Question

A bus starts moving with acceleration $2{\text{ms}}^{-2}$. A cyclist $96\text{m}$ behind the bus starts simultaneously toward the bus at a constant speed of $20\text{m/s}$. After some time, the bus will be left behind. If the bus continues moving with the same acceleration, after what time from the beginning, the bus will overtake the cyclist?

• A. $10\text{s}$
• B. $12\text{s}$
• C. $14\text{s}$
• D. $16\text{s}$

Answer 1

The correct option is B $12\text{s}$

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