Relative Velocity — Two Trains Moving in Opposite Directions

Relative velocity is one of those topics where the concept clicks in two minutes but students still lose marks because they flip the sign or use the wrong formula. The difference between same-direction and opposite-direction is massive — one gives you a slow relative speed, the other gives you a fast one. We'll nail both cases here.

Question

(Part A) Train A moves east at 72 km/h. Train B moves east at 36 km/h (same direction). Find the velocity of A relative to B.

(Part B) Train A moves east at 72 km/h. Train C moves west at 36 km/h (opposite direction). Find the velocity of A relative to C.

(Part C) Train A and Train C are 1800 m apart and moving toward each other (A at 72 km/h, C at 36 km/h). After how many seconds will they cross each other?

Solution — Step by Step

Step 1: Convert units to m/s.

$72 \text{ km/h} = \frac{72 \times 1000}{3600} = 20 \text{ m/s (east)}$

$36 \text{ km/h} = \frac{36 \times 1000}{3600} = 10 \text{ m/s}$

Part A: Same Direction

Taking east as positive: $v_A = +20$ m/s, $v_B = +10$ m/s

$v_{AB} = v_A - v_B = 20 - 10 = +10 \text{ m/s (east)}$

From B's frame: Train A moves east at 10 m/s. If you're sitting in Train B, Train A appears to creep forward slowly. Intuition confirmed.

Part B: Opposite Direction

Taking east as positive: $v_A = +20$ m/s, $v_C = -10$ m/s (moving west)

$v_{AC} = v_A - v_C = 20 - (-10) = +30 \text{ m/s (east)}$

From C's frame: Train A appears to rush toward you at 30 m/s. This is why head-on collisions are so devastating — relative speed is the sum of both speeds.

Part C: Time to Cross

The trains are approaching each other. Relative speed of approach:

$v_{rel} = v_A + v_C = 20 + 10 = 30 \text{ m/s}$

Distance $= 1800$ m

$t = \frac{d}{v_{rel}} = \frac{1800}{30} = 60 \text{ s}$

Answers:

• Velocity of A relative to B (same direction) $=$ 10 m/s east
• Velocity of A relative to C (opposite direction) $=$ 30 m/s east
• Time for trains A and C to meet $=$ 60 seconds

Why This Works

Relative velocity is simply a change of reference frame. When you subtract $v_B$ from everything, B appears stationary and A appears to move with velocity $v_A - v_B$. This is the core idea.

For two objects approaching each other, their separation decreases at the rate of their relative speed. That's why we divide distance by relative speed to find the meeting time — we're treating one as stationary and the other as moving at the combined speed.

Alternative Method — Ground Frame Analysis for Part C

Instead of using relative velocity, we can work in the ground frame:

Position of A at time $t$: $x_A = 0 + 20t$ (starting from west end)

Position of C at time $t$: $x_C = 1800 - 10t$ (starting from east end, moving west)

They meet when $x_A = x_C$:

$20t = 1800 - 10t$

$30t = 1800$

$t = 60 \text{ s}$

Same answer. The ground frame method is more visual — useful when the problem asks where they meet, not just when.

Common Mistake