In the water's frame, the raft stands still, so the boat takes equal time $\tau$ to go out and $\tau$ to come back, meaning the river carries the raft a distance $l$ in time $2\tau$, giving flow speed $v_0 = \dfrac{l}{2\tau} = 3.0\text{ km/h}$ ($0.83\text{ m/s}$)[cite: 1].
(ii) Problem Statement
"A motorboat going downstream overcame a raft at a point $A$; $\tau = 60\text{ min}$ later it turned back and after some time passed the raft at a distance $l = 6.0\text{ km}$ from the point $A$. Find the flow velocity assuming the duty of the engine to be constant."
Reference Answer Verification (from attached s1.png):
Our first-principles derivation agrees 100% with the supplied reference image (s1.png)[cite: 1].
In the reference solution, the ground-frame time equation for the raft and boat to reach meeting point $C$ is written as[cite: 1]:
$$\frac{l}{v_0} = \tau + \frac{(v_0 + v')\tau - l}{v' - v_0}$$
Solving this algebraic equation yields $v_0 = \dfrac{l}{2\tau} = \dfrac{6.0\text{ km}}{2 \times 1.0\text{ h}} = 3.0\text{ km/h}$[cite: 1]. Both the algebraic ground-frame method and the water-frame symmetry method produce the exact same expression and numerical value.
(iii) The Big Idea in Plain Words
Imagine walking on a long moving walkway at an airport. A suitcase is sitting on the walkway next to you at Point $A$.
Here, $\tau$ (tau) is the time you walk forward away from the suitcase ($60\text{ minutes}$). Constant engine duty means the boat pushes through the water at the same speed $v'$ (boat speed in calm water) whether going forward or backward[cite: 1].
Because the water carries both the boat and the raft together, the water does not change the distance between them. If you walk away from the suitcase for $60\text{ minutes}$ at constant effort, it takes you the exact same $60\text{ minutes}$ to walk back to the suitcase.
So the total trip takes $60 + 60 = 120\text{ minutes}$ ($2\text{ hours}$). During those $2\text{ hours}$, the floating raft simply drifted with the river for $l = 6.0\text{ km}$. Moving $6.0\text{ km}$ in $2\text{ hours}$ means the river flows at $3.0\text{ km/h}$.
(iv) The Solution Viewed in Each Reference Frame
1. Medium Frame (Water / Raft Frame) — Simplest View
Origin moves with water at $v_0$Why: We sit on the raft so the water and raft look completely still.
Let $v_0$ be the river flow speed and $v'$ be the boat speed relative to the water[cite: 1]. In the water frame, the raft stays at position $x_{\text{rel}} = 0$ for the whole trip.
Why: We find how far the boat travels away from the raft in time $\tau$.
The boat moves away from the raft at speed $v'$ for time $\tau = 1.0\text{ h}$. It reaches a separation distance $d_{\text{rel}} = v'\tau$.
Why: The engine works just as hard coming back, so the return speed in water is also $v'$.
The time $\tau_{\text{back}}$ to return the same distance $d_{\text{rel}}$ at the same speed $v'$ is: $$\tau_{\text{back}} = \frac{d_{\text{rel}}}{v'} = \frac{v'\tau}{v'} = \tau$$
Why: Now we use the total time $t_{\text{total}} = \tau + \tau = 2\tau$ to find how far the bank moved.
Relative to the water, Point $A$ on the riverbank drifts backward at speed $v_0$. In total time $2\tau$, Point $A$ is a distance $l = 6.0\text{ km}$ behind the raft: $$v_0 = \frac{l}{2\tau} = \frac{6.0\text{ km}}{2 \times 1.0\text{ h}} = 3.0\text{ km/h}$$
2. Ground or Lab Frame (Riverbank Observer)
Origin fixed at Point $A$ on bankWhy: A person standing on the riverbank adds the river speed downstream and subtracts it upstream.
Going downstream from $A$ to turnaround point $B$, the boat's ground speed is $v_{\text{down}} = v' + v_0$[cite: 1]. In time $\tau$, it reaches position $x_B$: $$x_B = (v' + v_0)\tau$$
Why: After turning at $B$, the boat fights the current to return upstream to meeting point $C$.
Going upstream, the boat moves toward $A$ with ground speed $v_{\text{up}} = v' - v_0$[cite: 1]. Since meeting point $C$ is at $x_C = l$ from $A$, the distance from $B$ back to $C$ is $x_B - l = (v' + v_0)\tau - l$[cite: 1].
Why: The total time taken by the raft to float from $A$ to $C$ equals the boat's downstream time plus upstream time[cite: 1].
The raft floats at speed $v_0$, taking total time $t = \dfrac{l}{v_0}$[cite: 1]. Setting the raft's time equal to the boat's total time (just as in s1.png)[cite: 1]:
$$\frac{l}{v_0} = \tau + \frac{(v_0 + v')\tau - l}{v' - v_0}$$
Why: We clear the fraction to cancel the unknown engine speed $v'$.
Combine the right-hand side over the common denominator $(v' - v_0)$: $$\frac{l}{v_0} = \frac{\tau(v' - v_0) + (v_0 + v')\tau - l}{v' - v_0} = \frac{2v'\tau - l}{v' - v_0}$$ Cross-multiply: $$l(v' - v_0) = v_0(2v'\tau - l) \implies lv' - lv_0 = 2v_0v'\tau - lv_0$$ Add $lv_0$ to both sides and divide by $v'$: $$lv' = 2v_0v'\tau \implies v_0 = \frac{l}{2\tau} = \frac{6.0\text{ km}}{2 \times 1.0\text{ h}} = 3.0\text{ km/h}$$
3. Centre-of-Mass (CM) Frame
This frame does not simplify this purely kinematic problem because the masses of the boat and raft are not given and external water drag acts when the boat turns; even if equal masses are assumed so the CM moves at $v_0 \pm \frac{1}{2}v'$, the separation and return still take equal times $\tau$, giving the exact same answer $v_0 = \dfrac{l}{2\tau} = 3.0\text{ km/h}$.
What Changes and What Stays the Same Across Frames
- What changes: The measured velocities of the boat and raft, the turnaround coordinate $x_B$, and the path length the boat travels.
- What stays the same (Invariants): Time intervals ($\tau = 60\text{ min}$ outbound and $\tau = 60\text{ min}$ inbound), the coincidence events (passing at $A$ and meeting again at $C$), and the final river flow speed $v_0 = 3.0\text{ km/h}$.
(v) The Solution by Each Method
Method 1: Relative-Motion Method
Fastest & Most Reliable-
Why: Subtract the river's velocity $v_0$ from both objects to work relative to the raft.
Relative to the raft, the boat moves downstream at speed $v'$ for time $\tau$, reaching relative distance $d = v'\tau$. -
Why: Constant engine power means the boat's relative return speed is also $v'$, so return time equals outbound time.
Return time is $\tau_{\text{return}} = \dfrac{v'\tau}{v'} = \tau$, making total trip time $t_{\text{total}} = 2\tau$. Since the raft drifts $l = 6.0\text{ km}$ at speed $v_0$ during $2\tau$, we get $v_0 = \dfrac{l}{2\tau} = 3.0\text{ km/h}$[cite: 1].
Method 2: Graph Method ($x$-$t$ & $v$-$t$ Diagrams)
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Why: On a relative velocity-time ($v_{\text{rel}}$ vs $t$) graph, the area under the graph equals the relative displacement between boat and raft.
From $t = 0$ to $t = \tau$, the graph is a rectangle of height $+v'$ and width $\tau$, giving positive area $+v'\tau$. -
Why: For the boat and raft to meet again, the negative area after turning must cancel the positive area.
After $t = \tau$, the relative velocity is $-v'$. To make the negative rectangle have area $-v'\tau$, its width must also be $\tau$, so the meeting happens at $t = 2\tau$. -
Why: On the ground position-time ($x$-$t$) graph, the slope of the raft's line is the flow speed $v_0$.
The raft's line passes through $(0,0)$ and $(2\tau, l)$. Its slope is $v_0 = \dfrac{l - 0}{2\tau - 0} = \dfrac{6.0\text{ km}}{2.0\text{ h}} = 3.0\text{ km/h}$.
Method 3: Calculus Method (Differential Equation & Integration)
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Why: Write the rate of change of position $\dfrac{dx}{dt}$ for both the boat and the raft.
For the raft: $\dfrac{dx_R}{dt} = v_0$. For the boat: $\dfrac{dx_B}{dt} = v_0 + v'$ for $0 \le t \le \tau$, and $\dfrac{dx_B}{dt} = v_0 - v'$ for $\tau < t \le t_m$, where $t_m$ is the meeting time. -
Why: Integrate both velocity equations from $t = 0$ to the meeting time $t_m$ where $x_B(t_m) = x_R(t_m) = l$.
$$x_R(t_m) = \int_0^{t_m} v_0\,dt = v_0 t_m = l$$ $$x_B(t_m) = \int_0^{\tau} (v_0 + v')\,dt + \int_{\tau}^{t_m} (v_0 - v')\,dt = v_0 t_m + v'\tau - v'(t_m - \tau)$$ -
Why: Set $x_B(t_m) = x_R(t_m)$ to solve for $t_m$ and then $v_0$.
$$v_0 t_m + v'(2\tau - t_m) = v_0 t_m \implies v'(2\tau - t_m) = 0 \implies t_m = 2\tau$$ Substitute $t_m = 2\tau$ into $v_0 t_m = l$: $$v_0 = \frac{l}{2\tau} = \frac{6.0\text{ km}}{2.0\text{ h}} = 3.0\text{ km/h}$$
Method 4 & 5: Force Method & Energy Method
Force Method: Because the boat moves at constant velocity in each leg, engine thrust $\vec{F}_{\text{thrust}}$ exactly balances water drag $\vec{F}_{\text{drag}}$ ($\sum F_x = 0, a_x = 0$), so Newton's second law reduces immediately to constant relative speed $v'$ and yields $v_0 = \dfrac{l}{2\tau} = 3.0\text{ km/h}$.
Energy Method: An energy conservation method does not suit this problem because the engine continuously supplies mechanical power that is dissipated by non-conservative water drag to maintain constant speed $v'$, giving the same kinematic result $v_0 = \dfrac{l}{2\tau} = 3.0\text{ km/h}$.
Exam Verdict: The Relative-Motion Method (Method 1) is both the fastest (2 lines of algebra) and the most reliable in an exam because it avoids algebraic fractions where sign errors happen.
(vi) Parameter Table
| Symbol | Plain Meaning | Given / Standard Value | SI Value |
|---|---|---|---|
| $\tau$ | Time boat travels downstream after passing raft at $A$ | $60\text{ min} = 1.0\text{ h}$[cite: 1] | $3600\text{ s}$ |
| $l$ | Distance from Point $A$ to second meeting point $C$ | $6.0\text{ km}$[cite: 1] | $6000\text{ m}$ |
| $v'$ | Boat speed relative to water (cancels out of final answer) | Constant ($> v_0$)[cite: 1] | Any constant |
| $2\tau$ | Total time between first pass ($A$) and second pass ($C$) | $120\text{ min} = 2.0\text{ h}$ | $7200\text{ s}$ |
| $v_0$ | River flow velocity (and raft speed) | $3.0\text{ km/h}$[cite: 1] | $0.833\text{ m/s}$ |
(vii) Common Exam Traps
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1.
Forgetting the Return Time ($\div \tau$ instead of $\div 2\tau$): Many students write $v_0 = \dfrac{l}{\tau} = 6.0\text{ km/h}$ by assuming the whole trip took $\tau = 60\text{ min}$. Remember $\tau$ is only the outbound leg; the return leg takes another $\tau$, making the total drift time $2\tau$.
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2.
Thinking You Are Missing the Boat's Speed $v'$: Students often panic because the problem does not give the motorboat's speed $v'$. As you can verify with the $v'$ slider above, $v'$ cancels out completely!
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3.
Upstream Velocity Sign Mix-up: In the ground frame, when the boat goes upstream toward $A$, its speed toward $A$ is $(v' - v_0)$[cite: 1], not $(v' + v_0)$. Mixing up signs in the ground-frame fraction makes $v'$ fail to cancel.