Question

A bus is beginning to move with an acceleration of $1{\text{m/s}}^2$. A boy who is $48\text{m}$ behind the bus starts running with constant speed of $10\text{m/s}$. The earliest time when the boy can catch the bus is

• A. 8 sec
• B. 10 sec
• C. 12 sec
• D. 14 sec

Answer 1

The correct option is A 8 sec


Initial velocity of boy with respect to bus $=10{\text{ms}}^{-1}$
Acceleration of boy with respect to bus $=-1{\text{ms}}^{-1}$
Solving with respect to bus,
$s=ut+\frac{1}{2}a{t}^2$
$48=10t-\frac{1}{2}{t}^2$
$t^2-20t+96=0$
$t^2-12t-8t+96=0$
$(t-8)(t-12)=0$
$t=8\text{s}$ and $12\text{s}$
$\therefore$ (a)