Question

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

• A. 4 s
• B. 8 s
• C. 12 s
• D. 16 s

Answer 1

Answer: (B)

Let at time $t$ , the cyclist overtakes the bus, then $96 +$ (distance traveled by bus in time $t$) = (distance traveled by cyclist in time $t$)
$\Rightarrow \frac{1}{2} \times 2\times 2 \times t^2 + 96 = 20 \times t$
$\Rightarrow t^2 - 20 t + 96 = 0$
This gives, $t = 8 \,s$ or $12 \,s$.
Hence, the cyclist will overtake the bus in $8 \,s$