A locomotive is moving at a speed of v_1 = 54 km/h . A train…

A locomotive is moving at a speed of $v_1 = 54\ \mathrm{km/h}$. A train of length $l = 150\ \mathrm{m}$ is approaching it from the opposite direction at a speed of $v_2 = 36\ \mathrm{km/h}$. How long does it take for the train to pass the locomotive's driver completely?

Explanation of Method

A train passing a locomotive's driver.

Step 1: Convert the speeds to m/s

The length is given in metres, so we work in m/s. Divide km/h by 3.6: \[v_1 = \frac{54 \cdot 1000\ \mathrm{m}}{3600\ \mathrm{s}} = 15\ \mathrm{m/s}, \qquad v_2 = \frac{36 \cdot 1000\ \mathrm{m}}{3600\ \mathrm{s}} = 10\ \mathrm{m/s}.\]

Step 2: Relative velocity

Look at the motion from the driver's point of view. The vehicles move toward each other, so the train approaches the driver with the sum of the two speeds:

\[ v_{\text{rel}} = v_1 + v_2 = 15\ \mathrm{m/s} + 10\ \mathrm{m/s} = 25\ \mathrm{m/s}. \]

Step 3: Time to pass

Passing "completely" starts when the front of the train reaches the driver and ends when its tail goes past him. Relative to the driver, the train moves exactly its own length $l$:

\[ t = \frac{l}{v_{\text{rel}}} = \frac{150\ \mathrm{m}}{25\ \mathrm{m/s}}. \]

\[t = 6\ \mathrm{s}\]

Check from the ground

$ \text{locomotive: } v_1 t = 15\ \mathrm{m/s} \cdot 6\ \mathrm{s} = 90\ \mathrm{m} $

$ \text{train: } v_2 t = 10\ \mathrm{m/s} \cdot 6\ \mathrm{s} = 60\ \mathrm{m} $

$ 90\ \mathrm{m} + 60\ \mathrm{m} = 150\ \mathrm{m} = l\ \checkmark $

$t = 0$ $t = 6\ \mathrm{s}$ $v_1$ $v_2$ $\text{driver}$ $l = 150\ \mathrm{m}$ $15 \cdot 6 = 90\ \mathrm{m}$ $10 \cdot 6 = 60\ \mathrm{m}$

locomotive (driver = red dot)
train

Top: the front of the train reaches the driver. Bottom: 6 s later the tail of the train passes him. In that time the locomotive moves 90 m to the right and the train 60 m to the left; together they cover exactly the train length, 150 m.

Result

\[\boxed{\,t = 6\ \mathrm{s}\,}\]

Remark

What if they moved in the same direction? Then the relative speed would be the difference, $v_1 - v_2 = 5\ \mathrm{m/s}$, and the locomotive would need $150/5 = 30\ \mathrm{s}$ to get past the train.