Q 11-02-027NEETJEE MainHard
A girl running at her top speed of $5$ m/s is $10$ m behind the door of a bus when the bus starts from rest with a constant acceleration of $1\ \text{m/s}^2$ away from her. How long does she take to reach the door?
Answer: (D) $(5 - \sqrt{5})$ s
Positions from the girl's starting point: girl $x_G = 5t$; door $x_D = 10 + \dfrac{1}{2}t^2$.
$$5t = 10 + \frac{t^2}{2} \;\Rightarrow\; t^2 - 10t + 20 = 0 \;\Rightarrow\; t = 5 \pm \sqrt{5}$$
She first reaches the door at the smaller root, $t = (5 - \sqrt{5})$ s $\approx 2.76$ s.
Check with the discriminant: $100 - 80 > 0$, so she does catch it.
Solution by Sreeraj P, M.Sc Physics