Question

A bus starts moving with acceleration $2{\text{ms}}^{-2}$. A cyclist $96\text{m}$ behind the bus starts simultaneously towards the bus at a constant speed of $20\text{m/s}$. After what time will he be able to overtake the bus?

• A. $4\text{s}$
• B. $8\text{s}$
• C. $12\text{s}$
• D. $16\text{s}$

Answer 1

The correct option is B $8\text{s}$

Let us assume that the cyclist overtakes the bus after time $t$ seconds. The bus will cover a distance ‘$S$’ in time $t$ seconds. Then the cyclist has to cover a distance $(96+S)\text{m}$ in order to overtake the bus.
96 + (distance travelled by bus in time $t$) = (distance travelled by cyclist in time $t$)
$\Rightarrow \frac{1}{2}\times 2\times t^2+96=20\times t$
$\Rightarrow t^2-20t+96=0$
This gives, $t=8\text{s}$ or $12\text{s}$. Hence, the cyclist will overtake the bus at $t=8\text{s}$.
After $t=12s$ bus will again overtake the cyclist as bus has some acceleration and cyclist moves with constant speed.

Alternate Solution:
Relative displacement, $S_{rel}=96\text{m}$
Relative velocity of cyclist w.r.t bus, $u_{rel}=20-0=20\text{m/s}$
Relative acceleration, $a_{rel}=0-2=-2{\text{m/s}}^2$
Using second equation of motion,
$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$
$96=20t-\frac{1}{2}\times 2\times t^2$
$\Rightarrow t^2-20t+96=0$
This gives, $t=8\text{s}$ or $12\text{s}$
Hence, the minimum time required by the cyclist to overtake the bus $8\text{s}$