Q 11-02-090JEE MainJEE Main 2023 (1 Feb, Shift 2)Easy
For a train engine moving with speed of $20\ \text{m s}^{-1}$, the driver must apply brakes at a distance of $500$ m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed $\sqrt x\ \text{m s}^{-1}$. The value of $x$ is ______. (Assuming same retardation is produced by brakes)
Numerical value type. Enter your answer.
Answer: 200
$a=\dfrac{400}{2\times500}=0.4\ \text{m s}^{-2}$. Braking over $250$ m: $v^2=400-2\times0.4\times250=200$, so $x=200$.
Solution by Sreeraj P, M.Sc Physics