Time, Speed & Distance • Topic 4 of 4
Boats & Streams
If a boat rows at b km/h in still water and the stream flows at s km/h, downstream speed = b + s and upstream speed = b - s. Hence b = (down + up)/2 and s = (down - up)/2. A round trip over distance d takes d/(b+s) + d/(b-s). These two averages crack almost every boat problem.
✅ Solved examples
1. Boat 10 km/h still water, stream 2 km/h. Downstream and upstream speeds?
Down 12, up 8 km/h.
2. Downstream 15 km/h, upstream 9 km/h. Boat and stream speed?
Boat = (15+9)/2 = 12; stream = (15-9)/2 = 3 km/h.
3. A boat goes 24 km downstream in 2 h and returns in 3 h. Boat speed in still water?
Down 12, up 8; boat = (12+8)/2 = 10 km/h.
4. Still-water speed 8, stream 2. Time for 30 km upstream?
Up speed 6; time = 30/6 = 5 h.
✏️ Practice — try these, take hints as needed
1. Boat 12, stream 3. Downstream speed?
b + s.
12 + 3.
—
15 km/h
2. Down 18, up 12. Stream speed?
(down-up)/2.
6/2.
—
3 km/h
3. Down 20, up 10. Boat speed?
(down+up)/2.
30/2.
—
15 km/h
4. Boat 9, stream 3. Time for 24 km downstream?
Down 12.
24/12.
—
2 h
5. 36 km downstream in 3 h, stream 2. Boat speed?
Down 12.
b = 12 - 2.
—
10 km/h
📝 Topic test — 8 questions
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Formula Reference Sheet
This chapter
Core relations
| Basic | distance = speed x time |
|---|---|
| Unit conversion | km/h x 5/18 = m/s ; m/s x 18/5 = km/h |
| Average speed (equal distance) | 2ab/(a+b) |
| Inverse rule | fixed distance: speed up x% -> time down by ratio |
Relative speed, trains, boats
| Same direction | relative speed = a - b |
|---|---|
| Opposite direction | relative speed = a + b |
| Train crossing object of length L | time = (train + L) / speed |
| Downstream / Upstream | b + s and b - s (boat b, stream s) |
SSC reference
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