Relative motion: express catches up when positions equal

Steps:

• Goods train position: $s_g = 10t$ (constant velocity, position 0 at t=0).
• Express train position: $s_e = \frac{1}{2} \times 1 \times t^2 = 0.5 t^2$ (from rest, position 0 at t=0).
• Set $s_e = s_g$: $0.5 t^2 = 10 t$.
• Solve: $t^2 - 20 t = 0$, $t(t - 20) = 0$, so t=20 s (discard t=0).

Why C is correct:

• Kinematic equation for catch-up: time $t = \frac{2v}{a} = \frac{2 \times 10}{1} = 20$ s, where v is initial speed difference, a is relative acceleration.

Why the others are wrong:

• A (6s): Underestimates time; express velocity only 6 m/s, still behind.
• B (10s): Time for express to match speed (v=at=10 m/s), but positions differ by 50 m.
• D (40s): Doubles the time, perhaps confusing a=0.5 m/s², but a=1 m/s².

Final answer: C