Question

In a car race, car

A

takes a time of

t s,

less than car

B

at the finish and passes the finishing point with a velocity

v

more than the car

B .

Assuming that the cars start from rest and travel with constant accelerations

a_{1}

and

a_{2}

respectively, show that

v = \sqrt{a_{1} a_{2} t} .

Accepted Answer

In the car race, the cars start from zero velocity, i.e., $u = 0$ and accelerate with constant accelerations. Here we'll discuss an important concept of uniformly accelerated motion. If a body starts from rest, i.e., with zero initial velocity and accelerates with acceleration $a$ after travelling distance $s$ , its velocity can be given by speed equation $v^{2} = u^{2} + 2 a s$

As we have $u = 0 \Rightarrow v = \sqrt{2 a s}$

For the time taken to travel this distance $s$ , we use equation as $s = u t + \frac{1}{2} a t^{2}$

Here again, we have

$u = 0, s = \frac{1}{2} a t^{2}$

$t = \sqrt{\frac{2 s}{a}}$

In the questions given, car $A$ starts with acceleration $a_{1}$ and car $B$ with acceleration $a_{2} .$ If car $B$ reaches the finishing point at time $T$ and with speed $u,$ car $A$ will reach at time $T - t$ and with speed $u + v$ as given in the question. As both the cars start from rest and cover same distance, say $s,$ we have

For car $A$ : $v + u = \sqrt{2 a_{1} s}$ and $T - t = \sqrt{\frac{2 s}{a_{1}}}$

For car $B$ : $u = \sqrt{2 a_{2} s}$ and $T = \sqrt{\frac{2 s}{a_{2}}}$

From above equations, eliminating the terms of $u$ and $T,$ we get

$v = \sqrt{2 a_{1} s} - \sqrt{2 a_{2} s}$

$t = \sqrt{\frac{2 s}{a_{2}}} - \sqrt{\frac{2 s}{a_{1}}}$

Dividing the above equations, we get, $v = \sqrt{a_{1} a_{2} t}$ .

B M Sharma Solutions for Chapter: Kinematics I, Exercise 23: ILLUSTRATION 4.22

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 23: ILLUSTRATION 4.22

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 23: ILLUSTRATION 4.22 with Hints & Solutions

Learn and Prepare for any exam you want !