Relative Motion of Two Trains Crossing Each Other

We have two trains moving toward each other on parallel tracks.

• Train A: length $L_A = 150\,\text{m}$, speed $v_A = 60\,\text{km/h}$
• Train B: length $L_B = 100\,\text{m}$, speed $v_B = 40\,\text{km/h}$

They are moving in opposite directions, so for the crossing time, we use relative speed.

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1. Total Distance to Completely Pass Each Other

To completely pass each other, the back end of one train must clear the front end of the other. That means the total distance that must be covered between their front ends is simply the sum of their lengths:

$$D_{\text{total}} = L_A + L_B = 150 + 100 = 250\,\text{m}$$

So the combined length to be cleared is 250 meters.

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2. Relative Speed

They move in opposite directions, so their relative speed is the sum of their speeds:

$$v_{\text{rel}} = v_A + v_B = 60 + 40 = 100\,\text{km/h}$$

We convert this to $\text{m/s}$:

$$100\,\text{km/h} = 100 \times \frac{1000}{3600}\,\text{m/s} = \frac{100000}{3600}\,\text{m/s} \approx 27.78\,\text{m/s}$$

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3. Time to Completely Cross Each Other

Time is distance divided by relative speed:

$$t = \frac{D_{\text{total}}}{v_{\text{rel}}} = \frac{250\,\text{m}}{27.78\,\text{m/s}} \approx 9.0\,\text{s}$$

So:

• Total distance to be covered to completely pass: $250\,\text{m}$
• Time taken to completely cross: about $9\,\text{s}$

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How the Visualization Works

This interactive visualization shows two colored trains moving in opposite directions on parallel tracks:

• The top track (cyan) is Train A (150 m).
• The bottom track (pink) is Train B (100 m).
• You can adjust:
  • Their speeds (in km/h),
  • Their lengths (in meters),
  • And a time scale (to slow down or speed up the animation).

The visualization:

1. Converts each train's speed from km/h to m/s.
2. Uses the sum of their lengths to find the crossing distance.
3. Uses the sum of speeds (relative speed) to compute the theoretical crossing time.
4. Animates both trains starting just as their front ends meet.
5. Draws a vertical reference line where they first meet.
6. Continues the animation until the tail of one train has just cleared the other, marking that crossing is complete.

Under the hood, it repeatedly computes:

$$D_{\text{cross}} = L_A + L_B,$$ $$v_{\text{rel}} = v_A + v_B,$$ $$t_{\text{cross}} = \frac{D_{\text{cross}}}{v_{\text{rel}}},$$

then uses the elapsed time $t$ to update each train's position so you can see how relative motion shortens the crossing time even though the total distance to be cleared is simply the sum of lengths.