Question

A train of length $L$ move with a constant speed $V_t$. A person at the back of the train fires a bullet at time $t = 0$ towards a target which is at a distance of $D$ (at time $t =0$) from the front of the train (on the same direction of motion). Another person at the front of the train fires another bullet at time $t\, = \,T$ towards the same target. Both bullets reach the target at the same time. Assuming the speed of the bullets, $V_b$, are same, the length of the train is

• A. $T \times (V_b + 2 V_t)$
• B. $T \times (V_b + V_t)$
• C. $2 \times T \times (V_b + 2V_t)$
• D. $2 \times T \times (V_b - 2 V_t)$

Answer 1

Answer: (B)

$T \times (V_b + V_t)$

Answer 2

Bullets from both person reaches target at same instant, so we equate time to get,
$\frac{L-D}{V_{b}-V_{t}}=\frac{D}{V_{b}-V_{t}}+T$
$\frac{L}{V_{b}-V_{t}}=\frac{D}{V_{b}+V_{t}}+\frac{D}{V_{b}-V_{t}}+T$
$=\frac{D\left(V_{b}-V_{t}+V_{b}+V_{t}\right)}{V_{b}^{2}-V_{t}^{2}}+T$
$\frac{L}{V_{b}-V_{t}}=\frac{2 V_{b} D}{V_{b}^{2}=V_{t}^{2}}+T$
$\Rightarrow L=\frac{2 V_{b} D}{V_{b}+V_{t}}+T\left(V_{b}-V_{t}\right)$