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

The rule is "subtract the seer" — the observer is B, so subtract ${\vec{v}}_B$: $v_{AB}=v_A-v_B=30-18=+12\text{ m/s}$ What it means: to a passenger on B, train A creeps forward at $12$ m/s. The $+$ sign says "in the $+$ (rightward) direction." ✓

$4$ m/s is the engine's effort through the water → that is $==v_{bw}==$. The bank-measured speed is $v_{bg}$ (generally different), and $v_w$ is the current — a separate number. ✓

Point straight across means the entire engine speed sits on the $x$-axis (across), so the across-speed is the full $6$ m/s. (a) $t_{\min }=\frac{d}{v_{bw}}=\frac{120}{6}=20\text{ s}$ (b) The drift is downstream, i.e. along the $y$-axis, so we label it $y_{\text{drift}}$. During those $20$ s the current slides the boat downstream at $2$ m/s: $y_{\text{drift}}=v_w{t}_{\min }=2\times 20=40\text{ m}$ Why the current didn't slow the crossing: the current is purely along $y$, perpendicular to the crossing motion along $x$. Perpendicular velocity can't touch the $x$-distance. ✓

In words, the figure shows: the cyan engine vector aimed upstream-and-across, a white current vector chained tip-to-tail onto it pointing downstream, and the amber resultant standing straight up the page (pure across) — proof the sideways parts cancelled. A small white arc marks the steering angle $\theta$. To kill all downstream drift, the upstream component of the boat must exactly cancel the current: $v_{bw}\sin \theta =v_w\Rightarrow \sin \theta =\frac{v_w}{v_{bw}}=\frac{2}{6}=\frac{1}{3}$ (a) $\theta =\arcsin (1/3)\approx 19.47^{\circ }$ upstream from straight-across (the white arc marked "theta" in the figure). (b) The leftover component points across. Why we may write $v_{bw}\cos \theta =\sqrt{v_{bw}^2-v_w^2}$: the engine vector $v_{bw}$ is the hypotenuse of a right triangle whose two legs are the upstream part $v_{bw}\sin \theta$ and the across part $v_{bw}\cos \theta$. We already forced the upstream leg to equal $v_w$. By Pythagoras the across leg is $\sqrt{v_{bw}^2-(v_{bw}\sin \theta {)}^2}=\sqrt{v_{bw}^2-v_w^2}$ — the same identity as $\cos \theta =\sqrt{1-{\sin }^2\theta }$, just scaled by $v_{bw}$. Hence: $v_{\text{across}}=v_{bw}\cos \theta =\sqrt{v_{bw}^2-v_w^2}=\sqrt{36-4}=\sqrt{32}\approx 5.657\text{ m/s}$ (c) $t=\frac{d}{v_{\text{across}}}=\frac{120}{\sqrt{32}}\approx 21.21\text{ s}$ Notice: $21.21>20$ — zero drift costs time, exactly as the parent note warned. ✓

First test the zero-drift condition: $\sin \theta =\frac{v_w}{v_{bw}}=\frac{2.0}{1.5}\approx 1.333>1$ No angle has a sine bigger than $1$, so landing directly opposite is impossible — the current always wins.

Now the least-drift heading. Look at the figure. In words, the figure shows: a white current vector along the downstream axis; a cyan circle of radius $v_{bw}$ centred at the current's tip (the boat's engine tip lands somewhere on this circle as the heading swings); and the amber ground-velocity arrow from the origin to a point on that circle. The least drift angle $\delta$ occurs where the amber arrow is tangent to the circle.

Why "tangent" is the answer: as we swing the heading, the amber tip runs around the circle. The angle $\delta$ that ${\vec{v}}_{bg}$ makes with the across-direction is smallest exactly when the amber line just grazes (is tangent to) the circle — push past that and the arrow swings back, increasing $\delta$. A radius drawn to a tangency point is perpendicular to the tangent line. The radius here is ${\vec{v}}_{bw}$ and the tangent line lies along ${\vec{v}}_{bg}$, so at the optimum ${\vec{v}}_{bg}\perp {\vec{v}}_{bw}$.

The right triangle at the optimum. With ${\vec{v}}_{bg}\perp {\vec{v}}_{bw}$, the three vectors form a right triangle in velocity space: hypotenuse $v_w$ (the current), one leg $v_{bw}$ (the engine, perpendicular to $v_{bg}$), and the other leg $v_{bg}$ itself. So: $v_{bg}=\sqrt{v_w^2-v_{bw}^2}=\sqrt{4-2.25}=\sqrt{1.75}\approx 1.3229\text{ m/s.}$ Heading. The angle between the current and the engine in that triangle satisfies $\cos {\theta }_{\text{best}}=v_{bw}/v_w$. Measured upstream from straight-across (same reference as $\theta$ throughout), the heading works out to ${\theta }_{\text{best}}=90^{\circ }-\arccos (v_{bw}/v_w)$. Numerically $\arccos (1.5/2.0)=41.41^{\circ }$, so ${\theta }_{\text{best}}=90^{\circ }-41.41^{\circ }=48.59^{\circ }$ upstream from straight-across.

Now split ${\vec{v}}_{bg}$ into its across and downstream parts — these decide the drift. The drift angle $\delta$ is the tilt of ${\vec{v}}_{bg}$ off the across-axis, and from the same right triangle $\sin \delta =v_{bw}/v_w$, $\cos \delta =\sqrt{v_w^2-v_{bw}^2}/v_w$. Therefore:

• across-speed $v_x=v_{bg}\cos \delta =\frac{v_{bw}}{v_w}\sqrt{v_w^2-v_{bw}^2}=\frac{1.5}{2.0}\sqrt{1.75}\approx 0.9922$ m/s,
• downstream-speed $v_y=v_{bg}\sin \delta =\frac{(v_w^2-v_{bw}^2)}{v_w}...$ — but we only need the drift ratio, which is $\tan \delta$: $\frac{v_y}{v_x}=\tan \delta =\frac{\sqrt{v_w^2-v_{bw}^2}}{v_{bw}}=\frac{\sqrt{1.75}}{1.5}\approx \mathrm{0.8819.}$

Crossing time uses the across-speed: $t=\frac{d}{v_x}=\frac{60}{0.9922}\approx 60.47\text{ s.}$ Least drift = (drift ratio) $\times$ width: $y_{\min }=\frac{v_y}{v_x}d=\tan \delta \cdot d=0.8819\times 60\approx 52.92\text{ m downstream.}$ So the swimmer cannot avoid drifting; the best they can do is land about $52.9$ m downstream. ✓

Subtract the observer (Q), slot by slot (tuple order is across, downstream): ${\vec{v}}_{PQ}={\vec{v}}_{Pg}-{\vec{v}}_{Qg}=(4-4,3-(-1))=(0,4)\text{ m/s}$ To Q, boat P slides purely downstream at $4$ m/s and shows no across-motion. Speed $=\sqrt{0^2+4^2}=4$ m/s. See Vectors — addition, components, unit vectors for the component subtraction. ✓

(a) Min-time landing drift $=v_{w}d/v_{bw}=3\cdot 90/5=54$ m downstream. The jetty is only $30$ m downstream, i.e. upstream of the min-time point, so we must aim somewhat upstream — but not fully (that would land at $0$ m).

(b) Let the boat heading make angle $\theta$ upstream from straight-across. Across-velocity: $v_x=v_{bw}\cos \theta$. Net downstream velocity: $v_y=v_w-v_{bw}\sin \theta$. Cross time $t=d/v_x$. Required drift: $v_{y}t=30$. $\frac{(v_w-v_{bw}\sin \theta )d}{v_{bw}\cos \theta }=30$ Plug numbers ($d=90,v_{bw}=5,v_w=3$): $\frac{(3-5\sin \theta )\cdot 90}{5\cos \theta }=30\Rightarrow 18(3-5\sin \theta )=30\cos \theta$ $54-90\sin \theta =30\cos \theta \Rightarrow 9-15\sin \theta =5\cos \theta .$ Solve: test $\theta \approx 16.26^{\circ }$ upstream, where $\sin \theta =0.28$ ($5\sin \theta =1.4$, so $3-1.4=1.6$) and $5\cos \theta =4.80$; then $\frac{1.6\cdot 90}{4.80}=30$ ✓.

(c) $v_x=5\cos 16.26^{\circ }=4.80$ m/s, so $t=\frac{90}{4.80}=18.75\text{ s}.$ ✓

This is the Rain-man umbrella problem riding on top of the river problem — layer the frames. Tuple order is (across, downstream, vertical) as fixed at the top of the page. New notation (define on first use): the subscript $r$ means rain. So ${\vec{v}}_{rg}$ = velocity of rain relative to ground (what a still bystander sees), and ${\vec{v}}_{rb}$ = velocity of rain relative to the boat (what the moving passenger sees). By the same "subtract the observer" rule, ${\vec{v}}_{rb}={\vec{v}}_{rg}-{\vec{v}}_{bg}$.

Step 1 — boat's ground velocity. Min-time: heading straight across, so ${\vec{v}}_{bg}=(v_{bw},v_w,0)=(5,4,0)\text{ m/s (across, downstream, vertical)}.$ Step 2 — rain relative to boat. Vertical is $z$, up positive, so falling rain is negative: ${\vec{v}}_{rg}=(0,0,-8)$. Subtract the observer (the boat), slot by slot: ${\vec{v}}_{rb}={\vec{v}}_{rg}-{\vec{v}}_{bg}=(0-5,0-4,-8-0)=(-5,-4,-8)\text{ m/s}.$ Step 3 — speed and direction. Horizontal part magnitude $=\sqrt{5^2+4^2}=\sqrt{41}\approx 6.40$ m/s. Total speed $=\sqrt{5^2+4^2+8^2}=\sqrt{105}\approx 10.25$ m/s. Angle from the vertical: $\tan \alpha =\frac{\text{horizontal}}{\text{vertical}}=\frac{\sqrt{41}}{8}\Rightarrow \alpha =\arctan \frac{\sqrt{41}}{8}\approx 38.66^{\circ }.$ The passenger tilts the umbrella $\approx 38.7^{\circ }$ from vertical, leaning it toward their own direction of motion (into the apparent rain). ✓

L1 :: read/name the velocities and subtract the seer. L2 :: apply min-time ($t=d/v_{bw}$) and zero-drift ($\sin \theta =v_w/v_{bw}$) formulas. L3 :: judge feasibility ($v_{bw}>v_w$) and find least-drift via the tangent-circle. L4 :: solve the coupled drift$=v_{y}t$ equation for a target landing point. L5 :: stack a second relative-motion problem (rain) on the boat frame.