Question

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


c) How much time did it take to stop after the brakes were applied?.

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Answer 1

Explanation:

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

Answer 2

Answer:

first u have to find out acceleration and then find time by using v=u+at