Problems Based on Trains — Extra Practice

Passing a pole or a standing man covers exactly the train’s length.

1. See what the train passes — for a pole / standing man / signal the distance is the train’s length $L$.
2. If the speed is in km/h and lengths in metres, convert the speed to m/s ($\times \frac{5}{18}$).
3. Length $=$ speed $\times$ time; speed $=$ length $÷$ time.
4. If the speed is asked, convert back to km/h ($\times \frac{18}{5}$).

Passing a platform covers $L+P$; solving the unknown from two situations.

• Platform / bridge: distance $=$ train $+$ platform. From the engine reaching the start to the last coach leaving the end, that is the distance.
• Pole and bridge both given: the speed is equal, so $\frac{L}{t_2}=\frac{L+B}{t_1}$ gives $L=\frac{B{t}_2}{t_1-t_2}$.
• Two bridges given: the train’s length cancels; speed $=\frac{B_1-B_2}{t_1-t_2}$.
• A platform as long as the train: total distance $=2L$.
• Time for one platform known: get the speed, then the time for another.

Distance $=x+y$; relative speed: $u+v$ in opposite directions, $u-v$ in the same direction.

• Imagine sitting in one of them — what is the other’s speed relative to me, and how far must it go to clear me (the two lengths together).
• Equal lengths, pole times $t_1,t_2$: time to cross each other $\frac{2t_1{t}_2}{t_1+t_2}$ (opposite) or $\frac{2t_1{t}_2}{|t_1-t_2|}$ (same direction).
• The view of a passenger or driver inside one train: the relative speed is the same ⟹ the lengths are in the ratio of the times.

The man’s length is negligible, so the distance is the train’s length; the speed is relative.

• Man in the train’s direction: relative speed $=v-w$ (the train passes him slowly).
• Man coming towards the train: relative speed $=v+w$ (passed quickly).
• If the times for two men differ, the train’s length is the same in both — two equations give both the speed and the length.
• Bring all speeds to one unit (m/s) and use time $=\frac{L}{\text{relative speed}}$.

When one starts late, first find the other’s lead by the time he starts.

• Straight meeting: time $=\frac{D}{a+b}$; meeting point $=a\times \text{time}$.
• One starts late: the first’s lead $=a\times t_0$; remaining distance $=D-a{t}_0$; then time at the joint speed.
• Same station, same direction, faster one starts late: distance $=\frac{ab{t}_0}{b-a}$ ($t_0$ in hours).
• One covers $d$ km more at the meeting: distance between stations $=\frac{d(a+b)}{a-b}$.
• Clock time of meeting: add the duration to the start time.

If two trains start together and the times to finish after meeting are given, find the ratio of speeds.

Both trains run for the same time $T$ before meeting. What the P-train covered by then is what the Q-train still needs $t_2$ for, and vice versa. Writing both with the speeds gives $\frac{a}{b}=\frac{t_2}{T}=\frac{T}{t_1}$, so $T^2=t_1{t}_2$ and $\frac{a}{b}=\sqrt{\frac{t_2}{t_1}}$.

The distance is the same in every case; that gives the equation.

• Speed reduced by $a$, time changes from $t_1$ to $t_2$: $t_{1}v=t_2(v-a)$ ⟹ $v=\frac{a{t}_2}{t_2-t_1}$; distance $=v{t}_1$.
• After an accident the speed becomes $\frac{p}{q}$: if the accident site shifts $x$ km on, exactly that stretch is now run at normal speed, saving its time.
• Stoppage minutes: $\frac{s_1-s_2}{s_1}\times 60$ per hour.

The slips that cost marks.

Habits for speed.

• 72 km/h = 20 m/s, 54 = 15, 36 = 10, 90 = 25 — memorise these four; most questions use them.
• Imagine sitting in the other train — the relative speed and the distance to clear become obvious.
• Equal lengths: time $t=\frac{2L}{\text{relative speed}}$.
• Check: a train speed of 30–150 km/h and a length of 100–800 m look right.

One-rule questions.

Exam-level questions.

Questions at the hard end of competitive exams.