Time, Speed & Distance

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Boats & Streams

⏱ 5 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

Two speeds to track: downstream (with the current) $u = b + s$ , and upstream (against it) $v = b - s$ , where $b$ is the boat in still water and $s$ the stream.

Recover them: $b = \dfrac{u + v}{2}$ and $s = \dfrac{u - v}{2}$ .

The bigger of two given leg speeds is always downstream — the current helps that way. A floating object moves at exactly $s$ .

Downstream and upstream

A stream of speed $s$ pushes a boat whose own still-water speed is $b$ :

u = b + s \ \text{(downstream)}, \qquad v = b - s \ \text{(upstream)}

The recovery formulas route through almost every answer:

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

Downstream 15 and upstream 10 → boat $\dfrac{25}{2} = 12.5$ km/h, stream 2.5 km/h. Since the boat must beat the stream to move forward, $u > v$ always.

Rule: The LARGER of the two given leg speeds is downstream. Label the legs before anything else.

Single legs

Each leg is plain $D = S \times T$ with that leg's effective speed. 42 km upstream with $b = 18$ , $s = 6$ : speed $12$ km/h → time $\dfrac{42}{12} = 3.5$ h.

Name the two leg speeds first, then divide. Never merge the distances — the two directions carry different speeds.

Round trips

The same distance down and back takes LONGER than in still water: the stream steals more time going up than it gives going down.

Both times given: $d = u\,t_1 = v\,t_2$ . Downstream in 4 h at 6 km/h → $d = 24$ km; back in 6 h → $v = 4$ ; stream $s = 1$ .
Total time given: $\dfrac{d}{u} + \frac{d}{v} = T$ .

For equal distances both ways the average speed is $\dfrac{2uv}{u+v}$ — the harmonic mean again, always less than $b$ .

Watch: Upstream time is always the longer one. If your answer says otherwise, u and v are swapped.

Two journeys, two equations

The hardest common shape: two journeys mixing up- and downstream distances, with total times.

Substitute $x = \frac{1}{v}$ (up) and $y = \frac{1}{u}$ (down) — each journey becomes linear:

\dfrac{8}{v} + \frac{16}{u} = 4 \Rightarrow 8x + 16y = 4, \qquad \frac{3}{v} + \frac{8}{u} = \frac{7}{4} \Rightarrow 3x + 8y = \frac{7}{4}

Double the second and subtract the first: $2x = \frac{1}{2}$ → $x = \frac{1}{4}$ , so $v = 4$ ; then $y = \frac{1}{8}$ , $u = 8$ → $b = 6$ , $s = 2$ .

Tip: Scale one equation so one variable matches, then subtract. The reciprocals make it plain linear work.

Drifting objects

A log, a cork or a swimmer who stops rowing moves at exactly the stream's speed $s$ .

Time to drift a distance $d$ : $\dfrac{d}{s}$ .
A rower crossing a 1 km river straight across, with $b = 4$ and $s = 3$ : crossing time $\dfrac{1}{4}$ h = 15 min, drifting $3 \times \frac{1}{4} = 0.75$ km downstream.

Careful: A floating object's speed is $s$ , not zero. And the crossing time uses only $b$ , since the stream acts sideways.

Average speed with a stream

A round trip at $u$ and $v$ over equal distances averages $\dfrac{2uv}{u+v}$ . In still-water terms that simplifies to $\dfrac{b^2 - s^2}{b}$ .

With $b = 5$ , $s = 1$ : $u = 6$ , $v = 4$ → average $\dfrac{48}{10} = 4.8$ km/h $= \frac{25 - 1}{5}$ . The stream always drags the round-trip average below the still-water speed.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Boat and stream from the two leg speeds

How to spot it:

Downstream and upstream speeds are both given; the still-water speed or the stream's speed is asked.

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

Method

Identify the downstream speed — the larger one.
Average the two for the boat; take half the difference for the stream.
Sanity check: b must exceed s.

Why it works:

u = b + s and v = b − s add and subtract into b and s directly.

Try this

A boat goes downstream at 15 km/h and upstream at 10 km/h. Find the speed of the stream.

Show solution

$s = \frac{15 - 10}{2} = 2.5$ km/h.
Boat: $\dfrac{15 + 10}{2} = 12.5$ km/h.

Answer

2.5 km/h (boat 12.5 km/h)

Type 2very common4 practice Q

Single leg time or distance

How to spot it:

Boat and stream speeds are known; the time for one upstream or downstream leg (or the distance in a given time) is asked.

t = \frac{D}{b \pm s}

Method

Fix the direction: downstream b + s, upstream b − s.
Time = distance ÷ that speed (or distance = speed × time).
Keep the units consistent before dividing.

Why it works:

Each leg is straight D = ST with the current added or removed.

Try this

A boat's speed in still water is 18 km/h and the stream runs at 6 km/h. Find the time it takes to go 42 km upstream.

Show solution

$v = 18 - 6 = 12$ km/h.
$t = \frac{42}{12} = 3.5$ hours.

Answer

3.5 hours

Type 3common2 practice Q

Round trip with equal distances

How to spot it:

The same distance is covered both ways; the total time, the stream or the distance is asked from partial data.

\frac{d}{u} + \frac{d}{v} = T, \qquad d = u\,t_1 = v\,t_2

Method

Equate the two expressions for d when the two times are given.
Or use the total-time equation when one time is missing.
Finish with b = (u+v)/2, s = (u−v)/2.

Why it works:

The same d anchors both legs, giving one equation per unknown.

Try this

A boat covers a certain distance downstream in 4 hours and returns upstream in 6 hours. If its downstream speed is 6 km/h, find the speed of the stream.

Show solution

$d = 6 \times 4 = 24$ km.
$v = \frac{24}{6} = 4$ km/h.
$s = \frac{6 - 4}{2} = 1$ km/h.

Answer

1 km/h

Type 4common2 practice Q

Drifting objects and two double trips

How to spot it:

A log or cork drifts with the stream, or two journeys with different up/down mixes and total times are given.

\text{drift time} = \frac{d}{s}; \qquad \frac{A}{v} + \frac{B}{u} = T_1, \ \frac{C}{v} + \frac{D}{u} = T_2

Method

Drift: the object moves at s; time = distance ÷ s.
Two trips: substitute x = 1/v, y = 1/u to make both equations linear.
Scale one equation to match a coefficient, subtract, then convert back to u, v.

Why it works:

Reciprocals turn the messy pair into plain linear equations.

Try this

A man can row 8 km upstream and 16 km downstream in 4 hours; he can also row 3 km upstream and 8 km downstream in 1 hour 45 minutes. Find his speed in still water.

Show solution

Equations: $8x + 16y = 4$ and $3x + 8y = \frac{7}{4}$ with $x = \frac{1}{v}, y = \frac{1}{u}$ .
Double the second, subtract: $2x = \frac{1}{2}$ → $v = 4$ ; then $u = 8$ .
$b = \frac{4 + 8}{2} = 6$ km/h.

Answer

6 km/h

Type 5occasional

Average speed of a round trip with stream

How to spot it:

A boat rows a stretch down and back; the average speed for the whole trip is asked.

\bar{S} = \frac{2uv}{u + v} = \frac{b^2 - s^2}{b}

Method

Write u = b + s and v = b − s.
Use the harmonic mean of u and v over the equal distances.
Or apply the shortcut $\dfrac{b^2 - s^2}{b}$ directly.

Why it works:

Equal distances average by the harmonic mean; the stream drags it below b.

Try this

A boat rows 24 km downstream and back. Its still-water speed is 5 km/h and the stream runs at 1 km/h. Find the average speed for the whole trip.

Show solution

$u = 6$ , $v = 4$ km/h.
$\bar{S} = \frac{2 \times 6 \times 4}{10} = 4.8$ km/h.
Check: $\dfrac{5^2 - 1}{5} = 4.8$ .

Answer

4.8 km/h

Formula sheet

Effective speeds

u = b + s, \quad v = b - s

Boat and stream from legs

b = \frac{u + v}{2}, \quad s = \frac{u - v}{2}

Time for two legs

t = \frac{d_1}{b + s} + \frac{d_2}{b - s}

Round-trip average

\bar{S} = \frac{2uv}{u+v} = \frac{b^2 - s^2}{b}

Drift

\text{drift} = s \times \text{time}

Shortcuts that save time

⚡ Halve the sum, halve the difference

Down and up speeds give the boat and the stream in one line each.

Example

A boat's downstream speed is 16 km/h and its upstream speed is 12 km/h. Find its still-water speed and the stream's speed.

Show solution

$b = \frac{16 + 12}{2} = 14$ km/h.
$s = \frac{16 - 12}{2} = 2$ km/h.

Answer

Boat 14 km/h; stream 2 km/h

⚡ Extract u and v from trip times

Each trip is one equation; the pair is linear in the reciprocals.

Example

A boat covers 30 km downstream in 2 hours and returns in 3 hours. Find the speed of the stream.

Show solution

$u = \frac{30}{2} = 15$ , $v = \frac{30}{3} = 10$ km/h.
$s = \frac{15 - 10}{2} = 2.5$ km/h.

Answer

2.5 km/h

⚡ Two double-trip equations

Two journeys pin down both leg speeds exactly.

Example

A boat goes 12 km upstream and 18 km downstream in 3 hours, and 24 km upstream and 12 km downstream in 4 hours. Find its speed in still water.

Show solution

With $x = \frac{1}{v}, y = \frac{1}{u}$ : $12x + 18y = 3$ and $24x + 12y = 4$ .
Solving gives $x = \frac{1}{12}, y = \frac{1}{8}$ → $v = 12, u = 8$ .
$b = \frac{12 + 8}{2} = 10$ km/h (stream 2).

Answer

10 km/h

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Swapping u and v.

Downstream is b + s — always the FASTER leg.

Mistake 02

Recovering b as (u − v)/2.

Boat = (u + v)/2; stream = (u − v)/2.

Mistake 03

Adding both distances before dividing.

The legs have different speeds — handle each separately.

Mistake 04

Reporting the boat's speed when the stream was asked.

Check the question's last line before choosing the option.

Mistake 05

Giving a floating log a speed of zero.

A drifting object moves at exactly the stream's speed s.

Quick revision

Read this the night before the exam.

u = b + s (down), v = b − s (up); b = (u+v)/2, s = (u−v)/2.
Leg time = leg distance ÷ leg speed; label fast/slow legs first.
Equal-distance round trip: average = 2uv/(u+v) = (b² − s²)/b.
Two journeys → equations in 1/u and 1/v; scale and subtract.
Drifting object speed = s; crossing time uses only b.

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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