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
Concept. Write both positions from the girl's start and solve for when they are equal.
Step 1. Positions. Girl: $5t$. Door: $10 + \dfrac{t^2}{2}$.
Step 2. Meeting.
$$t^2 - 10t + 20 = 0 \;\Rightarrow\; t = 5 \pm \sqrt5$$
Step 3. First meeting. The smaller root, $(5 - \sqrt5)$ s $\approx 2.76$ s.
Answer. $(5 - \sqrt5)$ s (Option D).
Solution by Sreeraj P, M.Sc Physics