Relative motion — 1D and 2D; river-boat problems: Worked examples

Before anything, name the three players first, then the two rules we lean on the whole page (built in the parent note).

The scenario matrix

Every problem in this topic lands in exactly one of these cells. The last column names the worked example that covers it.

#	Cell (case class)	The tricky bit	Covered by
C1	1D, both positive (same direction)	closing/opening speed = difference	Ex 1
C2	1D, opposite signs (toward/away)	subtracting a negative ⇒ speeds add	Ex 2
C3	2D min-time crossing	current is perpendicular ⇒ ignore it for time	Ex 3
C4	2D zero-drift crossing (land opposite)	aim upstream, $sin θ = v_{w} / v_{b w}$	Ex 4
C5	Degenerate: current ≥ boat speed	$sin θ \geq 1$ ⇒ cannot land opposite	Ex 5
C6	Zero current ( $v_{w} = 0$ )	both routes collapse into one	Ex 6
C7	General aim angle (neither route)	full vector add, drift + time from one heading	Ex 7
C8	Real-world word twist (rain–man cousin)	switch to the moving frame to "see" a wind/rain	Ex 8
C9	Exam twist: minimum drift when $v_{w} > v_{b w}$	can't reach opposite, but minimize drift	Ex 9

We will cover C1–C9. Cells C3, C4, C7, C9 get figures because their geometry is the whole point.

Ex 1 — C1: 1D, same direction

Worked example Two trains on parallel tracks

Train A runs at $+ 30$ m/s, Train B at $+ 18$ m/s, both toward the right. How fast does a passenger in B see A move?

Forecast: guess before reading — is it $48$ , $12$ , or something else, and in which direction?

Step 1 — Pick a sign convention. Right $= +$ . Both velocities are already signed: $v_{A} = + 30$ , $v_{B} = + 18$ . Why this step? Relative motion in 1D is just signed arithmetic; without a fixed "positive direction" the signs are meaningless.

Step 2 — Apply "subtract the observer". The observer is B, so $v_{A B} = v_{A} - v_{B} = 30 - 18 = + 12 m/s .$ Why this step? The rule $v_{A B} = v_{A} - v_{B}$ removes B's own motion — we ask "what's left of A once we stop moving with B?"

Step 3 — Read the sign. $+ 12$ means A still drifts to the right as seen from B, slowly. Why this step? A sign in 1D is a direction; a positive answer means "same way as our +axis".

Verify: From B's window, A pulls ahead by $12$ m every second — matches the everyday feeling that a slightly-faster car creeps forward. Units: m/s − m/s = m/s ✓.

Ex 2 — C2: 1D, opposite directions (subtracting a negative)

Worked example Head-on approach

Car A: $+ 25$ m/s. Car B drives toward A at $20$ m/s, so $v_{B} = - 20$ m/s. Find $v_{A B}$ (how fast B sees A coming).

Forecast: will subtracting change the $20$ into a $+ 20$ or a $- 20$ inside the formula?

Step 1 — Sign the velocities. Right $= +$ . A moves right: $v_{A} = + 25$ . B moves left toward A: $v_{B} = - 20$ . Why this step? "Toward A" must be turned into a signed number before we subtract, or the direction is lost.

Step 2 — Subtract the observer B. $v_{A B} = v_{A} - v_{B} = 25 - (- 20) = 25 + 20 = + 45 m/s .$ Why this step? Subtracting a negative adds — this is exactly why closing speeds are the sum of speeds for objects moving toward each other.

Step 3 — Interpret. B sees A rushing at $45$ m/s. The mirror check: $v_{B A} = v_{B} - v_{A} = - 20 - 25 = - 45$ , and indeed $v_{A B} = - v_{B A}$ .

Verify: If they start $450$ m apart they meet in $450/45 = 10$ s — sensible. Sign/units consistent ✓.

Ex 3 — C3: minimum-time crossing (figure)

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

Worked example Fastest crossing

River width $d = 120$ m. Boat-through-water speed $v_{b w} = 6$ m/s. Current $v_{w} = 4$ m/s. Point the boat straight across. Find crossing time and downstream drift.

Forecast: does the $4$ m/s current make the crossing take longer than the no-current time $120/6 = 20$ s, or exactly the same?

Step 1 — Use the fixed axes. As defined above, $x$ = across, $y$ = downstream (look at the figure). "Straight across" is $θ = 0$ , so the boat's engine points fully along $+ x$ . Why these axes? Splitting velocity into "across" and "downstream" makes the two motions independent — same trick as projectile motion in Kinematics in 2D — projectile motion.

Step 2 — Crossing uses only the across-component. Aiming straight across puts the entire engine speed on $x$ : $v_{x} = v_{b w} cos 0 = v_{b w} = 6$ m/s. The current has no $x$ -part (it's along $y$ ). $t = \frac{d}{v _{x}} = \frac{120}{6} = 20 s .$ Why this step? The orange $x$ -velocity is what shortens the across-distance; the current (magenta, along $y$ ) is perpendicular and cannot touch the crossing — see how the magenta arrow in the figure is sideways to the path.

Step 3 — Drift. The current pushes downstream for the whole $20$ s: $x_{drift} = v_{w} t = 4 \times 20 = 80 m .$ Why this step? $y$ -motion is uniform at $v_{w}$ ; distance = speed × time.

Verify: Crossing time equals the current-free time $120/6 = 20$ s — confirming current never affects min-time crossing. Ground speed $= 6^{2} + 4^{2} = 52 \approx 7.21$ m/s, angled downstream, but crossing still 20 s ✓.

Ex 4 — C4: zero-drift crossing, land opposite (figure)

Worked example Land directly opposite

Same river: $d = 120$ m, $v_{b w} = 6$ m/s, $v_{w} = 4$ m/s. Now aim upstream so the boat lands straight across (no drift). Find the heading angle and the crossing time.

Forecast: will this crossing be quicker or slower than the 20 s of Ex 3?

Step 1 — Demand zero downstream velocity. Using the heading angle $θ$ (upstream from straight-across, as defined above), the upstream component of the boat must exactly cancel the current: $v_{b w} sin θ = v_{w} .$ Why this step? If any $y$ -velocity survives, the boat drifts. Only the boat's upstream part $v_{b w} sin θ$ can fight the current, so we set it equal.

Step 2 — Solve for the angle. $sin θ = \frac{v _{w}}{v _{b w}} = \frac{4}{6} = 0.6667 \Rightarrow θ \approx 41. 8^{\circ} upstream .$ Why this step? $sin$ turns the angle-of-aim into the fraction of engine speed spent going upstream; $arcsin$ answers "which angle gives that fraction?"

Step 3 — Across-speed is what's left. The remaining component crosses the river: $v_{across} = v_{b w} cos θ = v_{b w}^{2} - v_{w}^{2} = 36 - 16 = 20 \approx 4.472 m/s .$ Why $v_{b w}^{2} - v_{w}^{2}$ ? The engine vector, its upstream part $v_{w}$ , and its across part form a right triangle (figure) — Pythagoras gives the across leg.

Step 4 — Crossing time. $t = \frac{d}{v _{across}} = \frac{120}{20} \approx 26.83 s .$ Why this step? Only the across-component $v_{across}$ shrinks the river's width $d$ ; time = distance ÷ that across-speed, exactly as in Ex 3 but with the reduced (cosine-cut) speed.

Verify: $26.83 > 20$ — zero drift costs time, exactly as the parent's trade-off predicts. And the ground velocity here is purely across ( $4.472$ m/s, $0$ downstream) ✓.

Ex 5 — C5: degenerate case, current beats the boat

Worked example Can't win

A weak swimmer: $v_{b w} = 2$ m/s. Fast current $v_{w} = 3$ m/s, river width $d = 50$ m. Can they land directly opposite? If not, what happens?

Forecast: is there any angle that gives zero drift?

Step 1 — Try the zero-drift condition. With $θ$ measured upstream from straight-across as always, $sin θ = \frac{v _{w}}{v _{b w}} = \frac{3}{2} = 1.5.$ Why this step? This is the only condition that removes drift; if it can't be met, the goal is impossible.

Step 2 — Check feasibility. $sin θ$ can never exceed $1$ (the sine of any real angle lives in $[- 1, 1]$ ). Since $1.5 > 1$ , no real angle $θ$ exists. Why this step? The largest upstream push the swimmer can produce is their full speed $v_{b w} = 2$ m/s (aiming dead upstream, $θ = 9 0^{\circ}$ ), still short of the $3$ m/s current.

Step 3 — Physical consequence. Whatever the heading, some downstream velocity always survives → the swimmer is always carried downstream. They can still reach the far bank, just never at the point opposite. Why this step? We must translate the "no real $θ$ " algebra into what actually happens in the water — an impossible equation means the target is unreachable, not that motion stops. Naming the physical outcome is what makes the maths trustworthy.

The boundary $v_{w} = v_{b w}$ (exactly tied). If current equals the swimmer speed, then $sin θ = v_{w} / v_{b w} = 1$ , so $θ = 9 0^{\circ}$ : the swimmer must aim dead upstream. Then the across-component is $v_{b w} cos 9 0^{\circ} = 0$ — the swimmer makes zero progress across and the crossing time $t = d /0$ is infinite. So $v_{w} = v_{b w}$ is the knife-edge: at the tie you can just barely stop the drift but you never actually cross; you need $v_{b w} > v_{w}$ strictly to reach the opposite point in finite time.

Verify: Condition to reach opposite is $v_{b w} > v_{w}$ (strict); here $2 < 3$ fails, and even the tie $v_{w} = v_{b w}$ gives $v_{across} = 0$ ✓. (Ex 9 shows what the swimmer should do instead.)

Ex 6 — C6: zero current (degenerate the other way)

Worked example Still water

Lake, no current: $v_{w} = 0$ . Boat $v_{b w} = 5$ m/s, width $d = 80$ m. Compare min-time and zero-drift routes.

Forecast: should the two routes give different answers when there's no current?

Step 1 — Min-time route. Aim straight across ( $θ = 0$ ): $t = d / v_{b w} = 80/5 = 16$ s, drift $= v_{w} t = 0 \times 16 = 0$ . Why this step? Straight-across always minimizes time; with no current the drift term vanishes on its own.

Step 2 — Zero-drift route. $sin θ = v_{w} / v_{b w} = 0/5 = 0 \Rightarrow θ = 0^{\circ}$ (aim straight across). Then $t = d / v_{b w}^{2} - v_{w}^{2} = 80/ 25 - 0 = 80/5 = 16$ s. Why this step? With $v_{w} = 0$ the two conditions demand the same heading — the routes fuse.

Step 3 — Conclusion. No current ⇒ min-time and zero-drift are the identical straight-across trip, $16$ s, zero drift. Why this step? Checking that both formulas collapse to the same answer at $v_{w} = 0$ is our sanity limit: any correct river formula must reduce to the ordinary "no-current" crossing when the current is switched off.

Verify: Both formulas return $16$ s; drift $0$ ✓. This is the sensible limit of every earlier formula as $v_{w} \to 0$ .

Ex 7 — C7: general heading (neither special route) (figure)

Worked example Arbitrary aim

A boat with $v_{b w} = 8$ m/s aims at $θ = 3 0^{\circ}$ upstream from straight-across. Current $v_{w} = 3$ m/s, width $d = 90$ m. Find the crossing time, the downstream drift, and the ground speed.

Forecast: will there be drift, and will it be upstream or downstream?

Step 1 — Recall the axes and split the engine velocity. As fixed above, $x$ = across, $y$ = downstream ( $+$ ), and $θ$ is measured upstream from the $+ x$ (straight-across) direction. The figure shows this heading. Split the engine velocity into components: $v_{x} = v_{b w} cos θ = 8 cos 3 0^{\circ} = 8 (0.8660) = 6.928 m/s (across, + x),$ $v_{b y} = - v_{b w} sin θ = - 8 sin 3 0^{\circ} = - 8 (0.5) = - 4 m/s (upstream, so - y) .$ Why this step? Re-stating the axes keeps every sign honest; components let us handle across and downstream separately, and the minus marks the aim being against the current. In the figure the orange arrow is this engine vector, tilted upstream.

Step 2 — Add the current to get ground velocity in $y$ . The magenta arrow in the figure is the current, added tip-to-tail: $v_{g y} = v_{b y} + v_{w} = - 4 + 3 = - 1 m/s (net upstream) .$ Why this step? Chain rule $v_{b g} = v_{b w} + v_{w}$ acts component-wise; the current only touches $y$ . The violet resultant in the figure is $v_{b g}$ .

Step 3 — Crossing time (uses $v_{x}$ only). $t = \frac{d}{v _{x}} = \frac{90}{6.928} \approx 12.99 s .$ Why this step? Crossing is an $x$ -only affair; the aim's across-component $v_{x}$ (unchanged by the current) is what eats the width $d$ , so time = width ÷ $v_{x}$ .

Step 4 — Drift. Signed drift $= v_{g y} t = (- 1) (12.99) \approx - 12.99$ m — i.e. about $13$ m upstream of the start. Why this step? Drift is how far the net downstream velocity carries the boat during the crossing time; since $v_{g y}$ is negative (net upstream), the boat lands upstream of opposite — matching the violet resultant in the figure tilting slightly upstream. The aim over-fought the current.

Step 5 — Ground speed. $∣ v_{b g} ∣ = v_{x}^{2} + v_{g y}^{2} = 6.92 8^{2} + 1^{2} = 48 + 1 = 49 = 7 m/s .$ Why this step? The two perpendicular ground-velocity components ( $v_{x}$ across, $v_{g y}$ along-stream) form a right triangle; Pythagoras gives the true speed the bank observer measures — the length of the violet arrow in the figure.

Verify: Negative drift confirms upstream landing (the aim beat the current). $49 = 7$ exactly ✓. Uses full vector addition — see Vectors — addition, components, unit vectors.

Ex 8 — C8: real-world word twist (rain-man cousin)

Worked example Wind on a cyclist

A cyclist rides due north at $6$ m/s. To her, the wind seems to blow from the north-west, straight at $4 5^{\circ}$ into her face, at $62$ m/s. What is the wind's true velocity over the ground?

Forecast: is the real wind blowing faster or slower than what she feels?

Step 1 — Name the frames. $c$ = cyclist, $g$ = ground, $W$ = wind. She measures $v_{W c}$ (wind relative to her). We want $v_{W g}$ . Why this step? "The wind she feels" is a relative velocity; the flag on a pole shows the ground wind. They differ by her motion — exactly the Rain-man umbrella problem setup.

Step 2 — Use the chain rule. $v_{W g} = v_{W c} + v_{c g} .$ Why this step? Inner index $c$ cancels, giving wind-over-ground: the wind she feels plus her own motion.

Step 3 — Choose axes and put in components. Let east $= + x$ , north $= + y$ . "Wind from the north-west" means it blows toward the south-east, i.e. at $- 4 5^{\circ}$ measured from the $+ x$ (east) axis. Its components are $v_{W c} = 62 (cos (- 4 5^{\circ}), sin (- 4 5^{\circ})) = 62 (\frac{1}{2}, - \frac{1}{2}) = (+ 6, - 6) m/s,$ i.e. $+ 6$ east, $- 6$ south (into her face and pushing her back). Her own velocity is due north: $v_{c g} = (0, + 6)$ m/s. Why this step? Turning the words "from the north-west at $4 5^{\circ}$ " into signed components is the only way to add the two velocities cleanly.

Step 4 — Add the vectors. $v_{W g} = v_{W c} + v_{c g} = (6, - 6) + (0, 6) = (6, 0) m/s .$ Why this step? Adding her northward $+ 6$ cancels the felt southward $- 6$ , leaving a purely eastward vector — her forward motion was manufacturing the "head-on" feeling.

Step 5 — Interpret and get the magnitude. The true wind blows due east at $∣ v_{W g} ∣ = 6^{2} + 0^{2} = 6 m/s .$ Why this step? We must convert the component answer $(6, 0)$ back into plain language — a due-east wind of $6$ m/s — because that is what "the wind's true velocity" asked for.

Verify: The true wind ( $6$ m/s) is slower than the felt $62 \approx 8.49$ m/s ✓ — motion always inflates the wind you feel, matching everyday cycling experience.

Ex 9 — C9: exam twist — minimize drift when you can't land opposite (figure)

Worked example Best you can do against a strong current

Back to Ex 5's swimmer: $v_{b w} = 2$ m/s, $v_{w} = 3$ m/s, width $d = 60$ m. Landing opposite is impossible. What heading gives the least drift, and how much drift is that?

Forecast: should the swimmer aim as far upstream as possible, or something in between?

Step 1 — Set up drift as a function of heading. Aim at angle $θ$ upstream from straight-across (the same baseline as always). Across-speed $= v_{b w} cos θ$ , net downstream speed $= v_{w} - v_{b w} sin θ$ . Crossing time $t = d / (v_{b w} cos θ)$ , so $drift (θ) = (v_{w} - v_{b w} sin θ) \cdot \frac{d}{v _{b w} c o s θ} .$ Why this step? We can't force drift to zero, so we treat it as a quantity to minimize over the one free choice we have — the heading $θ$ .

Step 2 — Minimize over $θ$ . Differentiating the drift expression and setting it to zero (or using the standard result) gives the drift smallest exactly when $cos θ = \frac{v _{b w}}{v _{w}} = \frac{2}{3} = 0.6667 \Rightarrow θ \approx 48.1 9^{\circ} upstream .$ Why this step? This heading is the sweet spot that balances two opposing effects: aiming more upstream fights the current harder but leaves less across-speed (so more time for the leftover current to push you). The minimum sits where those trade off. Note it's $cos$ , not $sin$ — different from the zero-drift rule of Ex 4.

Step 3 — Compute the minimum drift. With $cos θ = 2/3$ we get $sin θ = 1 - (2/3)^{2} = 5 /3 \approx 0.7454$ . Substituting back: $drift_{m i n} = \frac{d v _{w}^{2} - v _{b w}^{2}}{v _{b w}} = \frac{60 9 - 4}{2} = \frac{60 5}{2} = 305 \approx 67.08 m .$ Why this step? Plugging the optimal angle into $drift (θ)$ collapses the messy trig into this tidy closed form, which is the smallest downstream distance physically achievable — the violet point at the bottom of the curve in the figure.

Step 4 — Sanity on the heading. The optimal $θ \approx 48. 2^{\circ}$ is less upstream than aiming dead upstream ( $9 0^{\circ}$ ): pointing fully upstream would kill the across-speed and strand the swimmer. So the best plan is a partial upstream aim. Why this step? It guards against the tempting wrong answer "aim as far upstream as possible" — the forecast trap — by showing why the extreme heading is actually bad. In the figure the drift curve shoots up steeply as $θ \to 9 0^{\circ}$ .

Verify: The clean formula $d v_{w}^{2} - v_{b w}^{2} / v_{b w}$ requires $v_{w} > v_{b w}$ (here $3 > 2$ ) — exactly the regime where opposite-landing failed. $305 \approx 67.08$ m downstream ✓. This is the "consolation prize" heading the strong current leaves you.

Recall Which cell was hardest for you?

Trace each example back to its matrix row. If any cell still feels shaky, the fastest fix is to re-derive its heading rule ( $sin θ$ for zero drift, $cos θ$ for min drift) from the component picture. Zero-drift heading rule ::: $sin θ = v_{w} / v_{b w}$ (aim upstream) Min-drift heading when $v_{w} > v_{b w}$ ::: $cos θ = v_{b w} / v_{w}$ Min-drift value when opposite is impossible ::: $d v_{w}^{2} - v_{b w}^{2} / v_{b w}$ Boundary case $v_{w} = v_{b w}$ ::: $sin θ = 1$ , $θ = 9 0^{\circ}$ , across-speed $0$ , crossing time infinite — you need $v_{b w} > v_{w}$ strictly

Connections

Vectors — addition, components, unit vectors — every 2D example is a component split + add.
Kinematics in 2D — projectile motion — the "independent perpendicular axes" trick reused.
Frames of reference & Galilean transformation — the formal basis of the subtraction rule.
Newton's laws — inertial frames — why constant relative velocity keeps physics the same.
Rain-man umbrella problem — Ex 8 is its twin.
Relative motion — 1D and 2D; river-boat problems — parent topic.