Question

In a car race 'car A' takes 't' less than 'car B' and passes the finishing point with velocity 'v' more than the velocity with which 'car B' passes the point, assuming that the cars start from rest and travels with constant acceleration 'a1 & a2', show that v = t root a1a2.

Answer 1

Answer:

v = t √a1a2

Step-by-step explanation:

Let say Car A takes  t1  time  

Then Velocity of Car A  =  0 + a1t1

Car B takes  t1+t  time  

Then Velocity of Car B  =  0 + a2(t1 + t)

a1t1 = a2(t1 + t) + v

=> t1(a1 - a2) = a2t + v

=> t1 = (a2t + v)/(a1 - a2)

Distance covered by Car A = (1/2)a1t1²

Distance covered by Car B = (1/2)a2(t1 + t)²

(1/2)a1t1² = (1/2)a2(t1 + t)²

=> a1t1² = a2(t1 + t)²

=> a1( (a2t + v)/(a1 - a2))²  = a2((a2t + v)/(a1 - a2)  + t)²

=> a1(a2t + v)² = a2(a2t + v + a1t - a2t)²

=>  a1(a2t + v)² = a2(v + a1t)²

=> a1(a2²t² + v² + 2a2tv) = a2(v² + a1²t² + 2va1t)

=> a1a2²t² + a1v² + 2a1a2tv = a2v² + a2a1²t² + 2a1a2tv

=>  a1a2²t² + a1v² = a2v² + a2a1²t²

=> v²(a1 - a2)  = a1a2t²(a1 - a2)

=> v² = a1a2t²

=> v = t √a1a2

Answer 2

Step-by-step explanation:

Solution :

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

. <br> As we have

<br> For the time taken to travel this distance

, we use eqution as <br>

<br> Hence again we have

or

<br> In the questions given, car

starts with acceleration

and car

withe speed

, car

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

.