Question

In a car race, car A takes 2 s less than car B and passes the finishing point with a velocity v more than the velocity with which car B passes the point. Both the cars start from rest and travels with constant acceleration a1=4m/s^2 &a2=1m/s^2 respectively,find the velocity of v in ms^-1

Answer 1

Answer:

In the car race, the cars start from zero velocity an daccilerate with constant accelerations. Here we' ll discuss an imprtant concept of uniformly accelerated motion. If a body starts from rest, i.e., with zero initial velocity an daccelerates with acceleration (a) travellin distance

, its velocity can be given by speed equation

s

. <br> As we have

v2u2+2as

<br> For the time taken to travel this distance

s

, we use eqution as <br>

s=ut+1/2a

<br> Hence again we have

u=0.s1/2at

or

t=√2s/a

<br> In the questions given, car

starts with acceleration. A

and car withe speed, car a¹

B

will reach at time

and with speed

as given in the question. <br> As both the cars start from rest and cover same distance, say

, we have <br> For car A :

and

<br> for car B:

and

<br> From above equations, elinatint the terms of

and

, we get <br>

<br>

. <br> Diveding the above equations , we get