Irodov 1.1: The Raft Frame Cancels the Unknown – Amit & Deepali Sinha Foundation

Problem Irodov 1.1

A motorboat travels downstream and overtakes a raft drifting freely with the current at a point $A$ . A time $\tau = 60$ min after overtaking the raft, the boat turns back; some time later it meets the raft again, at a distance $l = 6.0$ km downstream of $A$ . The engine runs at constant duty throughout, so the boat’s speed through the water is the same constant value both going downstream and after it reverses. Find the speed of the current.

Position from A against time. The boat's track (upper curve) bends at t=τ, where it turns back; the raft's is the straight line through the origin. They meet again at M, a distance l from A. — Position from $A$ against time. The boat’s track (upper curve) bends at $t=\tau$ , where it turns back; the raft’s is the straight line through the origin. They meet again at $M$ , a distance $l$ from $A$ .

Given

$\tau = 60$ min
$l = 6.0$ km

Find

The flow (current) velocity

v

Solution

Step 1Move into the raft’s own frame

The raft drifts with the current, so a frame moving with the raft is a frame moving with the water itself: in it the water is at rest, and the boat’s speed relative to this frame is exactly its speed relative to the water. That speed is fixed by “the duty of the engine,” so it is the same constant $u$ whichever way the boat points, upstream or downstream.

Step 2The outward and return legs take equal time

At the instant the boat overtakes the raft, the two coincide, so this instant is $t=0$ in every frame. In the raft’s frame the boat moves away at the constant speed $u$ for a time $\tau$ , then reverses and returns at the same speed $u$ over the same distance it had opened up. Equal speed, equal distance, so the return leg also takes a time $\tau$ , and the boat is back at the raft (which sits fixed at the raft-frame origin) at raft-frame time

T = \tau + \tau = 2\tau. \tag{1}

Step 3Read the meeting time back in the ground frame

A Galilean change of frame does not change elapsed time, so the boat meets the raft again at ground time $T = 2\tau$ after the first encounter as well. In that same ground time the raft, drifting at the current’s speed $v$ , has moved a distance $l$ from $A$ , which is precisely how $l$ was defined in the statement:

l = v\cdot 2\tau

(2)

The same graph with the meeting time now read off: M sits at t=2τ, because the return leg took as long as the outward leg (Step 2). The raft's straight line then fixes l as the drift over 2τ, the relation boxed above. — The same graph with the meeting time now read off: $M$ sits at $t=2\tau$ , because the return leg took as long as the outward leg (Step 2). The raft’s straight line then fixes $l$ as the drift over $2\tau$ , the relation boxed above.

Step 4Solve for the current

Solving (2) for $v$ , then converting $\tau = 60$ min $=1.0$ h,

v = \frac{l}{2\tau} = \frac{6.0\text{ km}}{2\times 1.0\text{ h}}.

v = \dfrac{l}{2\tau} = 3.0\text{ km/h}

(3)

Check

The result carries no trace of the boat’s own speed $u$ , exactly as Step 4 argues: doubling or halving the engine’s power changes how far out the boat gets before turning back, but not the time $2\tau$ it takes to rejoin a raft that only ever drifts at $v$ . Units also check: km divided by h is km/h, a speed.

Remark

Any constant engine setting, not just ”constant duty” specifically, gives the same water-relative speed $u$ both downstream and upstream, so the outward and return legs (Step 2) take equal time by pure kinematic symmetry. That is exactly the Galilean invariance that makes the raft-frame shortcut work: shaft power itself is frame-independent, and the drag the engine fights depends on the boat’s speed through the water, which is the same symmetric quantity in either direction. Only a drag that depended on speed relative to the ground instead, from the river bed or the air, could break that symmetry and make the two legs take unequal time.