4.8 Boats & Streams and Problems on Trains

Part 1: Boats and Streams

The two basic relations

Let $b$ be the speed of the boat in still water and $s$ the speed of the stream.

$\text{Downstream speed } d = b + s \qquad \text{(with the current)}$

$\text{Upstream speed } u = b - s \qquad \text{(against the current)}$

Inverting these gives the two formulae that answer most questions directly:

$b = \frac{d + u}{2}, \qquad s = \frac{d - u}{2}$

In words: still-water speed is the average of downstream and upstream speeds, and stream speed is half their difference.

Worked example. A boat covers 30 km downstream in 2 hours and 18 km upstream in 3 hours. Find the speed of the boat and of the current.

$d = \frac{30}{2} = 15 \text{ km/h}, \qquad u = \frac{18}{3} = 6 \text{ km/h}$

$b = \frac{15+6}{2} = 10.5 \text{ km/h}, \qquad s = \frac{15-6}{2} = 4.5 \text{ km/h}$

Equal distance both ways

If the same distance $D$ is covered each way, the total time is

$T = \frac{D}{b+s} + \frac{D}{b-s}$

and the average speed for the round trip is

$v_{\text{avg}} = \frac{b^2 - s^2}{b}$

Note that this is always less than $b$: a current slows the round trip even though it helps one leg. This counter-intuitive result is a favourite objective item. It is the harmonic-mean effect, and the average speed is never the simple average of $b+s$ and $b-s$.

Worked example. A boat with still-water speed 10 km/h makes a round trip on a stream flowing at 2 km/h. Its average speed is

$\frac{100 - 4}{10} = 9.6 \text{ km/h}$

not 10 km/h.

Time-ratio questions

If a boat takes $k$ times as long upstream as downstream over the same distance, then

$\frac{b+s}{b-s} = k \quad \Longrightarrow \quad \frac{b}{s} = \frac{k+1}{k-1}$

Worked example. A boat takes thrice as long to go upstream as downstream. Then $k = 3$ and $b/s = 4/2 = 2$, so the boat's speed in still water is twice the current's speed.

Related variants

• A man swimming in a river follows exactly the same relations.
• If a boat's speed in still water is less than or equal to the current's speed, it cannot travel upstream at all — occasionally the intended answer.
• In still water ($s = 0$), upstream and downstream speeds coincide.

Part 2: Problems on Trains

Unit conversion first

Almost every train problem mixes km/h with metres and seconds, so fix the conversions:

$1 \text{ km/h} = \frac{5}{18} \text{ m/s}, \qquad 1 \text{ m/s} = \frac{18}{5} \text{ km/h}$

Converting to metres per second at the outset prevents most errors.

What distance does the train actually cover?

This is the conceptual heart of the topic.

A stationary point has no length; a platform does. Confusing the two is the commonest error in the topic.

Worked example. A 240 m train crosses a pole in 12 seconds. Find its speed.

$v = \frac{240}{12} = 20 \text{ m/s} = 20\times\frac{18}{5} = 72 \text{ km/h}$

Worked example. The same train crosses a 360 m platform. Time taken?

$t = \frac{240 + 360}{20} = \frac{600}{20} = 30 \text{ seconds}$

Relative speed

Worked example — crossing. Two trains of lengths 150 m and 200 m travel towards each other at 54 km/h and 36 km/h. How long to cross?

Convert: 54 km/h is 15 m/s and 36 km/h is 10 m/s, so the relative speed is 25 m/s. The distance is $150 + 200 = 350$ m, so

$t = \frac{350}{25} = 14 \text{ seconds}$

Worked example — overtaking. The same two trains travel in the same direction. The relative speed is $15 - 10 = 5$ m/s, so

$t = \frac{350}{5} = 70 \text{ seconds}$

The fivefold difference between crossing and overtaking times illustrates why identifying the direction is the first thing to do.

A useful ratio result

If two trains of the same length take $t_1$ and $t_2$ seconds respectively to cross a pole, and $t$ seconds to cross each other travelling in opposite directions, then

$\frac{v_1}{v_2} = \frac{t_2}{t_1}$

because equal distances mean speed is inversely proportional to time.

Shared Method

Both halves of this section are the same physics. Write down three things before any algebra: what distance is actually covered, what relative speed applies, and what units are in play. Nearly every error in these topics is a failure at one of those three points rather than in the arithmetic that follows.