A bus of $6m$ long starts moving with an acceleration of $2m{s}^{-2}$. A cyclist $90m$ behind the bus starts simultaneously towards the bus at $20m{s}^{-1}$. After what minimum time will he be able to overtake the bus?

The correct option is C $8s$

For boy to overtake,
In $t$ seconds bus travels a distance of $x$ meters and boy travels $(x+90+6)$ meters

Using second equation of motion for bus,
$x=ut+\frac{a{t}^2}{2},$
$x=\frac{a{t}^2}{2}=\frac{2\times t^2}{2}=t^2$...........................(1)

For boy, $Distance=speed\times time$
$x+96=vt,$
$x+96=20t$ .........................................(2)

using (1) and (2),
$t^2-20t+96=0$
we get $t=12and8$,
Since minimum time,
Hence, boy will catch the bus at $t=8seconds$.