Visual walkthrough — Relative motion — 1D and 2D; river-boat problems

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Step 1 — What is a velocity arrow?

WHAT. A velocity is drawn as an arrow. Its length = how fast (metres per second, written m/s), its direction = which way the motion goes.

WHY. Before we combine two motions, we must agree on how to draw one motion. An arrow is the honest picture: it holds a number (length) and a direction at the same time — which is exactly what velocity is.

PICTURE. In the figure below, one arrow starts at a dot (the boat) and points where the boat heads. We also set up our map: $x$ points across the river (from the near bank to the far bank), $y$ points downstream (the way the water flows). The river has width $d$ — the straight distance between the two banks.

The little arrow on top ($\stackrel{⃗}{}$) just means "this is a vector — it has a direction, not only a size."

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Step 2 — Why the boat's ground velocity is a SUM

WHAT. The boat's ground velocity is the two arrows laid tip-to-tail: $\underset{\text{what the bank sees}}{\underbrace{{\vec{v}}_{bg}}}=\underset{\text{boat through water}}{\underbrace{{\vec{v}}_{bw}}}+\underset{\text{water over ground}}{\underbrace{{\vec{v}}_w}}$

WHY this and not subtraction? Read the subscripts as a chain: $bw+wg=bg$. The inner letter $w$ appears on the right of the first and the left of the second — it cancels, leaving $bg$. This is the chain rule of frames from the parent note. Physically: the boat swims through the water, and the whole sheet of water slides over the ground — so the two motions genuinely stack.

PICTURE. Put the tail of ${\vec{v}}_w$ at the tip of ${\vec{v}}_{bw}$. The arrow from the very start to the very end is ${\vec{v}}_{bg}$. This is vector addition — tip-to-tail.

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Step 3 — Split every arrow into across + downstream

WHAT. Break each velocity into two independent pieces: an $x$-piece (across) and a $y$-piece (downstream). For an arrow of length $v$ tilted by angle $\theta$ from the $x$-axis: $v_x=v\cos \theta ,v_y=v\sin \theta$

WHY split? Because crossing the river only cares about the $x$-piece. The far bank is a fixed distance $d$ away across; how fast you close that gap is your across-speed alone. The downstream piece slides you sideways but never brings you closer to the far bank. Splitting lets us treat the two directions as two separate 1D problems — the same trick used in projectile motion.

$\cos$ and $\sin$ are the tools that read off "how much of this arrow points along $x$" and "how much along $y$." $\cos \theta$ is the adjacent-over-hypotenuse fraction (the across share); $\sin \theta$ is opposite-over-hypotenuse (the downstream share).

PICTURE. The tilted boat arrow drops a shadow onto the $x$-axis (its across part) and onto the $y$-axis (its downstream part). Those two shadows are the right-triangle legs of the arrow.

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Step 4 — Case A: aim STRAIGHT ACROSS (fastest crossing)

WHAT. Point ${\vec{v}}_{bw}$ entirely along $+x$ (angle $0$ from the across-direction). Then the boat's whole engine speed is across: $v_{bg,x}=v_{bw},v_{bg,y}=v_w$ Crossing time uses only the across-speed: $t_{\min }=\frac{d}{v_{bw}}$

Term by term: $d$ = river width (the distance to cover across); $v_{bw}$ = the full boat speed, now 100% aimed across; their ratio is time = distance ÷ speed.

WHY this is the minimum. The current only adds a $y$-velocity — it is perpendicular to the crossing. Perpendicular motion cannot speed up or slow down how fast you close the $x$-gap. So spending 100% of the engine on the across-direction is the fastest possible crossing. Nothing you do can beat $d/v_{bw}$.

The cost — drift. The drift is a downstream displacement, so it lives on the $y$-axis — call it $y_{\text{drift}}$. While you cross for time $t_{\min }$, the current carries you downstream: $y_{\text{drift}}=v_w{t}_{\min }=\frac{v_{w}d}{v_{bw}}$ Here $v_w$ = current speed, $t_{\min }$ = the crossing time from above; multiply to get the downstream ($y$) distance dragged. (It is along $y$, not $x$ — the $x$-axis was defined as across, and crossing is finished the instant you reach the far bank.)

PICTURE. The boat arrow points dead across; the current arrow adds a downstream nudge; the actual path over the ground is the slanted resultant, landing you downstream of straight-across.

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Step 5 — Case B: aim UPSTREAM to land directly opposite

WHAT. Tilt ${\vec{v}}_{bw}$ upstream by an angle $\theta$ (measured from the straight-across direction) so its upstream piece exactly cancels the current. "Zero drift" means the total downstream velocity is $0$: $\underset{\text{upstream piece of boat}}{\underbrace{v_{bw}\sin \theta }}=\underset{\text{current}}{\underbrace{v_w}}\Rightarrow \boxed{\sin \theta =\frac{v_w}{v_{bw}}}$

Term by term: $v_{bw}\sin \theta$ is the slice of the boat's velocity pointing against the flow; setting it equal to $v_w$ makes the two cancel so the net downstream velocity is zero. Solving for $\theta$ gives $\sin \theta =v_w/v_{bw}$.

WHY only the $\sin$ piece fights the current. The current runs purely along $y$. Only the boat's $y$-piece can oppose it, and that piece is $v_{bw}\sin \theta$. The $x$-piece ($v_{bw}\cos \theta$) does the crossing and never touches the fight.

The leftover across-speed. After giving up the $\sin$ slice to fighting the current, what's left crosses the river: $v_{\text{across}}=v_{bw}\cos \theta =\sqrt{v_{bw}^2-v_w^2},t=\frac{d}{\sqrt{v_{bw}^2-v_w^2}}$ The square root comes straight from the right triangle: the boat arrow (hypotenuse $v_{bw}$) has an upstream leg $v_w$, so the across leg is $\sqrt{v_{bw}^2-v_w^2}$ by Pythagoras.

PICTURE. The boat arrow leans upstream; its downstream shadow exactly matches the current arrow and they annihilate; the only surviving motion is straight across — the ground path is a clean perpendicular line to the far bank.

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Step 6 — Two edge cases: exact tie, and when the current wins

WHAT. The "land opposite" recipe needs $\sin \theta =v_w/v_{bw}$. But $\sin$ of a real angle can never exceed $1$. Three regimes appear as $v_w$ grows toward $v_{bw}$:

$\underset{\text{lands opposite, finite time}}{\underbrace{v_{bw}>v_w}}\underset{\text{tie: }\theta =90^{\circ },\text{never crosses}}{\underbrace{v_{bw}=v_w}}\underset{\text{no real }\theta ,\text{impossible}}{\underbrace{v_{bw}<v_w}}$

The exact tie $v_w=v_{bw}$ (treat it on its own). Here $\sin \theta =1$, so $\theta =90^{\circ }$ — you aim fully upstream. That leaves the across-speed $v_{\text{across}}=v_{bw}\cos 90^{\circ }=0$. A real angle exists, but with zero across-speed the crossing time $t=\frac{d}{\sqrt{v_{bw}^2-v_w^2}}=\frac{d}{0}\to \infty$ blows up: you point perfectly upstream, exactly hold your ground against the current, and never actually reach the far bank. So "a steering angle exists for $v_{bw}\ge v_w$" is technically true but misleading — at equality you can hold position opposite the start but not cross. Genuine crossing to the opposite point needs the strict inequality $v_{bw}>v_w$.

When the current wins, $v_w>v_{bw}$. The most upstream push the boat can ever give is its whole speed $v_{bw}$ (aim fully upstream, $\theta =90^{\circ }$). If the current $v_w$ is bigger than that, even a full-upstream boat still slips downstream. No real angle solves $\sin \theta =v_w/v_{bw}>1$ — you cannot reach the point directly opposite at all.

PICTURE. Two panels side by side: on the left ($v_{bw}>v_w$) the boat's upstream reach covers the current and a real triangle closes. On the right ($v_{bw}<v_w$) the boat arrow is too short to reach across the current arrow — the triangle can't close, so no valid heading exists. The tie sits exactly between: the boat arrow becomes vertical (fully upstream), the across-leg shrinks to zero.

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Step 7 — Put numbers on it

Recall Quick self-test

Fastest crossing time here? ::: $20$ s (aim straight across) Downstream ($y$) drift in that fastest crossing? ::: $60$ m Angle to land directly opposite? ::: $37^{\circ }$ upstream from straight-across Zero-drift crossing time? ::: $25$ s Could a boat with $v_{bw}=2$ m/s land opposite here? ::: No — $2<3$, so $\sin \theta =3/2>1$, impossible What if $v_{bw}=v_w=3$ exactly? ::: You can hold position opposite the start, but across-speed is $0$ → infinite time, so you never actually cross

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The one-picture summary

Everything at once: one boat arrow of fixed length $v_{bw}$, free to swing. Aim it straight → fastest crossing, but you drift downstream (along $y$). Aim it upstream → the downstream shadow cancels the current, so you land opposite, but the shorter across-leg means a slower crossing. Push $v_w$ up to $v_{bw}$ → the across-leg vanishes (infinite time); beyond that the triangle can't close at all.

Recall Feynman retelling — say it to a friend

Picture swimming on a moving carpet. Your arms give you a fixed speed relative to the carpet — that's your one arrow, and you get to point it wherever you like. The carpet itself always slides you downstream at a fixed rate. Point straight at the far wall: you cross in the least possible time, because all your effort goes into crossing — but the carpet drops you off downstream. Point a bit against the carpet's slide: part of your effort now cancels the carpet's push, so you land exactly across — but there's less effort left for crossing, so it takes longer. If the carpet slides exactly as fast as you swim: pointing fully against it, you hold your spot but make zero headway across — you never reach the far wall. If the carpet slides faster than you can swim: even pointing fully against it, you still lose ground — you can never land straight across. Same picture, four answers, one adjustable arrow.

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Connections

• Relative motion — 1D and 2D; river-boat problems — the parent this page visualises.
• Vectors — addition, components, unit vectors — Steps 2–3 are pure vector addition and components.
• Kinematics in 2D — projectile motion — same "independent perpendicular components" idea.
• Frames of reference & Galilean transformation — why the boat/water/ground velocities add.
• Rain-man umbrella problem — the twin of this problem with rain instead of a current.