Science > Physics > Motion in a Straight Line > Numerical Problems on Uniformly Accelerated Bodies Set – 05
Example 01:
A body moving in a straight line has its motion described by s = 2 – 3t + 8 t2 Find the equations for its velocity and acceleration as functions of time.
Solution:
Displacement is given by
s = 2 – 3t + 8 t2 …………. (1)
Differentiating both sides of equation (1) w.r.t. time t
Velocity = v = (ds/dt) = -3 + 16t = 16t – 3 …………. (2)
Differentiating both sides of equation (2) w.r.t. time t
Acceleration = a = (dv/dt) = -3 …………. (3)
Example 02:
A body moving along a straight line has its displacement in metres given by s = 3 + 2t + 4 . Find its velocity after 2 s and the acceleration.
Solution:
Displacement is given by
s = 3 + 2t + 4t2 …………. (1)
Differentiating both sides of equation (1) w.r.t. time t
Velocity = v = (ds/dt) = 2 + 8t …………. (2)
Differentiating both sides of equation (2) w.r.t. time t
Acceleration = a = (dv/dt) = 8 …………. (3)
Velocity after 2 seconds
v = 2 + 8t = 2 + 8(2) = 2 + 16 = 18 ms-1
Acceleration after 2 seconds
a = 8 ms-2
Ans: After 2 s, velocity of the body is 18 ms-1and its acceleration is 8 ms-2.
Example 03:
The equation of motion of a particle moving along a straight line is given by s = 63 t – 6t2 – t3 cm. Find its velocity and acceleration after 2 s. Also find the distance covered by it before coming to rest.
Solution:
Displacement is given by
s = 63 t – 6t2 – t3 …………. (1)
Differentiating both sides of equation (1) w.r.t. time t
Velocity = v = (ds/dt) = 63 – 12t – 3t2 …………. (2)
Differentiating both sides of equation (2) w.r.t. time t
Acceleration = a = (dv/dt) = -12 – 6t …………. (3)
Velocity after 2 seconds
v = 63 – 12t – 3t2
∴ v = 63 – 12(2) – 3(2)2
∴ v = 63 – 24 -12 = 27 cm s-1
Acceleration after 2 seconds
a = -12 – 6t
∴ a = -12 – 6(2)
∴ a = -12 -12 = -24 cm s-2
Negative sign indicates retardation
Now body coming to rest thus v = 0
63 – 12t – 3t2 = 0
∴ t2 + 4t -21 = 0
∴ (t + 7)(t – 3) = 0
∴ t = 7 = 0 and t – 3 = 0
∴ t = – 7 or t = 3
∴ t = -7 not possible
∴ t = 3
Thus displacement at end of 3 s
s = 63 t – 6t2 – t3
∴ s = 63 (3) – 6(3)2 – (3)3
∴ s = 189 – 54 – 27 = 108 cm
Ans: After 2 s, velocity of the body is 27 cms-1and its acceleration is – 24 cms-2(retardation). The body will travel a distance of 108 cm before coming to rest.
Example 04:
The displacement of a particle along a straight line from a fixed point on the line is given by s = 120t – 27t2 + 2t3 metres after t s. Find its velocity and acceleration after 2 s. Find the instants when velocity becomes zero and calculate the distance travelled between these instants.
Solution:
Displacement is given by
s = 120t – 27t2 + 2t3…………. (1)
Differentiating both sides of equation (1) w.r.t. time t
Velocity = v = (ds/dt) = 120 – 54t + 6t2 …………. (2)
Differentiating both sides of equation (2) w.r.t. time t
Acceleration = a = (dv/dt) = – 54 + 12t …………. (3)
Velocity after 2 s
v = 120 – 54t + 6t2
∴ v = 120 – 54(2) + 6(2)2
∴ v = 120 – 108 + 24 = 36 ms-1
Acceleration after 2 s
a = – 54 + 12t
∴ a = – 54 + 12(2)
∴ a = = – 54 + 24 = – 30 ms-2
When velocity is zero v = 0
120 – 54t + 6t2 = 0
∴ t2 – 9t + 20 = 0
∴ (t – 5) (t – 4) = 0
∴ t – 5 = 0 and t -4 = 0
∴ t = 5 and t = 4
Thus at t = 5 s and t = 4 s, the velocity of the body is zero
To Find displacement at t = 5 s
s = 120t – 27t2 + 2t3
∴ s = 120(5) – 27(5)2 + 2(5)3
∴ s = 600 – 675 + 250 = 175 m
To Find displacement at t = 4 s
s = 120t – 27t2 + 2t3
∴ s = 120(4) – 27(4)2 + 2(4)3
∴ s = 480 – 432 + 128 = 176 m
Thus displacement during this period = 176 -175 = 1m
Ans: After 2 s velocity is 36 ms-1 and the acceleration is – 30 ms-2 (retardation). The distance travelled between the instances when velocity becomes zero is 1 m.
Example 05:
A particle moves in a straight line such that its displacement x at a time t is given by s = t3 – 6t2 + 5t + 4 metres. Find its velocity when its acceleration is zero.
Solution:
Displacement is given by
s = t3 – 6t2 + 5t + 4…………. (1)
Differentiating both sides of equation (1) w.r.t. time t
Velocity = v = (ds/dt) = 3t2 – 12t + 5 …………. (2)
Differentiating both sides of equation (2) w.r.t. time t
Acceleration = a = (dv/dt) = 6t – 12 …………. (3)
Acceleration is zero. a = 0
∴ 6t – 12 = 0
∴ t = 12/6 = 2 s
Velocity at end of 2 s
v = 3t2 – 12t + 5
∴ v = 3(2)2 – 12(2) + 5
∴ v = 12 – 24 + 5 = – 7 ms-1
Ans: The velocity when acceleration is zero is – 7 ms-1.
Example 06:
Two particles starting from the same point O at the same time and along the same straight line are at distances s1 = 5t2 + 11t metres and s2 = 7t2 + 3t metres respectively from O after t s from start.
- When will they have the same speed?
- When will they meet?
- What will be their velocities, accelerations and distance from O when they meet?
Solution:
Consider first particle
Displacement is given by
s1 = 5t2 + 11t…………. (1)
Differentiating both sides of equation (1) w.r.t. time t
Velocity = v1 = (ds1/dt) = 10t + 11 …………. (2)
Differentiating both sides of equation (2) w.r.t. time t
Acceleration = a1 = (dv1/dt) = 10 …………. (3)
Consider second particle
Displacement is given by
S2 = 7t2 + 3t…………. (4)
Differentiating both sides of equation (4) w.r.t. time t
Velocity = v2 = (ds2/dt) = 14t + 3 …………. (5)
Differentiating both sides of equation (5) w.r.t. time t
Acceleration = a2 = (dv2/dt) = 14 …………. (6)
When two particles have the same speed
v1 = v2
∴ 10t + 11 =14t + 3
∴ 4t = 8
∴ t = 2 s
When will they meet?
As the two particles are meeting
s1 = s2
∴ 5t2 + 11t = 7t2 + 3t
∴ 2t2 – 8t = 0
∴ 2t (t – 4) = 0
t = 0 or t – 4 = 0
t = (at start) not practical answer
∴ t = 4
Find their velocities, accelerations and displacements at t = 4 s
For first particle
Displacement
s1 = 5t2 + 11t = 5(4)2 + 11 (4)
∴ s1 = 80 + 44 = 124 m
Velocity
v1 = 10t + 11 = 10(4) + 11
∴ v1 = 40 + 11 = 51 ms-1
Acceleration a1 = 10 ms-2
For second particle
Displacement
s2 = 7t2 + 3t = 7(4)2 + 3 (4)
∴ s2 = 112 + 12 = 124 m (obviously = s1)
Velocity
v2 = 14t + 3 = 14(4) + 3
∴ v2 = 56 + 3 = 59 ms-1
Acceleration a1 = 14 ms-2
Ans:
- At t = 2 s both particles will have same velocity
- At t = 4s both particles will meet
- When they meet the displacement of first particle is 124 m, its velocity is 51 ms-1, and its acceleration is 10 ms-2 and the displacement of second particle is 124 m, its velocity is 59 ms-1, and its acceleration is 14 ms-2