Question

A bus starts from rest moving with an acceleration of $2m/s^2$. A cyclist, $96m$ behind the bus starts simultaneously towards the bus at $20m/s$. After what time will he be able to overtake the bus:-

• A. $8s$
• B. $10s$
• C. $12s$
• D. $1s$

Answer 1

The correct option is A $8s$

Velocity of bus $V_b=0$ , Velocity of cyclist $V_c=20m/s$,

Acceleration of bus $a_b=2m/s^2$ Acceleration of cyclist $a_c=0$

Relative velocity $V_{cb}=V_c-V_b=20m/s$

Relative acceleration $a_{cb}=a_c-a_b=-2m/s^2$

Relative separation $=96m$

Using $s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow 96=20t-\frac{1}{2}2t^2$ $\Rightarrow t^2-20t+96=0$

Solving the equation , we get $t=8s$ and $t=12s$

Hence, after $8s$, cyclist will overtake the bus.