A vehicle of 1 tonne travelling with a speed of 60 m/s notices a cow on the road 9 m ahead applies brakes. It stops just in front of the cow.

(i) Find out the KE of the vehicle before applying brakes.

(ii) Calculate the retarding force provided by the brakes.

(iii) How much time did it take to stop after the brakes were applied?

(iv) What is the work done by the braking force?

Given, mass of the vehicle, m = 1 tonne = 1000 kg

Initial speed, u = 60 m/s

Distance between vehicle and the cow, s = 9m Final velocity, v = 0

(i) KE of vehicle before applying brakes is given by

1/2 m ${{u}^{2}}$ = 1/2x 1000 x 60 x 60 = 1800000 J

(ii) As we know, from the third equation of motion,

${{v}^{2}}$ - ${{u}^{2}}$ = 2as

0-3600 = 2 x ax 9

=> a = 200 m${{s}^{-2}}$

So, retarding force provide by the brakes

= ma= 1000 kg x (-200)m${{s}^{-2}}$ = -200000N

(iii) Now, again from the second equation of motion,

s = ut + 1/2 a${{y}^{2}}$

=* 9 = 60t + 1/2 x (-200) ${{t}^{2}}$

10t -3 = 0

t =3/10 =0.3 s

(iv) So, work done by the braking force is given by

= Fs = -200000 Nx 9 m =-1800000 J