Q153 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Physics
A train accelerates from rest at a constant rate 'p' for some time (attains velocity 'v') and then retards to rest at the constant rate 'q'. If the total distance covered by the train is 's', then the velocity 'v' of the train is
- A$\left( \left( \dfrac{p+q}{2pq} \right) \times s \right)^{\frac{1}{2}}$
- B$\left( \left( \dfrac{p-q}{2pq} \right) \times s \right)^{\frac{1}{2}}$
- Correct answerC$\left( \left( \dfrac{2pq}{p+q} \right) \times s \right)^{\frac{1}{2}}$
- D$\left( \left( \dfrac{2pq}{p-q} \right) \times s \right)^{\frac{1}{2}}$
Solution
Let $s_1$ be the distance covered during acceleration and $s_2$ be the distance covered during retardation.
Using the third equation of motion for the acceleration phase:
$v^2 = u^2 + 2as_1$
$v^2 = 0 + 2ps_1 \Rightarrow s_1 = \dfrac{v^2}{2p}$
For the retardation phase:
$0 = v^2 - 2qs_2 \Rightarrow s_2 = \dfrac{v^2}{2q}$
Total distance $s = s_1 + s_2$
$s = \dfrac{v^2}{2p} + \dfrac{v^2}{2q}$
$s = \dfrac{v^2}{2} \left( \dfrac{1}{p} + \dfrac{1}{q} \right)$
$s = \dfrac{v^2}{2} \left( \dfrac{p+q}{pq} \right)$
Solving for $v$:
$v^2 = \dfrac{2pq}{p+q} s$
$v = \left( \left( \dfrac{2pq}{p+q} \right) \times s \right)^{\dfrac{1}{2}}$
Answer: $\left( \left( \dfrac{2pq}{p+q} \right) \times s \right)^{\frac{1}{2}}$
Using the third equation of motion for the acceleration phase:
$v^2 = u^2 + 2as_1$
$v^2 = 0 + 2ps_1 \Rightarrow s_1 = \dfrac{v^2}{2p}$
For the retardation phase:
$0 = v^2 - 2qs_2 \Rightarrow s_2 = \dfrac{v^2}{2q}$
Total distance $s = s_1 + s_2$
$s = \dfrac{v^2}{2p} + \dfrac{v^2}{2q}$
$s = \dfrac{v^2}{2} \left( \dfrac{1}{p} + \dfrac{1}{q} \right)$
$s = \dfrac{v^2}{2} \left( \dfrac{p+q}{pq} \right)$
Solving for $v$:
$v^2 = \dfrac{2pq}{p+q} s$
$v = \left( \left( \dfrac{2pq}{p+q} \right) \times s \right)^{\dfrac{1}{2}}$
Answer: $\left( \left( \dfrac{2pq}{p+q} \right) \times s \right)^{\frac{1}{2}}$
Hence, the correct answer is option: 'C'.