Train Meeting Problem Solver
Solve classic two-train word problems step by step. Handles trains meeting head-on, overtaking in the same direction, two trains crossing each other (with lengths), a train passing a pole, and a train crossing a platform or bridge — with animated track visualization, relative-speed math, and full LaTeX explanations.
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About Train Meeting Problem Solver
The Train Meeting Problem Solver tackles the five most common train word problems in one place: two trains meeting head-on, a faster train overtaking a slower one in the same direction, two trains of given lengths crossing each other, a train passing a single point such as a pole or signal post, and a train crossing a platform, bridge, or tunnel. Type in the speeds, distances, and lengths in any mix of metric (km/h, m/s, km, m) or imperial units (mph, ft/s, mi, ft) — the solver converts everything to consistent SI units, applies the right relative-speed rule, and shows a full LaTeX-formatted solution alongside an animated track that visualizes the motion in real proportion.
How to use this solver
- Pick the scenario that matches your problem from the dropdown — meeting, overtaking, two crossing, passing a pole, or crossing a platform.
- Choose the display units. You can still mix km/h with metres, or mph with feet — the solver handles unit conversion internally.
- Enter the speeds, lengths, and gap. For the meeting scenario you can optionally add a head-start time for Train A in minutes.
- Click Solve. The headline value shows the meeting, overtaking, or crossing time. Below it you get the relative speed, distance covered by each train, and a step-by-step LaTeX explanation.
- Watch the animated track at the side and in the result panel — the trains move in real proportion to the speeds and direction of motion.
The five formulas at a glance
1. Meeting head-on
Trains move toward each other. Relative speed adds.
\( t = \dfrac{D}{v_1 + v_2} \)
2. Overtaking (same direction)
Faster train catches slower one. Relative speed subtracts.
\( t = \dfrac{D}{v_B - v_A} \)
3. Two trains crossing
Total distance to clear = sum of train lengths.
\( t = \dfrac{L_1 + L_2}{v_{rel}} \)
4. Passing a pole
The pole is a point. Train moves its own length.
\( t = \dfrac{L}{v} \)
5. Crossing a platform
Distance = train length + platform length.
\( t = \dfrac{L_{train} + L_{platform}}{v} \)
The relative-speed rule (the key idea)
Almost every train word problem reduces to one identity:
\[ \text{time} \;=\; \dfrac{\text{distance to cover}}{\text{relative speed}} \]
What changes between scenarios is the meaning of "distance" and the sign of the relative speed:
- Toward each other — the two trains close the gap together, so add their speeds: \( v_{rel} = v_1 + v_2 \).
- Same direction — only the difference in speed closes the gap: \( v_{rel} = v_B - v_A \). If both speeds are equal, the gap never closes.
- Crossing with lengths — the back of one train must clear the back of the other, so the distance to cover equals \( L_1 + L_2 \) instead of just the gap.
- Pole vs platform — a pole is a point (cover \( L_{train} \)); a platform has length (cover \( L_{train} + L_{platform} \)).
Worked example: meeting head-on
Two trains start 300 km apart and move toward each other. Train A travels at 60 km/h, Train B at 90 km/h. When and where do they meet?
- Convert speeds to m/s: \( v_1 = 60 \times \tfrac{5}{18} = 16.667 \) m/s; \( v_2 = 90 \times \tfrac{5}{18} = 25 \) m/s.
- Relative speed: \( v_{rel} = 16.667 + 25 = 41.667 \) m/s = 150 km/h.
- Time to meet: \( t = \dfrac{300\,000\;\text{m}}{41.667\;\text{m/s}} = 7200 \) s = 2 h.
- Distance covered by Train A: \( d_1 = 60 \times 2 = 120 \) km, so the meeting point is 120 km from A and 180 km from B.
Worked example: train crossing a platform
A 150 m train moving at 90 km/h needs to cross a 350 m platform.
- Convert the speed: \( 90 \times \tfrac{5}{18} = 25 \) m/s.
- Total distance to cover: \( 150 + 350 = 500 \) m.
- Crossing time: \( t = \dfrac{500}{25} = 20 \) s.
Common pitfalls and how to avoid them
- Mixing units — multiplying km/h by seconds gives a number that means nothing. Either convert km/h to m/s by multiplying by \(\tfrac{5}{18}\), or convert m/s to km/h by multiplying by 3.6. The solver does this automatically.
- Forgetting train length — when two trains cross or a train crosses a platform, the back of the train must clear the other end. Always add lengths to the distance.
- Wrong sign on relative speed — if you write \( v_1 + v_2 \) for a same-direction overtake, the time will be far too short. Add only when motion is opposing.
- Equal speeds in same direction — if the two trains have equal speed and travel the same way, the relative speed is zero and they never overtake.
- Head start vs distance — a head start is a time advantage, not a distance. Convert it to distance by multiplying the leader's speed by the head-start time.
Quick conversion reference
| From | To | Multiply by | Example |
|---|---|---|---|
| km/h | m/s | 5/18 ≈ 0.2778 | 72 km/h × 5/18 = 20 m/s |
| m/s | km/h | 18/5 = 3.6 | 25 m/s × 3.6 = 90 km/h |
| mph | m/s | 0.44704 | 60 mph × 0.44704 ≈ 26.82 m/s |
| m/s | mph | 2.2369 | 30 m/s × 2.2369 ≈ 67.1 mph |
| km | m | 1000 | 1.5 km = 1500 m |
| mi | m | 1609.344 | 2 mi ≈ 3218.7 m |
| ft | m | 0.3048 | 500 ft = 152.4 m |
Frequently asked questions
What is the formula for two trains meeting head-on?
When two trains move toward each other, their relative speed equals the sum of their individual speeds: \( v_{rel} = v_1 + v_2 \). The time to meet is the initial gap divided by this relative speed: \( t = D / (v_1 + v_2) \). Each train covers its own speed times \( t \). The meeting point is closer to whichever train is slower.
How do I solve an overtaking problem (same direction)?
When trains move in the same direction, the relative speed is the difference: \( v_{rel} = v_{faster} - v_{slower} \). The time for the faster train to catch the slower one is \( t = D / (v_{faster} - v_{slower}) \). If both speeds are equal, the faster never overtakes.
Why does train length matter when two trains cross each other?
Two trains finish crossing only when the last car of one has cleared the last car of the other. So the total distance their relative motion must cover equals the sum of their lengths: \( t = (L_1 + L_2) / v_{rel} \). Add the speeds for opposite-direction crossing, subtract them for same-direction crossing.
How long does a train take to pass a pole?
A pole, person, or signal post is a single point. The train clears it when its last car reaches the pole, so the train moves a distance equal to its own length. The time is simply train length divided by speed: \( t = L / v \).
How long does a train take to cross a platform or bridge?
A platform or bridge has length, so the train must travel its own length plus the platform length to fully clear the other end. The time is \( t = (L_{train} + L_{platform}) / v \).
How do I convert km/h to m/s?
Multiply by 1000/3600 = 5/18. So 72 km/h = 72 × 5/18 = 20 m/s. To go the other way, multiply m/s by 18/5 = 3.6. So 25 m/s = 25 × 3.6 = 90 km/h. The calculator does this conversion automatically before computing.
Reference this content, page, or tool as:
"Train Meeting Problem Solver" at https://MiniWebtool.com// from MiniWebtool, https://MiniWebtool.com/
by miniwebtool team. Updated: 2026-05-10
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