A classical mechanics problem by SRIJAN Singh

Required time $t$

$20t=96+\dfrac 12. 2t^2\implies t^2-20t+96=0$

This equation has two real roots : $8$ sec. and $12$ sec.

After $8$ sec., the cyclist will overtake the bus. Since the bus is speeding up, it will overtake the cyclist after $12$ sec. So the required answer is $\boxed 8$ sec.

[Everything is in S.I. units]

I just assumed the distance by cyclist to be $x$ and so that of bus will be $x+96$ .

Using equation of motion $s=ut+\dfrac{1}{2}at^2$ you can get the two equations relating time $t$ and $x$ .

In the end, you will get $x=20t-96=t^2$

Solving the quadratic, you get time $t=\{8,12\}$

Now since we are asked of overtaking, so we take the least time. $\boxed{t=8 \text{s}}$