Question

A car is standing 200m behind a bus which is also at rest.The two start moving at the same instant but with the different forward accelerations.The bus has an acceleration of 2m/s^2 and car has an acceleration of 4m/s^2 Car will catch up the bus after a time of ?

Answer 1

answer to this question is 10 seconds

Answer 2

Assume that x is the distance travelled by truck when car catch the bus.

From Equation of Motion,

We know that,

$d=v_{0} t+\frac{1}{2} a t^{2}$

Given:

Both are at rest,  

$Thus\ v_{0}=0$

Therefore, we get

$\mathrm{d}=\frac{1}{2} a t^{2}$

This implies,

$\mathrm{t}=\frac{\sqrt{2 d}}{a}$

Now,

$Time\ taken\ by\ car =\frac{\sqrt{2(x+200)}}{4}$

$Time\ taken\ by\ truck =\frac{\sqrt{2 x}}{2}$

Since both take same time,  

$\frac{\sqrt{2(x+200)}}{4}=\frac{\sqrt{2 x}}{2}$

This implies,

$\begin{array}{l}{x+\frac{200}{2}=x} \\ {\mathrm{x}=200} \\ {\mathrm{t}=\sqrt{2} \times \frac{200}{2}=\sqrt{200}=\sqrt{100(1+1)}=10 \sqrt{2}}\end{array}$

Therefore, The car will catch up with the bus after $10 \sqrt{2}\ s$.