Q 11-02-132JEE MainJEE Main 2017 (9 Apr)Easy
A car is standing $200$ m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration $2\ \text{m s}^{-2}$ and the car has acceleration $4\ \text{m s}^{-2}$. The car will catch up with the bus after time:
Answer: (D) $10\sqrt2$ s
Relative to the bus, the car starts from rest $200$ m behind with acceleration $4 - 2 = 2\ \text{m s}^{-2}$:
$$200 = \frac12(2)t^2 \;\Rightarrow\; t^2 = 200 \;\Rightarrow\; t = 10\sqrt2\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics