Q. A car starts with acceleration $2\,m s^{-2}$. A scooterist starts towards the car with a velocity of $20\,ms^{-1}$. If the scooterist was $96\,m$ behind the car at the start, the time in which he overtakes the car is
Solution:
Let $t$ be the required time.
Distance covered by car, $S =\frac{1}{2} \times 2 \times t^2=t^2$
Distance covered by scooter $=96+ S =20 t$
$\therefore 96+t^2=20 t$ i.e. $t^2-20 t+96=0$
Solving $t=8 s$.
