One train is l_1 = 500 m long and another is l_2 = 100 m…
One train is $l_1 = 500\ \mathrm{m}$ long and another is $l_2 = 100\ \mathrm{m}$ long. They move toward each other with speeds $v_1 = 16\ \mathrm{m/s}$ and $v_2 = 20\ \mathrm{m/s}$. How long does it take for the first train to pass completely by a window of the second train? How far does the first train travel in this time?
Explanation of Method
A train passing a window of another train.
Step 1: Relative speed
Seen from the window (a passenger in train 2), the first train comes toward him with the sum of the speeds, because the trains move toward each other: \[v_{\text{rel}} = v_1 + v_2 = 16\ \mathrm{m/s} + 20\ \mathrm{m/s} = 36\ \mathrm{m/s}.\]
Step 2: Distance to cover
The window is a single point. Passing "completely" means from the moment the front of train 1 reaches the window until its tail passes it, so relative to the window train 1 moves exactly its own length $l_1 = 500\ \mathrm{m}$. The length $l_2$ plays no role.
Step 3: Time
\[ t = \frac{l_1}{v_{\text{rel}}} = \frac{500\ \mathrm{m}}{36\ \mathrm{m/s}} = \frac{125}{9}\ \mathrm{s}. \]
\[t \approx 13.9\ \mathrm{s}\]
Step 4: Distance travelled by the first train
\[ s_1 = v_1\,t = 16\ \mathrm{m/s} \cdot \frac{125}{9}\ \mathrm{s} = \frac{2000}{9}\ \mathrm{m}. \]
\[s_1 \approx 222\ \mathrm{m}\]
Check from the ground
window: $ s_2 = 20 \cdot \tfrac{125}{9} \approx 277.8\ \mathrm{m} $
$ s_1 + s_2 = 222.2\ \mathrm{m} + 277.8\ \mathrm{m} = 500\ \mathrm{m} = l_1\ \checkmark $
- train 1
- train 2
- window
Result
\[\boxed{\,t \approx 13.9\ \mathrm{s}\,}\]
\[\boxed{\,s_1 \approx 222\ \mathrm{m}\,}\]
Remark
Note. This is the same idea as in Translational Kinematics 006. If the question were how long until the two trains have completely passed each other, both lengths would count: $t = \dfrac{l_1 + l_2}{v_1 + v_2} = \dfrac{600}{36} \approx 16.7\ \mathrm{s}$.