Question

A train of length $L$ moves 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. $T \times (V_b + 2V_t)$
• B. $T \times (V_b + V_t)$
• C. $2T \times (V_b + V_t)$
• D. $2T \times (V_b - V_t)$
• E. $T \times (V_b - V_t)$

Hint:

In relative motion problems, carefully track distances at different times and use consistent reference frames.

Answer 1

The Correct Option is B

Concept:
Relative motion is used here. Motion is observed from ground frame. Distance = speed $\times$ time.

Step 1: Bullet from the back (at $t=0$).
Initial distance to target: \[ D + L \] Time taken = $t$ \[ V_b t = D + L \quad ...(1) \]

Step 2: Bullet from the front (at $t=T$).
Distance at $t=T$ becomes: \[ D - V_t T \] Time taken = $t - T$ \[ V_b (t - T) = D - V_t T \quad ...(2) \]

Step 3: Solve equations.
From (1): \[ D = V_b t - L \] Substitute into (2): \[ V_b (t - T) = V_b t - L - V_t T \]

Step 4: Simplify.
\[ V_b t - V_b T = V_b t - L - V_t T \] \[ - V_b T = -L - V_t T \] \[ L = T(V_b + V_t) \]