Q 11-02-039NEETJEE MainHard
A train starts from rest at one station and accelerates at $1\ \text{m/s}^2$ until it reaches $20$ m/s. It runs at this speed and then decelerates at $2\ \text{m/s}^2$ to stop at the next station, $2$ km away. The total time of the journey is
Answer: (D) $115$ s
Concept. Split the journey into acceleration, uniform motion and deceleration.
Step 1. Accelerating. $t_1 = \dfrac{20}{1} = 20$ s, distance $\dfrac{20^2}{2} = 200$ m.
Step 2. Decelerating. $t_3 = \dfrac{20}{2} = 10$ s, distance $\dfrac{20^2}{4} = 100$ m.
Step 3. Uniform part. $2000 - 300 = 1700$ m at $20$ m/s: $t_2 = 85$ s.
Step 4. Total. $20 + 85 + 10 = 115$ s.
Answer. $115$ s (Option D).
Solution by Sreeraj P, M.Sc Physics