A train starts from a station with uniform acceleration. An…

A train starts from a station with uniform acceleration. An observer standing beside the track at the front of the train sees the first wagon pass him in $t_1 = 4\ \mathrm{s}$. How long does the 5th wagon take to pass him?

Step 1: When has the $n$-th wagon passed?

Let each wagon have length $l$. The train starts from rest, so after time $t$ it has moved $\tfrac12 a t^2$. The first $n$ wagons have passed the observer when the train has moved $nl$:

\[ nl = \tfrac12 a t_n^2, \]

\[ l = \tfrac12 a t_1^2. \]

Dividing the two equations: $\frac{t_n^2}{t_1^2} = n \;\Longrightarrow\; t_n = t_1\sqrt n$.

Step 2: Time for the $n$-th wagon alone

The $n$-th wagon passes between the moments $t_{n-1}$ (first $n-1$ wagons gone) and $t_n$: $\Delta t_n = t_n - t_{n-1} = t_1\left(\sqrt n - \sqrt{n-1}\right)$.

Step 3: The 5th wagon

\[ \Delta t_5 = 4\ \mathrm{s} \cdot \left(\sqrt5 - \sqrt4\right) = 4 \cdot (2.236 - 2)\ \mathrm{s}. \]

\[\Delta t_5 = 4(\sqrt5 - 2)\ \mathrm{s} \approx 0.94\ \mathrm{s}\]

Check

$ t_4 = 4\sqrt4 = 8\ \mathrm{s}, \quad t_5 = 4\sqrt5 \approx 8.944\ \mathrm{s} $

$ t_5 - t_4 \approx 0.944\ \mathrm{s}\ \checkmark $