Irodov 1.1: Why the Boat Speed Doesn't Matter

The problem

A motorboat going downstream passes a raft at point A. After $\tau = 60\,\text{min}$ the boat turns around and, after some further time, passes the raft again at a point $l = 6.0\,\text{km}$ from A. The engine speed (relative to water) is constant throughout. Find the flow velocity $u$.

The raft floats passively — it moves exactly as fast as the water. The problem gives you $\tau$ and $l$, but never specifies the engine speed $v$. That omission is the hint. The answer is

$u = \frac{l}{2\tau}$

and $v$ appears nowhere in it.

The frame argument

Switch to the reference frame of the water. In this frame the raft is stationary and there is no current — the water frame is inertial and there is no time dilation, so clocks in both frames agree.

In the water frame:

• Phase 1. The boat moves downstream at speed $v$ for time $\tau$, reaching a distance $v\tau$ from the raft.
• Phase 2. The boat turns around and moves upstream at the same speed $v$ (same engine, no current in this frame). It must cover the same distance $v\tau$ back to the raft, which takes exactly another $\tau$.

Total elapsed time: $2\tau$. During those $2\tau$ the raft has been carried downstream by the river at speed $u$, traveling a distance $l = u \cdot 2\tau$. Therefore

$\boxed{u = \frac{l}{2\tau} = \frac{6.0\,\text{km}}{2 \times 1\,\text{h}} = 3.0\,\text{km/h}}$

The quantity $v\tau$ — how far the boat got from the raft — appeared and disappeared. Phase 1 produced it; phase 2 consumed it exactly, because the speed and the target distance are the same. The engine speed has no route into the answer.

Verification in the lab frame

Let the current run in the $+x$ direction. The boat needs $v > u$ to make upstream headway after turning.

Phase 1 (downstream, duration $\tau$): boat velocity $= v + u$, raft velocity $= u$.

At $t = \tau$: boat is at $(v + u)\tau$, raft is at $u\tau$.

Phase 2 (upstream, duration $t_2$): boat velocity $= -(v - u) + 2u\tau$… rather than carry the algebra, set positions equal at $t = \tau + t_2$:

$\underbrace{(v+u)\tau - (v-u)t_2}_{\text{boat}} = \underbrace{u(\tau + t_2)}_{\text{raft}}$

Expand: $v\tau + u\tau - vt_2 + ut_2 = u\tau + ut_2$. The $u\tau$ and $ut_2$ terms cancel on both sides, leaving $v\tau = vt_2$, so $t_2 = \tau$.

The total time is again $2\tau$ and the raft displacement is $u \cdot 2\tau = l$. The algebra forced $v$ to cancel — the frame argument explains why it had to.

Interactive

The widget below fixes $u = 1.5$, $T = 1$, $l = 3.0$ and lets you vary $v$. The left panel shows the $x$-$t$ diagram from the river bank: the boat’s peak shifts up and down as $v$ changes, but the filled dot — where the boat meets the raft — stays fixed. The right panel shows the water frame: the raft is flat, the boat V grows and shrinks, and the V always closes back to zero at $t = 2T$.

The readout confirms: $u = l/2T = 1.5$, regardless of where you put the slider.

The river bank only ever measures two things: the time interval between meetings, and the raft’s displacement at the second meeting. The engine speed is invisible to any observer who stays on the bank — and the result shows it was never needed.