Time, Speed & Distance

Downstream and upstream

A current of speed $s$ pushes a boat whose own speed in still water is $b$: $u = b + s \ \text{(downstream)}, \qquad v = b - s \ \text{(upstream)}$ Memorise the recovery formulas — almost every answer routes through them: $b = \frac{u + v}{2}, \qquad s = \frac{u - v}{2}$ Downstream 15 and upstream 10 → boat 12.5 km/h, stream 2.5 km/h. Since $b > s$ for forward progress, $u > v$ always: the bigger of the two given speeds is downstream.

Trip questions

Each leg is plain $D = S \times T$ with the leg's effective speed. 42 km upstream in a boat with $b = 18$, $s = 6$: time $= \frac{42}{12} = 3.5$ h. Name the two leg speeds first, then divide — that is the whole method.

Round trips

A trip down and back the same distance takes longer than the same distance in still water, because the stream steals more time going up than it gives going down. Two shapes appear:

1. Both times given, equal distances: $\frac{d}{u} + \frac{d}{v} = T_{\text{total}}$, or if the times differ, $d = u\,t_1 = v\,t_2$. Downstream 4 h and upstream 6 h for the same distance with downstream speed 6 km/h → $d = 24$ km, $v = 4$, $s = 1$.
2. Stream given, one time known: work the missing leg speed out of $b \pm s$.

Also worth knowing: for equal distance both ways, the average speed is $\frac{2uv}{u+v}$ — the harmonic mean again, always less than $b$.

Two double trips → two linear equations

The hardest common shape: two journeys with different mixes of up- and downstream distances and their total times. With $a = \frac1v$ and $c = \frac1u$ as unknowns, each trip gives one linear equation — e.g. 8 km up and 16 km down in 4 h, 3 km up and 8 km down in 1¾ h: $\frac{8}{v} + \frac{16}{u} = 4, \qquad \frac{3}{v} + \frac{8}{u} = \frac74$ Doubling the second and subtracting gives $\frac{2}{v} = \frac12$ → $v = 4$, then $u = 8$ → $b = 6$, $s = 2$. Substituting reciprocals turns a messy pair into a clean one.

Still-water and drifting objects

A floating log, a cork or a swimmer who stops rowing moves at exactly the stream's speed. Two consequences:

• time to drift distance $d$ = $\frac{d}{s}$ (1 km wide river crossed straight with $b = 4$, $s = 3$ takes 15 min, drifting 0.75 km);
• downstream time = still-water time + drift time for the same stretch: $t_d = \frac{d}{b} + \frac{d}{s}$, and $t_u = \frac{d}{b} - \frac{d}{s}$.

Common traps

• Swapping $u$ and $v$ (downstream is always the FASTER leg).
• Reporting $b$ when the stream was asked, or vice versa — check the question.
• Adding speeds on the return leg.
• Missing that a floating object's speed is $s$, not zero.

Quick revision

• $u = b + s$, $v = b - s$; $b = \frac{u+v}{2}$, $s = \frac{u-v}{2}$.
• Leg time = leg distance ÷ leg speed; label each leg fast/slow first.
• Equal-distance round trip: $\bar S = \frac{2uv}{u+v}$; $t_u > t_d$ always.
• Two trips → equations in $\frac1u, \frac1v$.
• Log/cork speed = $s$; upstream and downstream times differ by $t_u - t_d = \frac{2ds}{b^2 - s^2}$.