Question

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

• A. $10\sqrt{2}s$
• B. $\sqrt{110}s$
• C. $\sqrt{120}s$
• D. $15s$

Answer 1

The correct option is A $10\sqrt{2}s$

Car and bus distance $=$ $200m$

car $a_c=4m/s^2$

bus $a_B=2m/s^2$

$a_c$ with respect to bus = $2m/s^2$ forward

So, $S=ut+\frac{1}{2}a{t}^2$

$200=0+\frac{1}{2}\times 2\times t^2$

$t^2=200$

$t=10\sqrt{2}s$