



= 2s t =

Using

ds dt s

3

or 2

 v we have

ds 3t 2  dt 2

 4t  2,

3

 3t 2  4t  2  dt  ds    2   1 0 









 t 3   2t 2  2t   1.5 s – 1 – 1 =   2  0



s = 2.5

3

Therefore the particle is 2.5 m from O when t = = 3 s. s.


KINEMATICS

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KINEMATICS

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PROFICIENCY TEST

I

The following questions deal with the basic concepts of this section. Answer the following followi ng brief br iefly. ly. G o to the nex t s ection ecti on only if your yo ur s cor e is i s at leas leas t 80%. D o not n ot cons c ons ult the S tudy Material Material while att attempt empting ing these thes e ques questions tions .

1.

2.

3.

4.

5.

Describe a physical situation, for example, with a ball or a car, for each of the following cases. (a) a = 0, v  0 ;

(b) v (b) v = 0, a  0;

(c) v < v < 0, a > 0;

(d) v = 0, a < 0.

True or false (a)

Positive slope of x of x versus versus t graph graph implies motion away from origin

(b)

Negative slope on the v versus versus t graph graph means the velocity of body is decreasing

(c)

Area enclosed between v versus versus t graph graph and t -axis -axis gives displacement.

(d)

Area enclosed between a versus t graph graph and t -axis -axis gives velocity.

The position time graph in figure depicts the journey of three bodies A bodies A,, B and C . (a)

At 1s, which has the greatest velocity?

(b)

At 2s, which has travelled the farthest?

(c)

When A A meets C , is B moving faster or slower than A than A? ?

(d)

Is there any time at which the velocity of A is equal to that of B?

A

x

B C

1s

2s

3s

4s

t

A ball is thrown up from the surface of earth and falls f alls back on it. Taking origin at ground and upward as positive direction, draw (a)

x – – t graph,

(b)

v – – t graph graph and

(c)

a – t graph graph

A particle starts with a velocity of 6 m/s and moves with uniform acceleration of 3 m/s2. Find its velocity after 10 seconds.


KINEMATICS

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6.

A body starts from rest and moves with uniform acceleration of 4 m/s2 for one minute. What is the velocity acquired by the body?

7.

An automobile moving with a velocity of 72 km/hr is brought to rest in a distance of 10 m by the application of brakes. Find the retardation assuming it t o be uniform.

8.

A train starting from rest accelerates uniformly for 100 s, runs at a constant speed for 5 minutes and then comes to a stop st op with uniform retardation in the next 150 seconds. During this motion it covers a distance of 4.25 km. Find its constant speed.

9.

A stone is dropped from a height of 100 100 m from a balloon which rises up from the ground. Find the velocity of the balloon at that moment if the stone reaches the ground 5 seconds after it was dropped.

10.

–2 A particle moves in straight line with acceleration 2 t ms ms –2 at time t . If it starts from rest of O on the line, find its velocity and displacement from O at time t = 3 s.


KINEMATICS

ANSWERS NSWERS TO T O PROFICIENCY TEST T EST 2.

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I

(a) True (b) True (c) True (d) False

3.

(a) B (b) C (c) Slower (d) Yes, in the interval 2 to 3 s

5.

The velocity at the end of 10 seconds is 36 m/s.

6.

240 m/s

7.

20 m/s2

8.

36 km/hr

9.

Velocity of the balloon = 5 m/s directed upward

10.

9 m/s, 9 m


KINEMATICS

SECTION

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II


KINEMATICS

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3. MOTION IN TWO DIMENSIONS Motion in a plane is called as motion in two dimensions e.g., projectile motion, circular motion etc. For the analysis of such motion our reference will be made of an origin and two co-ordinate axes x axes x and and y . Position of particle is known by knowing its co-ordinate ( x , y ). ). Velocity of particle will be resultant of velocities in x and and y direction direction v x and v y y . Similarly acceleration will be in the two directions. For analysis of such motion we analyse the motion along two axes independently, i.e., while dealing motion in x-direction we need not to think what is going on in -direction and vice versa. y -direction We have to study about projectile motion, circular motion and relative velocity under the head of motion in two dimensions. We shall discuss circular motion in next lesson.

4.

PROJECT PROJECTILE ILE MOTION MOT ION

A projectile is a particle, which is given an initial velocity, and then moves under the action of its weight alone. If the initial velocity is vertical, the particle moves in a straight line and such motion we had already discussed in ‘motion in one dimension as motion under gravity’. Here we are going to discuss the motion of particle which is projected obliquely near the earth’s surface. While discussing such motion we shall suppose the motion to the be within such a moderate distance from the earth’s surface, that acceleration due to gravity may be considered to remain sensibly constant. We shall also neglect the resistance of air and consider the motion to be in vacuum. 4.1

IMPORTANT TERMS USED IN PROJECTILE MOTION

When a particle is projected into air, the angle that the direction of projection makes with horizontal plane through the point of projection is called the angle of projection; projection; the path, which the particle describes, is called the trajectory ; the distance between the point of projection and the point where the path meets any plane draws through the point of projection is its range; range; the time that elapses in air is called as time of flight and and the maximum distance above the plane during its motion is called as maximum height attained by the projectile. 4.2

ANALYTICAL ANALYTICAL TREATMENT OF PROJECTILE MOTION

Consider a particle projected with a velocity u of an angle  with the horizontal from earth’s surface. If the earth did not attract a particle to itself, the particle would describe a straight line; on account of attraction of earth, however, the particle describes a curved path. This curve will be proved later to be always a parabola.


KINEMATICS

Let us take origin at the point of projection and x-axis x-axis and y -axis -axis along the surface of earth and perpendicular to it respectively as shown in figure.

OP-MI-P-24

y u

P

 O

x

Fig. 6 By the principle of physical independence of forces, the weight of the body only has effect on the motion in vertical direction. It, therefore, has no effect on the velocity of the body in the horizontal direction, and horizontal velocity therefore remains unaltered. Motion in x -direction: -direction: Motion in x in x –direction –direction is motion with uniform velocity. At,

t = 0, x 0 = 0 and u x = ucos

Position after time t , x = x = x 0 + u x t

 x = (u (u cos) t

… (i)

Velocity of any time t , v x = u x



v x = u cos

… (ii)

Motion in y -direction: -direction: Motion in y-direction is motion with uniformly acceleration. when, t = = 0, y 0 = 0, uy = u sin  and ay = – = –g g



After time ‘t ‘t ‘, ‘, v y uy + ay t y =



v y u sin  gt gt y = y = = y 0 + uy t + +

1 2

… (iii) ay t 2

… (iv)

2 2 Also, v y 2ay y y = uy + 2a

– 2gy gy  v y y2 = u2 sin2 – 2

… (v)

Time of Flight (T ): Time of flight is the time during which particle moves from O to O i.e., when t = = T , y = = 0


KINEMATICS

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 From equation (iv) O = usinT –



T =

1 2

gT 2

2u sin  , g

…(13 …(13))


KINEMATICS

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Range of Projectile (R ) : Range is horizontal distance traveled in time T , i.e.,

R = x = x in in time T

 From equation (ii) R = = ucos.T = ucos

2 u sin 

g

u 2 sin 2 R = g

…(14 …(14))

Maximum height reached (H): At the time particle reaches its maximum height velocity of particle becomes parallel to horizontal direction i.e., v y 0 when y = = H y =

 From equation (v) 0 = u2sin2 - 2gH 2gH



u 2 sin 2  H  2g

…(15 …(15))

Equation of trajectory: The path traced by a particle in motion is called trajectory and it can be known by knowing the relation between x between x and and y From equation (i) and (iv) eliminating eliminating time t we we get y = x = x tan tan



1 gx 2 2 u

2

sec2

…(16 …(16))

This is trajectory of path and is equation of parabola. So we can say the path of particle is parabolic. Velocity and direction of motion after a given ti me: After time ‘t ‘t ‘ v x = ucos and v y y = usin – gt Hence resultant velocity v = =

2

v x

 v y 2


KINEMATICS

=

u 2 cos 2

If direction of motion makes an angle tan  =



v y v x



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  (u sin   gt ) 2

 with horizontal.

u sin   gt u cos 

–1  u sin   gt   = tan –1    u cos  


KINEMATICS

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Speed at a given height At a height ‘h ‘h’, v x = ucos

And



v y y =

u 2 sin2   2gh

2

Resultant speed v = =

v x

v=

 v y 2

u2

 2gh

Note that this is the speed that a particle would have at height h if it is projected vertically from ground with u.


Question:

A particle is projected up from the ground with a velocity of 150 m/s at an angle of projection 30° with the horizontal. Find (a) the time of flight (b) the vertical velocity at a height of 70 m.

S olution: olution :

(a)

(b)

The time of flight T 

2u sin g



2  150 150  sin30 10

= 15 s

The vertical velocity at this height

2 2 (u sin )  2  10  70 = (150 sin 30 )  20 70 = 4225 v y 2 = (u

v = 65 m/s y

4.3

SOME IMPORTANT POINTS HORIZONTAL PLANE (i)

REGARDING

PROJECTILE

MOTION

OVER

A

For a given velocity of projection, the range of horizontal plane will be maximum when angle of projection is 45°.


KINEMATICS

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u 2 sin 2 We have range of projectile. R  g Therefore if we keep on increasing  range will increase and then decrease. Its value will be maximum when sin2  is maximum i.e.,  = 45° Also, maximum range range R max max =

(ii)

u2 g

For a given range and given initial speed of projection, there are two possible angle of projection which are complementary angle i.e., if one is  other will be (90° – ).

I llustrati llustratio on 7

Question: S olution: oluti on:

What is the least velocity with which a cricket ball can be thrown through a distance of 1000 m? Since the range is given, the least velocity of projection is that value when the angle of projection is 45 . For velocity u to be least u 2 sin2 g

= 1000 where  = 45

2

or,

u = 1000 g 2

u = 1000  10 = 10000 u = 100 m/s I llustrati llustratio on 8

Question:


Find the maximum horizontal range when the velocity of projection is 30 m/s. Find the two directions of projection to give a range of 45 m when projected with same –1 2 velocity of 30 ms . Take g= 10 m/s . (i) Maximum range R m

(ii) Now

u 2 sin2

or, sin 2 =

g



u 2 30 2 g



10

= 90 m

= 45

45  10 30  30



1 2

or, 2 = 30 or 150 [ (180  ) = sin ] or

 = 15 or 75

Therefore for a given velocity of projection and for a given range, two directions of projection are possible.


KINEMATICS

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In a projectile motion on a horizontal surface if we we consider cons ider air resistance, res istance, what what wil wil be effect on (a) time of flight; (b) maximum height; (c) range ?

4.4

RANGE OF PROJECTILE ON AN INCLINED PLANE THROUGH THE POINT OF PROJECTION

A particle particle is projected projected from a point point A A on an inclined plane, which is inclined at an angle

 to the

horizon with a velocity u at an elevation . The direction of projection lies in the vertical plane through AB through AB,, the line of the greatest slope of the plane. Let the particle strike the plane at B so that AB is AB is the range on the inclined plane. The initial velocity of projection u can be resolved into a component u cos (  ) along the plane and a component u sin (  ) perpendicular to the plane. The acceleration due to gravity g which acts vertically down can be resolved into components g sin  along the plane and g cos  perpendicular to the plane. By the principle of physical independence of forces the motion along the plane may be considered independent of the motion perpendicular perpendicular to the plane. Let T be be the time, which the particle takes to go from A to B. Then in this time the distance traversed by the projectile perpendicular perpendicular to the plane is zero.



0 = u sin (  ) T 



T = =

1 2

u B





C

Fig. 7.

g cosT 2

2usin() g cos

…(17 …(17))

During this time the horizontal velocity of the projectile ( u cos ) remains constant. Hence the horizontal distance described is given by AC = = u cos T = =

 

2u 2 sin()cos

g cos

AC 2u 2 sin()cos  AB = AB = cos g cos 2 

Range on the inclined plane 

2u 2 sin()cos

g cos 2 

...(18 ...(18))

Maximum range on the inclined plane


KINEMATICS

R 

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2u 2 sin (  ) cos g cos 2 

u2

=

g cos2 

[sin (2  )  sin ]

For given values of u and , R is maximum when sin (2  ) = 1 i.e., (2  ) = 90

 = (45 + /2) If R m represents the maximum range on the inclined plane,

R m

R m





u2 g cos 2 

(1  sin )

u2 g (1sin)

For a given velocity of projection, projection, it can be shown that t here are two directions of projection which are equally inclined to the direction of maximum range.

Now

R = =

u2 g cos2 

[sin (2  )  sin ]

For given values of u,  and R , sin (2  ) is constant. There are two values of (2  each less than 180° that can satisfy the above equation. Let (21  ) and (22  ) be the two values. Then 21   = 180  (22  )

1  /2 = 90  (2  /2) 1  (45 + /2) = (45 + /2)  2


 )

KINEMATICS

OP-MI-P-32

Since (45 + /2) is the angle of projection giving the maximum range, it follows that the direction giving maximum range bisects the angle between the two angles of projection that can give a particular range.


KINEMATICS

OP-MI-P-33


Question:

A particle is projected at an angle

with horizontal from the foot of a plane whose 0

inclination to horizontal is . If – = 45 and particle strikes the inclined plane perpendicular to it, then find the value of cot .

S olution:

Let u be the velocity of projection so that u cos (   ) and u sin (   ) are the initial velocities respectively parallel and perpendicular to the inclined plane. The acceleration in these two directions are ( g sin sin ) and (g cos cos  ). The initial component of velocity perpendicular to PQ is u sin (  ) and the acceleration in this direction is ( g cos ). If T is the time the particle takes to go from P to to Q then in time T the the space described in a direction perpendicular to PQ is zero. 0 = u sin (  )  T 

T = =

1 2

g cos

  T 2

u

 P

Q

 N

2u sin()

g cos

If the direction of motion at the instant when the particle hits the plane be perpendicular to the plane, then the velocity at that instant parallel to the plane must be zero.



u cos (    )  g sin sin T = = 0

u cos() g sin



T 

2u sin()

g cos

cot  = 2 tan (  ) = 2

5. RELAT ELAT IVE VELOCITY The terms ‘rest’ and ‘motion’ are only relative. For example, when we say that a train is moving with velocity 30 m.p.h, we mean is that it is the velocity with which the train moves with respect to an observer on the earth who is regarded as fixed. This is not true strictly since a person on the earth unconsciously partakes the rotatory motion of the earth round its axis and the motion of earth round the sun. In addition, he shares the motion of entire solar system through space with respect to certain fixed f ixed stars. Thus there is no absolutely fixed point on the earth about which we can measure motion. Hence a person on the earth can never realise absolute motion or absolute rest. Let us consider two motor cars A and A and B moving in the same direction on a road with equal speed. To a person seated in A, A, if he were unconscious of his motion, the car B would appear to


KINEMATICS

OP-MI-P-34

be at rest. The line joining the two cars will always remain constant in magnitude and direction. The velocity of B relative to A to A or or the velocity of A of A relative relative to B is zero.


KINEMATICS

OP-MI-P-35

On the other hand if A if A is is moving with 30 m.p.h and B with 40 m.p.h in the same direction, a person in A would observe the car B to be drawing away from him at the rate of 10 m.p.h. This represents the velocity of B relative to A. A. If, however, B is moving opposite to the direction of A with velocity 40 m.p.h., for a person in A in A,, B appears to draw away from him at the rate of 70 m.p.h. This, therefore, represents the velocity of B relative to A to A.. Definition of Relative velocity When the distance between two moving points A and A and B is altering, either in magnitude or in direction or both, each point is said to possess a velocity relative to the other. The velocity of one of the moving points, say, A, A, relative to the other point B is obtained by compounding with the velocity of A of A,, the reversed velocity of B. The velocity of A of A relative relative to B is the velocity with which A will appear to move to B, if B is reduced to rest. If velocity A is v A and that of B is v B with respect to a stationary frame, then from the definition, relative velocity of A A with respect to B, v AB is given by

v B

B

A

v A

Fig. 8. v AB = v A



…(19 …(19))

v B

If angle between v A and v B is  then | v AB | v A2

 v B2  2v A v B cos 

…(20 …(20))

Also angle  made by relative velocity with v A is given by tan 



v B sin  v A

…(21 …(21))

 v B cos 

From the above definition of relative velocity it follows that if we impress on both the moving points A points A and B, a velocity equal and opposite to that of B, then B would be reduced to rest and A will have two velocities (i) its own velocity and (ii) the reversed velocity of B. These two can be compounded into a single velocity by the parallelogram law, which will give the velocity of A relative to B. I llustrati llustratio on 10

Question:


Two trains A and B have lengths 100 m and 80 m respectively. They move in opposite directions along parallel tracks at 72 km/hr and 36 km/hr respectively. What is the time taken by one train to cross the other? Velocity of train A train A,, v A = 72 km/hr = 72



5 18

m/s = 20 m/s


KINEMATICS

Velocity of train B, v B =  36 km/hr =  36 

OP-MI-P-36

5 18

=  10 m/s

 ve sign indicates the oppositely directed velocity. Velocity of A relative to B = v A  v B = 20 + 10 = 30 m/s. The train A is A is now supposed to move with velocity 30 m/s while the train B is ‘stationary’. For the train A train A to to cross B or vice versa, the total distance to be crossed = Length of train B + Length of train A train A = 80m +100 m = 180 m Time taken

=

180m 30m/s

= 6 s


Question:


A monkey is climbing a vertical tree with a velocity of15 m/s while a dog runs towards the tree chasing the monkey with a velocity of 20 m/s. Find the direction of velocity of the dog relative to the monkey.

Velocity of the dog relative to the monkey = velocity of

 VM

dog  velocity of monkey = V D

 V M

= V D  ( V M )

VD

O

 VD

 VM

V

M

This velocity makes an angle

D

 with the horizontal, where tan  =

M 15 3



20 4

 3   = tan1  = 37  4  Two Two particles are undergoing projectile motion. What is the nature of path of one particle particle observed observed from from other other particle? particle?


or

KINEMATICS

PROFICIENCY TEST

OP-MI-P-37

II

The following questions deal with the basic concepts of this section. Answer the following followi ng brief br iefly. ly. G o to the nex t s ection ecti on only if your yo ur s cor e is i s at leas leas t 80%. D o not n ot cons c ons ult the Study S tudy Material Material while attem attempting pting the ques ques tions . 1.

If a person wants to hit a target, in what direction should he point his rifle? (Higher, lower or in the same direction as target)?

2.

A person sitting in a train moving with uniform horizontal velocity drops a ball vertically downward. downward. What is the path observed by (i)

a person in the train?

(ii)

a person in a second train moving in the opposite direction on a parallel track?

3.

Two balls are projected from the same point in directions inclined at 30° and 60  to horizontal. If they attain the same height, what is the ratio of squares of their velocities of projection? What is this ratio if they have the same horizontal range?

4.

At what angle should a body be projected with a velocity 8 m/s just to pass over an obstacle 16 m high at a horizontal distance of 32 m? (Assume g = = 10 m/s2,  =

10 )

5.

Two stones are projected from a point P in in the same direction with a velocity of 20 m/s one at angle  above the horizontal and the other at angle  to the vertical. Find the horizontal distance of each stone from P one second after projection and the horizontal separation between the stones. (Take cos  = 4/5; sin  = 3/5)

6.

A 100 m long train running with uniform velocity overtakes a man running in the same direction on the platform at a speed of 5 m/s in 10 seconds. Find the velocity of the train.


KINEMATICS

7.

OP-MI-P-38

N

A ship ‘ A’ A’ steams due north at 16 km/hr and a ship ‘ B’ due west at 12 km/hr. At a certain moment position of B is 10 i and of A of A is 0i . The distances are measured in kilometer. Find the magnitude of velocity of A relative to B. Find also the nearest distance of approach of the ships. ˆ

V

A

ˆ

O

V

B


E

KINEMATICS

ANSWERS NSWERS TO T O PROFICIENCY TEST T EST

3.

3 : 1; 1 : 1

4. 4.

45°

5.

16 m; 12 m; 4 m

6.

15 m/s.

7.

20 km/h, 8 km

OP-MI-P-39

II


KINEMATICS

OP-MI-P-40

SOLVED OBJECIVE EXAMPL EXAMPLES ES Example 1: A particle is travelling with velocity of 2 m/s and moves in a straight line with a retardation of 2 0.1 m/s . The time at which the particle is 15m from the starting point is (a) 10 s

(b) 20 s

(c) 25 s

(d) 40 s

S olution: oluti on: S

1 2

 ut  at 2;15  2t 

1  (0.1)t 2 2

2 2 40t – t or t – 40 – 40t t + + 300 = 0  20  15 = 40t –

(t – 30) – 30) (t (t – 10) – 10) = 0; t = = 30 s or

t = = 10 s

The particle is at a distance 15 m from starting point at t = = 10 s and s and also t = = 30 s. s. (a)

Example 2: 2

2

A particle moves along a straight line according to the law S = at + 2 bt + + c . The acceleration of the particle varies as 3

(a) S

(b) S

2/3

2

(c) S

(d) S

5/2

S olution: oluti on: 2

1/2

S = (at (at + 2bt 2bt + + c ) Differentiating,

d 2 S dt 2

=

(at b ) (at b )   at 2 2bt c   a    at 2 2bt c (at 2 2bt c ) a(at 2 2bt c )(at b ) 2



2

at



at b dS 1 2  (at 2bt c ) 1/ 2 (2at 2b) dt 2 at 2 2bt c

d 2 S 2

dt



2

2bt c (at 2bt c ) 1

S

3



(ac b 2 )

S S 2

 acceleration S –3

(a)

Example 3: A car accelerates from rest at a constant rate for sometime, after which it decelerates at a constant rate to come to rest. If the total time elapsed is t , the maximum velocity acquired by car is


KINEMATICS

(a)

αβ t α  β

(b)

 β) α  ( α αβ

t

(c)

OP-MI-P-41

α 2  β 2 αβ

t

(d)

α 2  β 2 αβ

t

S olution: oluti on: S1



u = V

A u=0

For motion from B to C , 0V t 2

or,

V V

t  t 1 t 2 

V

(



 

)



v = = 0

B v = V

For motion from A from A to to B, V t 1 or



S2



C

V t 1 



V or t 2 



V ()



t

(a)

Example 4: Two particles P and Q start simultaneously from A with velocities 15 m/s and 20 m/s respectively. They move in the same direction with different accelerations. When P overtakes Q at B , velocity of P is 30 m/s. The velocity of Q at B is (a) 30 m/s

(b) 25 m/s

(c) 20 m/s

(d) 15 m/s

S olution: olution : As displacement (in uniformly accelerated motion) = average velocity

 time

The average velocity is the same, when overtaking takes place. 15 + 30 = 20 + v or,

v = = 25 m/s (b)

Example 5: A stone A is dropped from rest from a height h above the ground. A second stone B is simultaneously thrown vertically up with velocity v . The value of v which would enable the stone B to meet the stone A midway between their initial positions is (a) 2 g h

(b) 2 g h

(c) g h

(d)

2 g h

S olution: oluti on: Time of travel of each stone = t


KINEMATICS

Distance travelled by each stone =

For stone A stone A,,

For stone B,



h 1

 gt 2

i .e.,t 

2 2

h 2

 ut 

1 2

gt 2

OP-MI-P-42

h 2

h g

u

h g



1 2

 h    g 

g 

h h h u  2 g 2

or, u

h h g

 uh

g h

 g h

(c)

Example 6: A body is dropped from rest from a height h . It covers a distance 9 h /25 in the last second of fall. The height h is (a) 102.5 m

(b) 112.5 m

(c) 122.5 m

(d) 132.5 m

S olution: oluti on: t is time to reach ground. 1  9  1 h  at 2 ;1 h  a(t 1) 2 2  25  2

4 t 1  1 9  (t 1) 2 ;16 (t 1) 2  2 or     2 t 5  25  t 25 t h =

1 2

t = = 5 sec

 9.8  52 = 122.5 m (c)

Example 7: A particle P is projected vertically upward from a point A . Six seconds later, another particle Q is projected vertically upward from A . Both P and Q reach A simultaneously. The ratio of maximum heights reached by P and Q = 64 : 25. Find the velocity of the projection of Q in m/s (a) 7 g

(b) 6 g

(c) 5 g

(d) 4 g

S olution: oluti on: Study More with www.puucho.com Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396

KINEMATICS

1 2

2

g(t + + 3) : 2

1 2

OP-MI-P-43

2

gt = 64 : 25

2

or (t (t + + 3) : t = 64 : 25; or (t ( t + + 3) : t = = 8 : 5 5t + + 15 = 8t 8t or 3t = 15; t = 5 seconds v = = g  t = = 9.8  5 = 49 m/s = 5g m/s (c)


KINEMATICS

OP-MI-P-44

Example 8: A stone is dropped from rest from the top of a cliff. A second stone is thrown vertically down with a velocity of 30 m/s two seconds later. At what distance from the top of a cliff do they meet? (a) 60 m

(b) 120 m

(c) 80 m

(d) 44 m

S olution: oluti on: The two stones meet at distance S from top of cliff t seconds seconds after first stone is dropped. For 1st stone S = 1

i.e.,

2

1 2

2

gt ; For 2nd stone S = u(t – 2) – 2) +

2

gt = ut – 2 – 2u u +

1 2

1 2

2

g (t – 2) – 2)

2

gt – 2 – 2gt gt + + 2g 2g

0 = (u (u – 2 – 2g g )t – 2( – 2(u u – g ); t ); t 

Distance S at which which they meet =

1 2

2(u g ) 2(3010)

u 2g

 gt 2 =

1 2



3020

= 4 s

 10  16

= 80 m from top of cliff (c)

Example 9: A particle is projected from a point O with velocity u in a direction making an angle upward with the horizontal. At P , it is moving at right angles to its initial direction of projection. Its velocity at P is (a) u tan

(b) u cot

(c) u cosec

(d) u sec

S olution: oluti on: v cos cos (90

P

sin  = u cos ; v = = u cot  ) = v sin

90°

(b)

 (90°

u O

 )

v



Example 10: In the previous example the time of flight from O to P is (a)

u cosec α g

(b)

u sin si n α g

(c)

u tan tan α g

(d)

u sec α g

S olution: oluti on: v = = g t = = (g (g cos cos)t = = u cot




KINEMATICS

t = =

OP-MI-P-45

u cot u 1 u cosec α    g cos g sin g (a)

Example 11: A particle is projected from O and is moving freely under gravity and strikes the horizontal plane through O at a distance R from it. Then which of the following is incorrect? (a) There will be two angles of projection if R g < u

2

(b) There will be more than two angles of projection if R g = u

2

(c) The two possible angles of projection are complementary (d) The products of the times of flight for two directions of projection is 2R / g g

S olution: oluti on: The range R = =

sin 2 =

sin 21

2

t 1t 2 =

g

Rg u

2

2 0 = u , then sin 2 = 1, so  = 45 .  Rg =

= sin 22

or, 2 cos ( 1 + If Rg < < u ,

2u 2 sincos

Rg u2

sin 21  sin 22 = 0

2) sin ( 1  2) = 0

2 = u , 1 = 2 = /2 1 + 2 = /2; if Rg =

2u sin 1 2u sin 2

g





g



4u 2 sin 1 cos1

g 2



2R

g

since

1 + 2 = /2

(b)

Example 12: The velocity of a particle P moving freely under gravity is 4.9 m/s, the direction being 30° with the downward normal (a) its acceleration normal to the direction of motion at P = 9.8 m/s

2

(b) the radius of curvature of P of the parabolic trajectory of particle is 4. 9 m (c) the particle has no acceleration normal to the direction of motion (d) the radius of curvature at P of the path depends upon the initial velocity of projection

S olution: oluti on: The component of acceleration perpendicular to velocity is g cos 60 and this provide the necessary centripetal acceleration.

60° 30°


v

KINEMATICS

g cos cos 60 =

g 2

= 4.9 m/s

2

The radius of curvature is

r = =

v 2 4.9

OP-MI-P-46

v 2 r

= 4.9;

= 4.9 m

(b)

Example 13: A boat which has a speed of 5 km/hr in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in kilometres per hour is (a) 1

(b) 3

(c) 4

(d)

41

S olution: oluti on: For shortest possible path

v 

1km 15 min

2 v w  v 2

v w

vw

 4 km/h

 v

vb

 v b2

–1  v b2  v 2  5 2  4 2  3kmh –1

(b)

Example 14: A man running at 6 km/hr on a horizontal road in vertically falling rain observes that the rain hits him at 30° from the vertical. The actual velocity of rain has magnitude (b) 6 3 km/hr

(a) 6 km/hr

(c) 2 3 km/hr

(d) 2 km/hr


O

Velocity of rain = Velocity of man + Relative velocity of rain OR gives the actual velocity.

VR OR

tan 30



or,

6 3

1



6

6 km

30° Relative velocity

3 OR V

R

(b)


M

KINEMATICS

OP-MI-P-47

Example 15: A boat which has a speed of 5 km/hr in still water crosses a river of width 3 km along the shortest possible path in t min. min. The river flows at the rate of 3 km/hr. The time taken t is is (a) 20 min

(b) 25 min

(c) 45 min

(d) 55 min

S olution: oluti on: t 



AB 5 2 3 2



3 = 45 minutes 4

(c)


KINEMATICS

OP-MI-P-48

SOLVED SUBJECT SUBJECTIVE IVE EXAMPLES EXAMPLES

Example 1: 2

A particle moves along a straight path A B C with a uniform acceleration of 0.5 m/s . While it crosses A its velocity is found to be 5 m/s. It reaches C with a velocity 40 m/s, 30 seconds after it has crossed B in its path. Find the distance A B .

S olution: oluti on: The velocity while it crosses the point A is A is 5 m/s

u = 5 m/s A t = = 0

= 40 m/s v =

B t = = t 1

S1

C t = t1 + 30

Considering the displacement AC displacement AC , initial velocity u = 5 m/s final velocity v = 40 m/s acceleration a

= 0.5 m/s

 time of motion t =

2

v u 40 5



a

0.5

= 70 s

For the displacement AB displacement AB,, initial velocity u = 5 m/s acceleration a

= 0.5 m/s

2

time of motion t = 70  30 = 40 s



AB = AB = S = ut + +

1 2

2

 1   0.5  40 2  = 200 + 400 = 600 m  2 

at = (5  40) + 

Example 2: A particle moving with uniform acceleration in a straight line covers a distance of 3 m in the 8th second and 5 m in the 16th second of its motion. What is the displacement (in cm) of the particle from the beginning of the 6th second to the end of 15th second?

S olution: oluti on: The distance traveled during the nth second of motion of a body is given by

Sn

u 

1 2

a (2n 1)


KINEMATICS

OP-MI-P-49

For the motion during the 8th second, 3 = u +

1 2

a (161)

 u

15a

... (i)

2

For the motion during the 16th second, 5 = u +

1 2

a ( 32  1)  u

31a



... (ii)

2

Subtracting equations (i) from (ii) 8a = 2 or acceleration a acceleration a =

1 4

ms

From equation (1), u = 3

2

 15 1  9     ms2  2 4  8

Now, the velocity at the end of 5 s (velocity at the beginning of 6th second) v 1 = u + 5a 5a The velocity at the end of 15th s, v 2 = u + 15a 15a Average velocity during this interval of 10 seconds =



(u 5a )(u 15a ) 2

v 1 v 2 2

= u + 10a 10a

Distance travelled during this interval S = average velocity x velocity x time time = (u (u + 10a) =

 t

290 290  9 10  = 36.25 m = 3625 cm     10  8  8 4 

Example 3: An automobile can accelerate or decelerate at a maximum value of

5 3

2

m/s and can attain a

maximum speed of 90 km/hr. If it starts from rest, what is the shortest time in which it can travel one kilometre, if it is to come to rest at the end of the kilometre run?

S olution: oluti on: In order that the time of motion be shortest, the automobile should attain the maximum velocity with the maximum acceleration after the start, maintain the maximum velocity for as along as possible and then decelerate with the maximum retardation possible, consistent with the condition that, the automobile should come to rest immediately after covering a distance of 1 km.


KINEMATICS

OP-MI-P-50

Let t 1 be the time of acceleration, t 2 be the time of uniform velocity and t 3 be the time of retardation. Now, maximum velocity possible = 90 km/hr = 90

t 1





5 18

m/s = 25 m/s

v u 250 = 15 s  5 a 3

Similarly, the time of retardation is also given by

t 3



0  25 = 15 s 5



3

During the period of acceleration, the distance covered = average velocity x velocity x time time =

25  0 2

 15 = 187.5 m

During the period of retardation, the distance covered is the same and hence = 187.5 m

 the total distance covered under constant velocity = 1000  375 = 625 m Time of motion under constant velocity, t 2 

625 25

 25s

= t 1 + t 2 + t 3 = 15 + 25 + 15 = 55 seconds  the shortest time of motion = t

Example 4: A stone is dropped into a well and the sound of the splash is heard 3

1 8

seconds later. If the 2

velocity of sound in air is 352.8 m/s, find the depth of the well (in cm). g = 9.8 m/s .

S olution: oluti on: Let x Let x metres metres be the depth of the well and t the the time taken by the stone to reach the surface of water. In this case u = 0, a = 9.8 m/s Now in the relation, S = ut + +

2

1 2

at 2 ,

we have x = = 0 +

1 2

2

(9.8) t


KINEMATICS

or

2

x = = 4.9t 4.9t

OP-MI-P-51

... (i)

Time taken by sound to travel the distance x up (Motion of sound wave is not affected by gravity) is

 1   3  t  seconds  8  Distance traveled = Velocity of sound

 Time taken

 25   t   8 

x = 352.8

... (ii)

From equations (i) and (ii), 4.9t 2

 25  352 t  352 .8  8  2

or,

4.9t 4.9t + 352.8t 352.8 t – 1102.5 – 1102.5 = 0

or,

t + 72t 72t – 225 – 225 = 0

2

Solving, t = 3 or – or – 75 75 s. Since the negative value of t has no meaning, t = = 3 s. 2

This gives x gives x = = 4.9t 4.9t = 4.9 x 9 = 44.1 m Hence the depth of the well = 44.1 m = 4410 cm


KINEMATICS

OP-MI-P-52

Example 5: A circus artist maintains four balls in motion making each in turn rise to a height of 5 m from his hand. With what velocity (in m/s) does he project them and the height (in cm) of the other 2 three balls at the instant when the fourth one is just leaving his hand? (take g = 10 m/s .)

S olution: oluti on: Obviously, to maintain proper distances, the artists must throw the balls after equal intervals of time. Let the interval of time be t , so that when the fourth ball is just leaving his hand, the first ball would have travelled for time 3t 3 t , the second for time 2t 2 t and and the third for time t . The second obviously would just have reached the maximum height of 5 m. If v be be the initial velocity of throw of each ball, then for the second ball we have, v 2 = 0 = v – – g (2t (2t ) and

(2t ) – s2 = 5 = v (2t

... (i) 1

2

g ( 2t ) 2

... (ii)

These gives, v = = 2gt 2gt 2

and

v  2t 2t = = 2t 2t + 5

or

v = = 20t 20t

and

v  2t 2t = = 20t 20t + 5

2

2

Solving for t , we get 20t 20t = 5 or t = =

Therefore, v = 20 

1 2

1 2

second.

= 10 m/s. Thus each ball is thrown up with initial velocity velocity of 10 m/s. m/s.

For the first ball, which would have come down for time (3t – 2t) – 2t) = t , we have S = 0 +

1 2

g t 2 2

5  1  =  10    = 125 cm 2 4  2  1

Therefore, it will be at a height of (5 – (5 – 1.25) 1.25) = 375 cm from cm from the hand and going downwards. For the third ball, which will have risen up for time t , 2

 1  1  1  S3  vt  gt  10     10     5  1.25  375 cm 2  2  2  2  1

2

Example 6: Study More with www.puucho.com Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016; Ph.: 2653 7392/93/94/95; Fax: 2653 7396

KINEMATICS

OP-MI-P-53

A stone is projected from the point on the ground in such a direction so as to hit a bird on the top of a telegraph post of height h and then attain the maximum height 2 h above the ground. If at the instant of projection the bird were to fly away horizontally with uniform speed





v = 2 2  1 m/s. Find the horizontal velocity of the stone if the stone still hits the bird while descending.


KINEMATICS

OP-MI-P-54

S olution: oluti on: The situation is shown in Figure. L et  be the angle of projection and u the velocity of projection. Maximum height MN = = 2h 2h

MN = = 2h 2h =

u 2 sin 2 2g



 u sin  = 2 gh

Y

... (i)

u

Let t be be the time taken by stone to attain the vertical height h above the ground.



h = (u (u sin )t 

t 2

1 2



gt 2

O

M

A h

B h N

 2u sin  2h  t  0  g  g 

usin u 2 sin2  t   g g 2



2h g

Substituting the value of u sin  from (i),

t 

2 gh

g

t 1 



4gh 2h



2h

4h

g

g 2



=

g

g

t 2



4h

g



2h

g

4h 2h  g g

where t 1 and t 2 are time to reach A reach A and and B respectively shown in the figure. If v is is the horizontal velocity of bird, then AB = AB = vt 2. AB is AB is also equal to u cos  (t (t 2  t 1), where u cos  is constant horizontal velocity of stone.

t 2 t 1 2



u cos

  2

2h g 2h = vt 2 g


X

KINEMATICS

v ucos

2



2h

g t 2

2h

2



2h g

g

(

v u cos 



OP-MI-P-55

2 1)



2 2 1

u cos   1 m/s


KINEMATICS

OP-MI-P-56

Example 7: Two particles are projected at the same instant from two points A and B on the same horizontal level where A B = 56 m, the motion taking place in a vertical plane through A B . The 5 particle from A has an initial velocity of 39 m/s at an angle sin 1 with A B and the 13 particle from B has an initial velocity of 25 m/s at an angle sin

1

3 with BA. Show that the 5

particles would collide in mid-air and find when and where the impact occurs.

S olution: oluti on: AB = AB = 56 m. At A At A,, a particle is projected with velocity u = 39 m/s. u1 and u2 are its horizontal and vertical components respectively. The angle u makes with AB is AB is 1. Given that sin

5

12

13

13

 1  cos 1 

u u2 A

.

v

1 u1

56 m

2

v

v1

B

Similarly, for the particle projected from B, with velocity v = = 25 m/s, v 1 and v 2 are the horizontal and vertical components respectively. sin

2 =

3 5

 cos 2 =

5

1 = 39 

Now u2 = u sin

v 2 = v sin sin

4

2 = 25 

3 5

. 5 13

= 15 m/s.

= 15 m/s.

The vertical components of the velocities are the same at the start. Subsequently at any other instant t their their vertical displacement are equal and have a value h = 15t 15t  5t 5t

2

which means that the line joining their positions at the instant t continues to be horizontal and the particles come closer to each other. Their relative velocity in the horizontal direction = 39 cos = 39

12



13

1 + 25 cos 2  25 

4 5

 36  20 = 56 m/s


KINEMATICS

Time of collision

OP-MI-P-57

AB 56 s, after they were projected.  = 1 s, 56 56



Height at which the collision occurs = ut 

1 2 1 at 15(1) (10)(1) 2 2 2

= 10 m The horizontal distance of the position of collision from A = 39 

12 13

 1 s = 36 m


KINEMATICS

OP-MI-P-58

Example 8: A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill is = 30 and the angle of barrel to the horizontal is = 60 . The initial velocity of shell is 21 m/s. Find the distance from the gun to the point at which the shell falls.

S olution: oluti on: We can write the equation of motion as x = ut cos

y = ut sin





gt 2 2

u OA = OA = 

A At the moment the shell falls to the ground

=60º =30º

x =  cos  =  cos 30 y =  sin  =  sin 30

 cos

 sin

 

= ut cos cos 

... (i)

gt 2 = ut sin sin   2

... (ii)

t =



sin



cos

u cos cossin

cos



g  2 cos 2

 2u 2 cos 2 





2u 2 sin()cos

g cos 2



Substituting u = 21 m/s,  = 30,  = 60 and g = = 9.8 m/s , we get 2

= 30 metres

Example 9: A man can swim at a velocity V 1 relative to water in a river flowing with speed V 2. Show that it will take him

V 1 V 1 2 V 2 2

times as long to swim a certain distance upstream and back as to

swim the same distance and back perpendicular to the direction of the stream ( V 1 > V 2). (given

24V 1  25V 2 )


KINEMATICS

OP-MI-P-59

S olution: oluti on: Suppose the man swims a distance x distance x up up and the same distance down the stream. Velocity of man upstream relative to the ground = V 1  V 2.

x Time taken for this, t 1  V 1 V 2 Velocity of man downstream relative to the ground = V 1 + V 2

x Time taken for this, t 2  V 1 V 2 Total time taken t 1 t 2



x



x



2V 1 x

V 1 V 2 V 1 V 2 V 12 V 22

Next the man intends crossing the river perpendicular to the direction of the stream. If he wants to cross the river straight across he must swim in a direction OM such such that the vector sum of velocity of man + velocity of river will give him a velocity relative to the ground in a direction perpendicular to

 the direction of the stream. In the Figure the velocity relative to the ground is O R and the magnitude



of OR  V 12 V 22 Now the man swims a distance x up and x down perpendicular to the river flow. Time taken for this,

t 

R

2 x

V 12 V 22

V1

Then the ratio,



V2

M

2V 1 x

 2

V 12 V 2

t 1 t 2

2 x

2V 1 x

 2

V 12 V 2

t

V 12 V 22





V 1



V1 + V 2

2 x

V 12 V 22

O

= 5 = 5

V 12 V 22

Example 10: A man walking eastward at 6 km/hr finds that the wind seems to blow directly from north. On doubling his velocity, the wind appears to come N 30° 30° E . Find the speed of the wind.

S olution: oluti on: Actual velocity of the man = 6 km/hr eastward. The direction of the relative velocity of the wind in this case is North to South.


KINEMATICS



OP-MI-P-60



If OA represents the velocity of man and AB represents the relative velocity of the wind, then

 velocity of man + relative relative velocity of wind = velocity velocity of wind = OB (say) It is also given that when the velocity of the man is doubled (i.e., 12 km/hr) the wind seems to blow

 from a direction N 30°E. 30°E. Representing this by vector OC = New velocity of man = 12 km/hr.

The direction of the relative velocity of the wind in this case is CB. CB. The two directions of the relative velocity

N W

O

12 km/h 6 km/h A 60º

N 60º

C

30º

 meet at B. Hence OB should give the real velocity of the wind. From the geometry of the Figure, it is clear that OBC is is an equilateral triangle. Hence the magnitude of the real velocity of the wind = 12 km/hr .

S

S Direction N 30º E B


E

KINEMATICS

OP-MI-P-61

MIND MAP

1.

Relation between kinematic

motion in one variables for motion dimension

 v  a



2.

Equations of motion in one dimension

 

Motion with uniform velocity S= vt Motion with uniform acceleration, 1 S  ut  at 2 2 v = u + at

dx dt

dv d 2 x vdv   dt dt 2 dx

v 2

 u 2  2as

Sn

 u  2n  1

a 2

3. Graphical representation of motion  Slope of tangent to position time graph gives velocity.

   

Slope of tangent to v  curve gives t t curve acceleration. Area enclosed between v   t t curve and time axis between an interval of time gives displacement. Slope of tangent to a t t curve curve gives rate of change of acceleration Area enclosed between a t t curve and time axis between an interval of time gives change in velocity.

KINEMATICS

4.

Projectile on horizontal plane



Time of flight, T =



Range R =

2u sin 

5.

g



u 2 sin 2

 

6.

Relative velocity



V AB

2u sin   

g cos 

 V A V B

2

Range, R =

2u sin    cos 

V B

2 cos2  



V A

Maximum Height, H =

u 2 sin2  2g

Equation of trajectory ,

y = x tan

 

Time of flight, T =

g

 u

Projectile on inclined plane





2

1 gx

sec

2



2 u2 For maximum range,  = 45 For a given speed and given range, there are two possible angles of projection;  and (90  ).

u







2 2 | V AB | V AV B cos  A  V B  2V



If relative velocity makes an angle  with V A then,

tan 



V B sin  V A

 V B cos 


KINEMATICS

OP-MI-P-62


KINEMATICS

OP-MI-P-63

EXERCISE – I AIEEE-SINGLE CHOICE CORRECT

1.

On a displacement-time displacement-time graph two straight lines make angles 30  and 60 with the time-axis. The ratio of the velocities represented by them is (a) 1 :

2.

3

(b) 1 : 3

(c)

3 :1

The velocity-time plot is shown in figure. Find the average speed in time interval t = = 0 to t = 40s during the period

(d) 3 : 1

v 5 m/s 0

(a) zero

10

(b) 2.5 m/s

40

t (s)

5 m/s (c) 5 m/s

3.

A block is released from rest on a smooth inclined inclined plane. If Sn denotes the distance Sn traveled by it from t = (n (n –1) –1) s to t = = n s, then the ratio is S n 1 (a)

4.

(d) none of these

2n 2n  1

(b)

2n  1 2n  1

(c)

2n  1 2n  1

Figure shows the graph of acceleration of particle as a function of time. The maximum speed of the particle is (particle starts from rest) (a) 7 m/s

(b) 8 m/s

(c) 4 m/s

(d) 16 m/s

(d)

) s / m (

2

2n 2n  1

y

n o i 4 t a r e l e c c A

0 2

–

4 5 2

x

time(s)


KINEMATICS

5.

If the displacement (s) and time (t) graph of two moving particles A particles A and and B in straight line is shown in the figure, then which of following is incorrect.

OP-MI-P-64

S A B

(a) A (a) A is is moving with constant velocity (b) B is moving with increasing speed (c) A (c) A is is moving with non-zero constant acceleration acceleration

t

(d) acceleration of B may be constant


KINEMATICS

6.

A body covers one-third of the distance with a speed v 1, the second one-third of the distance with a speed v 2 and the remaining distance with a speed v 3. The average speed is (a)

(c)

7.

v 1  v 2 v 1v 2

 v 2v 3  v 3v 1

v 1v 2

 v 2v 3  v 3v 1

v 1v 2v 3

(d)

3

A particle is falling freely under gravity. In first t second it covers s1 and in the next t seconds it covers s2, then t is is given by

s2

 s1

2g

s 2  s1

(b)

g

s 2 s1

(c)

g

s2

(d)

2

 s12 g

Two cars A A and B are traveling in the same direction with velocities v A and v B (v A > v B) when the car A A is at a distance s behind the car B, the driver of the car A A applies brakes producing a uniform retardation a and there will be no collision when

(c) s 

(v A

 v B ) 2 2a

(v A

 v B ) 2 2a

(b) s 

(d) s 

(v A

 v B ) 2 2a

(v A

only

 v B )2 2a

A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d /2. /2. Neglecting subsequent motion and air resistance, its velocity v varies varies with the height h above the ground as v

v

v

d

(a)

10. 10.

3v 1v 2v 3

(b)

3

(a) s 

9.

 v 3

3

(a)

8.

OP-MI-P-65

v d

h

(b)

d

h

(c)

h

d

(d)

The x The x – t graph graph shown in figure f igure represents (a) Constant velocity (b) Velocity of the body is continuously changing

t n e m e c a l p s i D

h

t 1 Time(t )

(c) Instantaneous velocity (d) The body travels with constant speed upto time t 1 and then stops


KINEMATICS

11.

12.

A velocity of 5 km/hr to the East is changed into 5 km/hr to the North. The change in velocity is (a) zero

(b) 5 2 km/hr due North-East

(c) 5 2 km/hr due North-West

(d) 10 km/hr due East

A particle particle is projected such that the the horizontal horizontal range and vertical height are the same. Then the angle of projection with horizontal is –1 (a) tan –1 (4)

13.

 1    4 

–1 (b) tan –1 

(c)

/4

(d) /3

–1 A cricket ball is hit with a velocity of 25 m s –1 at angle of 60° above the horizontal. How far above the ground, ball passes over a fielder 50 m from the bat (consider the ball is struck –2 very close to the ground) Take 3 = 1.7 and g = = 10 m s –2 .

(a) 6.8 m

14.

OP-MI-P-66

(b) 7 m

(c) 5 m

(d) 10 m

–1 –2 A body has an initial velocity of 3 ms –1 and has an acceleration of 1 ms –2 normal to the direction of the initial velocity. Then its velocity, 4 second after the start is –1 (a) 7 ms –1 along the direction of initial velocity –1 (b) 7 ms –1 along the normal to the direction of the initial velocity –1 (c) 7 ms –1 mid-way between the two directions –1 (d) 5 ms –1 at an angle of tan1

15.

3

with the direction of the initial velocity

A disc disc of of radius radius 1 m is rotating about its centre with angular velocity of 20 rad/s. A constant retardation of 2  rad/s2 starts to act on disc. The angular displacement (in radian) of the disc before it comes to stop is (a) 100

16.

4

(b) 100

(c) 10

A large rectangular rectangular box falls vertically with an acceleration a. A toy gun fixed at A A and aimed towards C fires fires a particle P . Which of the following statement is false?

(d) 1002

u

(a) P will will hit C if if a = g (b) P will will hit the roof BC if if a > g

C

B

P A

D

(c) P will will hit the wall CD if CD if a < g (d) may be either (a), (b) or (c), depending on the projection speed of P .


KINEMATICS

17.

OP-MI-P-67

Two particles are projected simultaneously simultaneously in the same vertical plane from the same point, with different speeds u1 and u2, making angles 1 and 2 respectively with the horizontal, such that u1 cos 1  u 2 cos  2 . The path followed by one, as seen by the other (as long as both are in flight), is (a) a horizontal straight line (b) a vertical straight line (c) a parabola (d) a straight line making an angle | 1 – 2| with the horizontal

18.

A particle starts from the origin of coordinates at time t = = 0 and moves in the xy the xy plane plane with 2 a constant acceleration  in the y -direction. -direction. Its equation of motion is y   x . Its velocity component in the x the x -direction -direction is (a) variable

19.

(b)

2



(c)

 2

(d)

 2

A man who can swim at a speed v relative relative to the water wants to cross a river of width d , flowing with a speed u. The point opposite to him across the river is P . Which of the following statement is false? (a) The minimum time in which he can cross the river is d /v (b) he can reach the point P in in time dv . (c) he can reach the point P in in time

d v 2

 u2

(d) he cannot reach P if if u > v .

20.

A particle is projected with velocity velocity u at angle  with the horizontal. The time after which the acceleration vector and velocity vector becomes perpendicular to each other during the path is (a)

21.

u sin  g

(b)

u cos  g

(c)

2u sin  g

(d)

2u cos  g

An aircraft flying horizontally horizontally at 360 km/hr releases a bomb bomb (with zero velocity relative to the aircraft) at a stationary tank 200 m away from a point just below the aircraft. W hat must be the height of the aircraft above the tank if t he bomb is to hit the tank? t ank? (a) 100 m

(b) 9.8 m

(c) 19.6 m

(d) 98 m


KINEMATICS

22.

OP-MI-P-68

ˆ A projectile is given an initial velocity of ( i ˆ  2 j )m/s. The cartesian equation of its path is ˆ (g = = 10 m/s2) (here i ˆ & j are the unit vectors along horizontal and vertical direction) direction)

(a) y = = 2 x – 5 – 5 x 2

23.

(d) y = = 2 x – 25 – 25 x 2

(b) 2.73 m/s

(c) 2.93 m/s

(d) 2.63 m/s

A shot is fired from a gun with a muzzle velocity velocity of 98 m/s at an angle of elevation 45 from the ground. Its range is found to be 900 m. The range decreased by air resistance is (a) 8 m

25.

(c) 4y = = 2 x – 5 – 5 x 2

A particle moves one quarter of a circular path of radius 20 m in 10 s. The magnitude of average velocity of the particle is (a) 2.83 m/s

24.

(b) y = x = x – 5 – 5 x 2

(b) 80 m

(c) 98 m

(d) 9.8 m

The velocity of projection of a particle if it does not rise more than 3 m in a range of 600 m is (a) 383.4 m/s

(b) 273 m/s

(c) 343 m/s

(d) 3.83 m/s


KINEMATICS

OP-MI-P-69

EXERCISE – II IIT-JEE- SINGLE CHOICE CORRECT

1.

A body A body A of of mass 4 kg is dropped from a height of 100 m. Another body B of mass 2 kg is dropped from a height of 50 m at the same time. (a) Both the bodies r each the ground simultaneously. simultaneously. (b) A (b) A takes takes nearly 0.7th of time required by B. (c) B takes nearly 0.7th of time required by A by A.. (d) A (d) A takes takes double the time required by B.

2.

3.

A man on the observation platform of a train moving with constant velocity drops a ball while leaning over the railing. The path of the ball as seen by an observer standing on the ground nearby is a (a) straight line vertically down

(b) horizontal straight line

(c) parabola

(d) circle

A particle starting from rest undergoes a rectilinear motion with with acceleration a. The variation of a with time t is shown in the figure. The maximum velocity attained by the particle during the motion is

(m/s ) a(m/s 10

11

4.

(a) 55 m/s

(b) 550 m/s

(c) 110 m/s

(d) 650 m/s

t (s) (s)



ˆ Velocity and acceleration of a particle at some instant of time are v  3i ˆ  4 j m/s and



a

  6i ˆ  8 j ˆ m/s2 respectively. At the same instant particle is at origin. Maximum

x-co-ordinate of particle will be (a) 1.5 m

5.

(b) 0.75 m

(c) 2.25 m

(d) 4.0 m

A man can swim in still water at 6 km/hr. He wants to reach a point just opposite to his starting point. In which direction he should swim, if the river is flowing at 3 km/hr? (a) 1200 with the river flow

(b) 1500 with the river flow


KINEMATICS

(c) 900 with the river flow

6.

(d) none

1 If a particle is projected from origin and its follows the trajectory y  x  x 2 , then the time 2 of flight is (g (g = = acceleration acceleration due to gravity) 1

(a)

7.

g

2

(b)

g

(c)

s

straight line

(b)

s

circle

t

9.

g

(d)

(c)

(a) 10 m

(b) 30 m

(c) 15 m

(d) 25 m

g

P x

s (d)

sinusoidal

s parabolic

t

t

4

y

A particle A particle A is is projected form the ground with an initial velocity of 10 m/s at an angle of 60 0 with horizontal. From what height h should an another particle B be projected horizontally with velocity 5 m/s so that both the particles collides at point C if both are projected simultaneously simultaneously (g = = 10 m/s2)

t

B

5 m/s

h 10 m/s

A

600

C

A bullet is fired into a fixed target looses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion? (a) 0.5 cm

10.

3

A particle moves on a circular path with uniform angular velocity. The displacement (s) of the particle from point P can can be shown on the graph as

(a)

8.

OP-MI-P-70

(b) 1 cm

(c) 1.5 cm

(d) none

A particle goes from A from A to B in 10 seconds with uniform acceleration. Its velocities at A and A and 2 B are 5 m/s and 25 m/s. Its acceleration (in m/s ) is (a) 0.5

(b) 2.5

(c) 2

(d) 3


KINEMATICS

11.

OP-MI-P-71

A trolley runs down a slope from rest with constant acceleration. acceleration. In the first second second of its its 2 motion it travels 1.6 m. Its acceleration acceleration (in m/s ) is (a) 3.2

(b) 1.6

(c) 0.8

(d) 2.4


KINEMATICS

12.

OP-MI-P-72

From the velocity-time graph given of a particle moving in a straight line, one one can conclude that

(m/s) v (m/s)

A

4 m/s

0

B

3

C 12

8

Time (in s )

(a) its average velocity during the 12 seconds interval is 24/7 m/s (b) its velocity for the first 3 seconds is uniform and is equal to 4 m/s (c) the body has a constant acceleration between t = = 3 s and t = = 8 s (d) the body has a uniform velocity from t = = 8 s to t = = 12 s

13.

Two trains each of length 90 m moving in opposite directions along parallel tracks meet when their speeds are 60 km/hr and 40 km/hr. If their accelerations are 0.3 m/s 2 and 0.15 m/s2 respectively, find the time they take to pass each other. (a) 8 s

14.

(b) 4 s

(d) 6.17 s

A particle moves with constant acceleration for 6 seconds after starting from rest. The distances travelled during the consecutive 2 seconds interval are in the ratio (a) 1 : 1 : 1

15.

(c) 2 s

(b) 1 : 2 : 3

(c) 1 : 3 : 5

(d) 1 : 5 : 9

The velocity-time graph of a body is given below.

m/s) v ( m/s)

C

60 40

A

20 0

10

20

B

30

40

50

60

70

time (s)

The maximum acceleration in m/s 2 is


KINEMATICS

(a) 4

16.

(b) 3

(c) 2

(d) 1

A particle has displacement displacement of 40 m along OX in 5 s and 30 m perpendicular to OX (parallel to OY ) in the next 5 s. The magnitude of the average velocity in m/s of the particle during the 10 s interval is (a) 6

17.

OP-MI-P-73

(b) 7

(c) 8

(d) 5

The velocity-time graph of a body moving along a straight line is as follows: 2 /s 1 0

1

2

3

4

5

6

7

time in s

1 2 The displacement of the body in 5 s is (a) 5 m

18.

19.

(c) 4 m

(d) 3 m

A particle is projected up an inclined as shown in figure. For maximum range over the inclined plane, the value of  should be (a) 45°

(b) 15°

(c) 30°

(d) 60°

u



30°

–1 A ball rolls off the top of a staircase with a horizontal velocity u ms –1 . If the steps are h m high and w m m wide the ball will hit the edge of the nth step if

(a) n = 20.

(b) 2 m

gw 2 2hu 2

(b) n =

2hu 2

gw 2

(c) n 

2u 2

gw 2 h

(d) n



2hw 2 u 2

g

A large number of bullets are fired in all directions with the same sped v . The maximum area on the ground on which these bullets will spread is (a)

v 2 g

(b)

v 4 g 2

(c)

 2 v 4 g 2

(d)

 2 v 2 g 2


KINEMATICS

21.

OP-MI-P-74

A person is standing in a stationary lift drops a coin from a certain height h. It takes time t to reach the floor of the lift. If the lift is rising up with a uniform acceleration a, the time taken by the coin (dropped from the same height h) to reach the floor will be

(a) t

(b) t

a g

 

(c) t 1 

a   g 



1 2

  

(d) t  1 

a  g 

1 2


KINEMATICS

22.

OP-MI-P-75

When a man moves down the inclined plane with a constant speed 5 m/s which makes an angle of 37 0 with the horizontal, he finds that the rain is falling vertically downward. When he moves up the same inclined plane with the same speed, he finds that the rain makes an

 7   with the horizontal. The speed of the rain is  8 

angle  = tan1 (a)

23.

(b)

32 m/s

(c) 5 m/s

(d)

73 m/s

A particle is moving in straight line. The velocity v of the particle varies with time t as v  t 2  4t , then the distance traveled by the particle in t = = 0 to t = = 6s (where t in second and v is is in m/s). (a)

24.

116 116 m/s

64 3

m

(b) zero

(c)

32 3

m

(d) none

The graph which represents the variation of slope m of the trajectory of a projectile with horizontal displacement s is

(a)

(b)

s

s

m (c)

(d)

s

s

25.

A radar observes an aeroplane moving horizontally with a velocity v at height h from ground. Ratio of angular acceleration to angular velocity of aeroplane, w.r.t. radar (in terms of v , h and ) is

(a)

v h

sin2 

(b)

v cos 2 h

(c)

v cos2  h

v h



(d)

Radar

v sin 2 h


KINEMATICS

OP-MI-P-76

EXERCISE – III ONE OR MORE THAN ONE CHOICE CORRECT

1.

The displacement-time displacement-time graph of a moving particle is a straight line. Therefore (a) its acceleration is constant (b) its velocity is constant (c) its displacement is constant (d) both its velocity and acceleration are uniform

2.

A train moving with a constant speed along a straight track takes a bend in a curve with the same speed. Due to this (a) its velocity is changed in magnitude (b) its velocity is not changed (c) its speed is not changed (d) its velocity is changed

3.

A particle moving in a straight line has velocity ( v ) and displacement (s ( s) related as is in m/s and displacement s is in metre. Then v  4 1  s , where velocity v is (a) acceleration of the particle is 8 m/s 2 (b) velocity of the particle at t = = 2s is 20 m/s (c) displacement of the particle at t = = 2s is 24 m (d) displacement of the particle at t = 1s is 8 m

4.

A ball is thrown upward from the ground with velocity u. It is at a height 100 m at two times t 1 and t 2 respectively. If g = = 10 m/s2, then (a) t 1 t 2= 20

5.

(b) t 1 + t 2= 20

(c) t 1t 2 

u 5

(d) t 1  t 2 

u 5

The maximum horizontal range and maximum height attained by a projectile are R and and H respectively. respectively. If a constant horizontal acceleration a = g /4 /4 is imparted to the projectile due to wind, then


KINEMATICS

(a) its horizontal range is R (c) its maximum height is 6.

OP-MI-P-77

(b) its horizontal range is R + + H

H

(d) its maximum height is H

2

Velocity-displacement Velocity-displacement graph of a particle moving in a straight line is as shown in the figure.

v

(a) magnitude of acceleration of particle is decreasing (b) magnitude of acceleration of particle is increasing (c) acceleration versus displacement graph is straight line

s

(d) acceleration versus displacement graph is parabola 7.

Starting from rest a particle is first accelerated for time t 1 with constant acceleration a1 and then stops in time t 2 with constant retardation a2. Let v 1 be the average velocity in this case and s1 the total displacement. In the second case it is accelerated for the same time t 1 with constant acceleration 2a 2 a1 and comes to rest with constant retardation a2 in time t 3. If v 2 is the average velocity in this case and s2 the total displacement. displacement. Then (a) v 2  2v 1

8.

(b) 2v 1  v 2  4v 1

(c) s 2  2s1

(d) 2s 1  s 2  4s 1 

Acceleration of a particle which is at rest at x = 0 is a

 4  2 x i ˆ . Select the correct

alternative(s)

9.

10.

(a) particle further comes to rest at x at x = 4

(b) particle oscillates about x about x = = 2

(c) maximum speed of particle is 4 units

(d) all of the above

A car is moving rectilinearly rectilinearly on a horizontal path with acceleration a0. A person sitting inside the car observes that an insect S is crawling up the screen with an acceleration a. If inclination of the screen with the horizontal, the acceleration of the insect.  is the inclination (a) parallel to screen is a +a +a0 cos 

(b) along the horizontal is a 0 – a cos 

(c) perpendicular to screen is a0 sin 

(d) perpendicular perpendicular to screen is a0 tan 

A particle is projected from a point P with with a velocity v at at an angle  with horizontal. At a certain point Q, it moves at right angles to its initial direction. Then (a) velocity of particle at Q is v sin sin  (b) velocity of particle at Q is v cot cot  (c) time of flight from P to to Q is (v (v /g) /g) cossec 


KINEMATICS

OP-MI-P-78

(d) time of flight from P to to Q is (v (v /g) /g) sec 

11.

A particle moving along a straight line with uniform acceleration has velocities 7 m/s at at P and 17 m/s at Q. R is is the mid point of PQ. PQ. Then (a) the average velocity between R and and Q is 15 m/s (b) the ratio of time to go g o from P to to R and and that from R to Q is 3 : 2 (c) the velocity at R is is 10 m/s (d) the average velocity between between P and R is 10 m/s

12.



River is flowing with a velocity v R 

v BR 

 4i ˆ m/s. A boat is moving with a velocity of

 2i ˆ  4 j ˆ m/s relative to river. The width of the river is 100 m along y-direction.

Choose the correct alternative(s) alternative(s) (a) the boat will cross the river river in 25s (b) absolute absolute velocity velocity of boat boat is 2 5 m/s (c) drift of the boat along the river river current is 50 m (d) the boat boat can never never cross the river. river.

13.





ˆ Velocity of a particle moving in a curvilinear curvilinear path varies with time as v  2t i ˆ  t 2 j m/s.

Here, t is in second. At t = = 1s (a) acceleration of particle is 8 m/s 2 6

(b) tangential acceleration of particle is

m/s2

5 (c) radial acceleration of particle is

2

(d) radius of curvature to the path is

14.

5

m/s2

5 5 2

m

During uniform circular motion of a particle, which of the following is incorrect? (a) distance-time graph is a straight line (b) distance-time graph is a parabola


KINEMATICS

OP-MI-P-79

(c) displacement-time graph is a straight line (d) displacement-time graph is a parabola

15.

A particle is projected vertically upward and it will will reach at a height from the ground in t 1 sec and time to reach the ground from that height is t 2 then (a) Initial velocity of projection is

g t 1

 t 2  2

(b) velocity at half of its maximum height will be

g t 1  t 2  2 2

(c) maximum distance travelled by particle in vertical direction is

g 2

t 1  t 2 2

(d) all of the above


KINEMATICS

OP-MI-P-80

EXERCISE – IV MATCH THE FOLLOWING Note: Each statement in column – column – I I has one or more than one match in column –II. –II. 1.

The displacement versus time graph of a particle is as shown in f igure. C

D

B

S A O

t

Column-I

Column-II

I.

In path OA

A.

Velocity increases with time.

II.

In path AB path AB

B.

Magnitude of acceleration is a non-zero.

III.

In path BC

C.

Velocity is a non-zero constant

IV.

In path CD

D.

acceleration is zero

E.

velocity is zero

Note: Each statement in column – column – I I has only one match in column – column –II II 2.

The displacement( displacement( x x ) of a particle depends on time (t ( t ) as x  t 2

 t 3

Column-I

Column-II

The particle will return to its starting point after time

A.

II.

The particle will be at rest at time

B.

III.

The average velocity of particle at any time t =t 0, taken over time from starting to t = = t 0 is equal to its instantaneous velocity at time t = = t 0. The value of t 0 is equal to

I.

IV.

No net force will act on the particle at t ime t is is equal to

C.

2 3

   3

D.

 2

E.

zero


KINEMATICS

3.

OP-MI-P-81

The path of the projectile is represented by y  Px  Qx 2 , where horizontal and vertical directions are chosen as positive x and positive y- axis respectively. Column-I

Column-II

I.

Range

A.

P /Q

II.

maximum height

B.

Q

III.

time of flight

C.

/4Q P 2/4Q

IV.

tangent of angle of projection is

2

D. E.

P Qg P

REASONING TYPE Directions: Read the following questions and choose (A) If both the statements are true and statement-2 statement-2 is the correct explanation of statement-1. statement-1. (B) If both the statements are true but statement-2 statement-2 is not the correct explanation of statement-1. statement-1. (C) If statement-1 is statement-1 is True and statement-2 is statement-2 is False. (D) If statement-1 is statement-1 is False and statement-2 is statement-2 is True.

1.

Statement-1: A body having having uniform speed in circular path path has a constant constant acceleration. Statement-2: Direction of acceleration is always towards the centre. (a) (A)

2.

(b) (B)

(c) (C)

(d) (D)

Statement-1: Two bodies of masses M and m (M > M > m) are allowed to fall from the same height if the force of air resistance for each be the same then both the bodies will reach the earth simultaneously. simultaneously. Statement-2: Acceleration due to gravity does does not depend on the mass of the body. body. (a) (A)

3.

(b) (B)

(c) (C)

(d) (D)

Statement-1: During the motion, the magnitude of displacement may be equal t o distance. Statement-2: When the particle is moving on the straight line then displacement will be equal to distance.


KINEMATICS

(a) (A)

4.

(b) (B)

OP-MI-P-82

(c) (C)

(d) (D)

Statement-1: Net acceleration of a particle, moving on a circle with constant tangential acceleration, remains constant. Statement-2:

aN

 ar 2  aT 2 , where aN is net acceleration, ar is centripetal acceleration

and aT is tangential acceleration. acceleration. (a) (A) 5.

(b) (B)

(c) (C)

(d) (D)



Statement-1: If v is the velocity of the particle in motion at time t , then

Statement-2:



(a) (A)



v dt  magnitude of displacement. displacement.

(b) (B)











v dt  | v | dt



v dt  distance covered.

(c) (C)

(d) (D)

LINKED COMPREHENSION TYPE

Two particles are moving on a circle of radius a with uniform speed v, one v, one is moving on a peripheral of the circle while other is oscillating on its diameter between the ends of the diameter. Let us suppose that T is time period of one rotation for first particle and assume both will starts from same point on circle then

1.

What is the magnitude of the relative velocity at t = 0 (a) v

2.

2v

(c) 2v

(d)

3v

What will be the magnitude of the relative velocity at t = = T /6 /6 (a) v 2  3

3.

(b)

(b) v 2  3

(c) v 4  3

What will the magnitude of the relative velocity at t   (a) v 2  3

(b) v 2  3

T 3

(d) v 2  2

.

(c) v 4  3

(d) v 2  2


KINEMATICS

4.

What will be the magnitude of the relative velocity at t  (a) v

(b)

2v

(c) 2v

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T 2

. (d)

3v


KINEMATICS

OP-MI-P-84

EXERCISE – V SUBJECTIVE PROBLEMS

1.

A train passes a station A A at 40 km/hr and maintains this speed for 7 km and is then uniformly retarded, stopping at B, which is 8.5 km from A. A. A second train starts from A at the instant the first train passes and being accelerated for part of the journey and uniformly retarded for the rest stops at B at the same time as the first train. What is the greatest speed of the second train?

2.

A particle is moving in a straight line and is observed to be at a distance ‘ a’ from a marked point initially, to be at a distance ‘b ‘ b’ after after an interval of n seconds, to be at a distance ‘c ‘ c ’ after 2n 2n seconds and to be at a distance ‘d ‘ d ’ after 3n 3n seconds. Prove that if the acceleration c a2b is uniform, d  a = 3(c 3(c  b) and that the acceleration is equal to . n2

3.

A particle projected with velocity u strikes at right angles a plane through the point of projection inclined at an angle  to the horizon. Show that the height of the point struck above the horizontal plane through the point of projection is time of flight up to that instant is, t 

4.

2u

g 13sin

2



2u 2  sin 2

    13sin 2   and that the g  

.

A particle is projected with a velocity 2 ag so that it just clears two walls of equal height ‘a’, which are at a distance 2a 2 a apart. Show that the time of passing between the walls is

2 a/g .

5.

A bottle was released from rest from a height of 60 m above the ground. Simultaneously, Simultaneously, a stone was thrown from a point on the ground 60 m distant horizontally from the bottle, with a velocity u at an angle of projection of , in a vertical plane containing the bottle. If the stone strikes the bottle 3 s after the instant of projection, find the velocity u and the angle  of projection.


KINEMATICS

6.

OP-MI-P-85

An aeroplane flies horizontally horizontally at height h at a constant speed v . An anti-aircraft gun fires a shell at the plane when it is vertically above the gun. Show that the minimum muzzle velocity of the shell required to hit the plane is

v 2 2gh at an angle tan -1 ( 2gh / v ) .


KINEMATICS

7.

OP-MI-P-86

Two particles start simultaneously simultaneously from the same point and move along two straight lines, one with uniform velocity u and the other with constant acceleration f . Show that their relative velocity is least after time

 ucos    and that the least relative velocity is u sin ,  f 

where  is the angle between between the lines.

8.

Two trains each having a speed of 30 km/hr are headed at each other on the same straight track. A bird that can fly at 60 km/hr flies off one train when they are 60 km apart and heads directly for the other train. On reaching the other train it flies directly back to the first and so forth. Find (a) how many trips can the bird make from one train to the other before the trains collide? (b) what is the total distance travelled by the bird?

9.

A projectile is projected with a velocity u at an angle  to the horizontal, in the vertical plane. If after time t , it is moving in a direction making an angle  with the horizontal, horizontal, prove that gt cos cos  = u sin (  )

10.

Two seconds after the projection, a projectile is moving in a direction direction at 30° to the horizon. horizon. After one more second, it is moving horizontally. horizontally. Determine the magnitude and direction of the initial velocity.


KINEMATICS

OP-MI-P-87

ANSWERS EXERCISE – I AIEEE-SINGLE CHOICE CORRECT

1. (b)

2.

(b)

3. (b)

4. (b)

5.

(c)

6. (b)

7.

(b)

8. (c)

9. (a)

10. (d)

11. (c)

12. (a)

13. (c)

14. (d)

15. (a)

16. (d)

17. (b)

18. (d)

19. (b)

20. (a)

21. (c)

22. (a)

23. (a)

24. (b)

25. (a)

EXERCISE – II IIT-JEE-SINGLE CHOICE CORRECT

1. (c)

2.

(c)

3. (a)

4. (b)

5.

(a)

6. (b)

7.

(c)

8. (c)

9. (b)

10. (c)

11. (a)

12. (c)

13. (d)

14. (c)

15. (a)

16. (d)

17. (d)

18. (c)

19. (b)

20. (b)

21. (c)

22. (b)

23. (a)

24. (c)

25. (d)

EXERCISE – III ONE OR MORE THAN ONE CHOICE CORRECT


KINEMATICS

OP-MI-P-88

1. (a,b)

2. (c,d)

3. (a,b,c,d)

4. (a,d)

5. (b,d)

6. (a,c)

7. (a,d)

8. (a,b)

9. (b,c)

10. (b,c)

11. (a,b,d)

12. (a,b,c)

13. (b,c,d)

14. (b,c,d)

15. (a,b)

EXERCISE – IV MATCH THE FOLLOWING 1.

I – A, II – C, C, D; III – III – B; B; IV – IV – D, D, E – A, B; II –

2.

I – B; – B; II – II – A, A, E; III – III – D; D; IV – IV – C C

3.

I – A; – A; II – II – C; C; III – III – D; D; IV – IV – E E

REASONING TYPE 1. (d)

2. (d)

3. (c)

4. (d)

5.

(a)

LINKED COMPREHENSION TYPE

1. (b)

2. (a)

3. (a)

4. (b)

EXERCISE – V SUBJECTIVE PROBLEMS

1.

68 km/hr

5.

28.2 m/s,  = 45


KINEMATICS

8.

(a) Infinity (b) 60 km

10.

2g 3 , 60

OP-MI-P-89


Brilliants Kinematics

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