(Rectilinear Motion with Constant Acceleration)
14. An automobile moving at a constant velocity of 45 ft per sec passes a gasoline station. Two seconds later, another automobile leaves the gasoline station and accelerates at the constant rate of 6 ft per sec 2. How soon will the second automobile overtake the first?
SOLUTION: Let the second automobile catch the first one after t sec, Distance covered by first in (t+2) sec = Distance covered by second in t sec
45(t+2) = 0 + 1/2 x 6 x t 2 t2 -15t - 30 = 0 −± −± (− (−−()(−) t= ()
t = [15 +/- 18.57]/2
t = 16.79 sec (positive sign is taken because time cannot be negative)
15. A balloon rises from the ground with a constant acceleration of 4 ft per sec2. Five seconds later, a stone is thrown vertically up from the launching site. What must be the minimum initial velocity of the stone for it to just touch the balloon? Note that the balloon and stone have the same velocity at contact.
SOLUTION: as g=32ft/s2 balloon: (5 seconds after the ground) h=1/2at2 = ½(4)(5)2 h= 50ft v=at = (4)(5) v= 20ft/s
= v0t-16t2 (stone height) v=v0+gt = v0-32t (stone velocity)
balloon=stone (velocity) 20+4t= v0-32t v0= 36t+20
t seconds after: h=h0+v0t+1/2at2 = 50+20t+2t2 h= 2t2+20t+50
balloon=stone (height) 2t2+20t+50= v0t-16t2 = (36t+20)t-16t2 = 20t2+20t 18t2= 50 t= 1.667s
(balloon height) v=20+4t (balloon velocity)
v0= (36)(1.667+20)
stone: h=v0t+1/2gt 2 = v0t+1/2(-32)t^2
V0= 80 ft/s (stone initial velocity)
Rectilinear Motion with Variable Acceleration Using Equations of Motions
1. The motion of a particle is given by the equation s
2t
4
1
t
6
3
2t
2
, where s is in feet and t in seconds.
Compute the values of v and a when t = 2 sec.
SOLUTION: S= 2t4- t3+ 2t2 v=ds/dt
v=8t3−1/2t2+4t a=dv/dt a=24t2−2t/2+4 When t = 2 sec v=8(23)−1/2(22)+4(2)
v=70 ft/s a=24(22)−2+4
a=98 ft/s2 Summary of answers:
v=70 ft/s a=98 ft/s2
2. A particle moves in a straight line according to the law s = t3 - 40 t where s is in feet and t in seconds. (a) When t = 5 sec, compute the velocity. (b) Find the average velocity during the fourth second. (c) When the particle again comes to rest, what is its acceleration?
SOLUTION:
a. S=t3-40t v= 3t2 – 40 a=6t @t= 5 sec v= 3(5)2-40 v= 35 ft/sec total ditance b. Vave= total time −−(−) = − − = Vave= -3m/sec
@ 4th seconds s= (4)3-40(40) s= -96 m c. If v=0 0=3t2-40 t= 3.65 seconds a=6t a=6(3.65) a= 21.9ft/sec2
Summary of answers: a. v= 35 ft/sec b. Vave= -3m/sec c. a= 21.9ft/sec2
3. A ladder of length L moves with its ends in contact with a vertical wall and a horizontal floor. If the ladder starts from a vertical position and its lower end A moves along the floor with a constant velocity vA, show that the velocity of its upper end B is vB = -vAtan θ where θ is the angle between the ladder and the wall.
SOLUTION: B θ
L Y Va A X=Vat g=√L X dy dy Vb= = × (-2X) dt y dt dx =V dy a x Therefore: Vb = - Va y Vb = -Vatan
But X = Vat +
When θ= 90°, Vb= ∞, which is impossible.
4. The rectilinear motion of a particle is given by s = v 2 - 9 where s is in feet and v in feet per second. When t = 0, s = 0 and v = 3 ft per sec. Find the s-t, v-t, and a-t relations.
SOLUTION: S+9=v2
t+c=ln[secθ+tanθ]
√ + ln[ + ] =t+c
v=√ 5 9 = ds/dt d ec^θdθ =dt= +tan^θ √ + ec^θdθ = ecθ
c=0 √ + t= ln[ + ] √ + e = + t
s= 3tanθ
(v −) + v − e= + t
ds=3sec2θdθ
dv d 2V = dt dt
dt=secθdθ
2Va=V a= 1/2
√ s 9
s
3
5. The velocity of a particle moving along the X axis is defined by v = x 3 - 4x2 + 6 x where v is in feet per second and x is in feet. Compute the value of the acceleration when x = 2 ft.
SOLUTION: v=(2)3-4(2)2+6(2) v=4ft/sec vdv=adx a=vdv/dx dv/dx=2 a=4(2)
a=8 ft/sec2
6. The motion of a particle is defined by the relation a = 4 t, where a is in feet per second 2 and t in seconds. It is known that s = 1 ft and v = 2 ft per sec when t = 1 sec. Determine the relations between v and t, s and t, v and s.
SOLUTION:
dv/dt=4t t^ v= +c-2t2+c c=0 ; if t=1 v=2t2 √ v t= √ ds/dt=2t2
c=1/3 t^ s= +
3s=2t3+1 √ v 3s= 2 ( )3+1 √
t^ s= +c
3s=
(√ v)^ +1 √
()^ 1= +c
3s=
(√ v)^ +1 √
3=2+3c
(3s-1)√ 2-v3/2
7. The motion of a particle is governed by the equation a
8
s
2
, where a is in feet per second2
and s is in feet. When t = 1 sec, s = 4 ft and v = 2 ft per Sec. Determine the relations between v and t, s and t, v and s.
SOLUTION: ads=vdv vdv=-8/s2 ds −8^− v2/2= +c −
v2=16s-1+c if s=4 ; v=2 ∴ c=0 v2=16/s s=16/v2 v= =ds/dt √ 8
4dt=√ s ds / 4t= +c 4= 2/3(4)3/2+c C= -4/3 4t =2/3 s3/2+(-4/3) 12t =2s3/2-4 3/2 12t = 2 ( ) -4 v^
12t =
8 -4 v^
1
8. The motion of a particle is given by a 6v 2 , where a is in feet per second 2 and v in feet per second. When t is zero, s = 6 ft and v = 0. Find the relations between v and t, s and t, v and s.
Given: 1
a
6v 2 ,
t=0, s=6, v=0 Required: the relations between v and t, s and t, v and s.
SOLUTION: dv/dt= 6v1/2 dv/6√ = dt dv/(v)-1/2 = 6dt 2v1/2 = 6t + C If t=0 and v=0 therefore C= 0 2v1/2 = 6t V1/2 =3t V= 9t2 T=√ /3 ds/dt = 9t2
s= + C
if s=6, t=0 therefore C= 6 s= + 6 3 =3t + 6
S= 3(√ /3 )3 + 6 ^ S= + 6 ()
9s = v3/2 + 54
9. The motion of a particle is governed by the relation a = 4t2, where a is n feet per second 2 and t is in seconds. When t is zero, v = 2 ft per sec and s = 4 ft. Find the values of v and s when t = 2 sec. Given: a=4t2 t=0 v=2 s=4 Required: v and s when t=2
SOLUTION: dv/dt = 4t2 v= 4t3/3 + C () 2= →C=2 Therefore: 2 V= V=12.67 ft/s answer
If t=0, s=4 ; C=4 2 4 S= () S=
2(2) 4
S=13.33 ft answer
4 = 2 3 2 S=
1. From the v-t curve shown, determine the distance traveled in 4 sec and also in 6 sec.
SOLUTION: At 4 seconds: S=1/2(4)(20) S=40ft At 6 seconds: S=1/2(6)(20) S=60ft
2. The motion of a particle starting from rest is governed by the a-t curve shown. Determine the displacement at t = 9 sec.
SOLUTION: Solving for v1 and v2 a1= 2t integrating: a=∫ 2 a=2 ∫ = + C v1 = t2 + C if v1=0, t=0 therefore C1=0 v1 = t2 = 62 = 36 ft/s 20 integrating:
a2= -
=
∫( 2 0)
20 = × 20 v2 =
if v1=v2=9 and t=6 36 =
20
() 36 =
C2 = -60 () v2 = v2 = 66 ft/s
20(6)
20(9) 60
Solving for s1 and s2 : v1 = t2 integrating: =∫ S1 = If s=0, t=0 therefore C1=0 S1 = S1 =
=
S1 = 72 ft v2 = integrating:
20
= ∫(
20 60)dt
S2= 60 S2 = 10 60 If s1=s2=72 and t=6 72 = (6) 10(6) 60(6) C2 = 120
When t=9 S2 =
10 60
S2 = (9)
10(9) 60(9) 120
S2 =228 ft answer
