Question

Starting form rest a object move with the accelaratin with 1m/s^2 for first 10sec and for the next 10sec move with the uniform velocity the find the total distance covered by the object

Answer 1

150 m

First we have to break the total journey into two - one is of the first 10 seconds and the second is of the next 10 seconds.

Consider the motion of the object in the first 10 second. Here t = 10 s.

Since the body is starting from rest, we have,

u = 0 m s^(-1)

Acceleration, a = 1 m s^(-2)

By first kinematic equation, final velocity,

v = u + a t

v = 1 × 10

v = 10 m s^(-1)

By the second kinematic equation,

s = u t + (a t²) / 2

s = (1 × 10²) / 2

s = 100 / 2

s = 50 m

Now consider the motion of the object in the next 10 seconds. Here also, t = 10 s.

But here the initial velocity is nothing but the velocity attained in the total journey at the 10th second, i.e., it is the final velocity of the object in the first 10 seconds, i.e.,

u = 10 m s^(-1)

Acceleration, a = 0 m s^(-2), since the body has uniform velocity at this time.

Again by second kinematic equation,

s = u t + (a t²) / 2

s = 10 × 10 + (0 × 10²) / 2

s = 100 m

Hence the total displacement is 50 + 100 = 150 m.

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