Question

A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. A further acceleration raises the speed to 90 km. p.h. in 10 seconds.The brakes are now applied to bring the car to rest under uniform retardation in 5 seconds. Find the distance travelled during braking.

(a) 200 m

(b) 300 m

(c) 225 m

(d) 335 m

I got this question during a job interview.

My question is taken from Numericals On Kinematics Of Motion topic in division Kinematics & Kinetics of Motion of Machine Kinematics

Answer 1

Correct option is (c) 225 m

To explain I would say: Given : u = 0 ; v = 72 km. p.h. = 20 m/s ; s = 500 m

First of all, let us consider the motion of the car from rest.

Acceleration of the car

Let a = Acceleration of the car.

We know that

v^2 = u^2 + 2as

or, 20^2 = 0 + 2a x 500 = 1000a

or, a = 20^2/1000 = 0.4 m/s^2

Let t = Time taken by the car to attain the speed.

We know that v = u + a.t

∴ 20 = 0 + 0.4 × t or t = 20/0.4 = 50 s

Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds.

Given : Initial velocity, u = 72 km.p.h. = 20 m/s ;

Final velocity, v = 96 km.p.h. = 25 m/s ; t = 10 s

Let a = Acceleration of the car.

We know that v = u + a.t

25 = 20 + a × 10 or a = (25 – 20)/10 = 0.5 m^2

We know that distance moved by the car,

s = ut + 1/2 at^2

  = 20 x 10 + 1/2 0.5(10)^2 = 225 m