Relative motion — 1D and 2D; river-boat problems: Question bank

Before we start, one reminder of the notation used throughout, so no symbol is unearned — read this against Figure 1, which draws every quantity as an arrow on the river.

Definition The symbols you will meet here

All velocities carry two frame labels: the first letter is what moves, the second is who watches. "With respect to" always names the watcher.

$r_{A}, r_{B}$ ::: position vectors — the arrow drawn from the ground origin $O$ to object A (or B). "From ground" is their frame; nothing about position is absolute until you pick $O$ .
$v_{A}, v_{B}$ ::: velocities of A and B each measured with respect to the ground — the rate at which $r_{A}$ and $r_{B}$ change. Whenever you see a bare $v_{A}$ in the subtraction rule, read " $A$ with respect to the ground."
$v_{A B}$ ::: velocity of A as seen by B — how fast and in what direction A moves in B's eyes.
$v_{b w}$ ::: velocity of boat with respect to the water (what the engine/arms produce). Its magnitude is fixed by the motor or swimmer.
$v_{w}$ ::: velocity of the water with respect to the ground (the current).
$v_{b g}$ ::: velocity of the boat with respect to the ground (what a person on the bank sees).
$d$ ::: the river's width — the across-the-stream distance you must cover.
$t$ ::: the crossing time — how long it takes to go from one bank to the other, always $t = (across distance) / (across speed)$ .
$θ$ ::: the steering angle, measured from the straight-across direction, tilted upstream (see Figure 3).

Figure 2 sketches the two geometric facts the whole bank leans on: how the min-time and zero-drift routes build their velocity triangles.

Figure 3 zooms into the zero-drift triangle so you can see where $sin θ = v_{w} / v_{b w}$ and $t = d / v_{b w}^{2} - v_{w}^{2}$ come from, rather than trusting a cited formula.

Spot the error

Each line states a flawed piece of reasoning. Reveal names the error and repairs it.

"Boat does 5 m/s, current 3 m/s, so it crosses the ground at 8 m/s."

Error: adding perpendicular vectors as scalars. They're at right angles (see Figure 2), so the ground speed is

5^{2} + 3^{2} \approx 5.83

m/s, and it points partly downstream, not straight across.

"Relative means combining, so

v_{A B} = v_{A} + v_{B}

Error: wrong operation. Relative velocity subtracts the observer:

v_{A B} = v_{A} - v_{B}

. Adding is only for the chain rule

v_{A C} = v_{A B} + v_{B C}

, where a shared middle index cancels.

"To land directly opposite, aim the boat straight across and the current will average out."

Error: nothing cancels the current in that setup. Aiming straight across leaves the full current uncancelled, so you drift. You must tilt upstream so a component of

v_{b w}

actively fights the current (Figure 3).

"A faster current always increases your crossing time."

Error: this conflates the two routes. For the min-time (straight-across) route, current changes only drift, not time. It only lengthens time on the zero-drift route, where you divert power upstream.

"Since the water carries the boat, the ground velocity is just the current velocity."

Error: ignores the engine. The boat also moves through the water; ground velocity is the sum

v_{b g} = v_{b w} + v_{w}

, not the current alone.

"For the shortest path we minimize

v_{b w} sin θ

, so we set

θ = 0

Error: shortest path means zero drift, which requires

v_{b w} sin θ = v_{w}

(matching the current), not minimizing it. Setting

θ = 0

gives maximum drift, i.e. the longest path.

"The chain rule

v_{A C} = v_{A B} + v_{B C}

needs the outer indices to match."

Error: it's the inner indices (

B

and

B

) that must match and cancel, like fractions; the outer letters

A

and

C

survive to name the result.

Why questions

Why is velocity always stated "with respect to" some frame?

Because motion has no absolute meaning — sitting in a train you can't tell if you or the next train moves. A velocity value is empty until you name what it's measured against.

Why does subtracting a negative velocity give a larger closing speed?

A negative velocity means "moving the other way (toward you)." Subtracting it,

v_{A} - (- v_{B}) = v_{A} + v_{B}

, adds the magnitudes — exactly the physical "they approach at the sum" intuition.

Why doesn't the current affect the crossing time in the min-time route?

The current is purely along the downstream (

y

) axis, perpendicular to the across (

x

) motion. Perpendicular components are independent, so the current can't touch the

x

-distance or the time to cover it.

Why is the zero-drift crossing time

d / v_{b w}^{2} - v_{w}^{2}

rather than

d / v_{b w}

Because you spend part of the boat's speed pointing upstream to cancel the current (Figure 3). Only the leftover across-component

v_{b w}^{2} - v_{w}^{2}

does the crossing, and a smaller across-speed means more time.

Why can we differentiate

r_{A B} = r_{A} - r_{B}

term-by-term to get the velocity rule?

Here

r_{A}, r_{B}

are the ground-frame positions of A and B (arrows from origin

O

), and

r_{A B} = r_{A} - r_{B}

is the arrow from B to A. In Newtonian physics time is universal and the frames don't rotate, so

t

is the same for everyone and

d / d t

distributes over the subtraction, giving

v_{A B} = v_{A} - v_{B}

Why does the resultant ground velocity point downstream in a straight-across crossing?

The across-component (

v_{b w}

) and the downstream-component (

v_{w}

) add tip-to-tail; the sum tilts toward the downstream side, so the true path is a diagonal, not a straight line across (Figure 2, left).

Why is aiming upstream the only way to fight the current, not aiming downstream?

Only an upstream velocity component points against the current; a downstream tilt would add to the drift. To net zero downstream motion you must oppose it directly.

Edge cases

What happens to both routes when the current is zero (

v_{w} = 0

They become identical:

sin θ = 0/ v_{b w} = 0

so you aim straight across in both, drift is zero, and both give

t = d / v_{b w}

. No current means no trade-off.

What is the steering angle when the current exactly equals the boat speed (

v_{w} = v_{b w}

sin θ = v_{w} / v_{b w} = 1

, so

θ = 90°

— you'd have to aim fully upstream, leaving zero across-speed. You'd never actually cross, so landing opposite is impossible (the limiting case).

What is the min-time drift when the current is faster than the boat (

v_{w} > v_{b w}

Min-time crossing still works — you always cross in

t = d / v_{b w}

— but the drift

v_{w} t = v_{w} d / v_{b w}

now exceeds the width

d

. You cross, but land far downstream; you just can't land opposite.

What does

v_{A B}

equal if A and B move with identical velocity?

Zero:

v_{A} - v_{B} = 0

. In each other's frame they appear at rest — this is why co-moving frames (both inertial) can't detect their shared motion.

In the limit

v_{w} \to v_{b w}

from below, what happens to the zero-drift crossing time?

The across-speed

v_{b w}^{2} - v_{w}^{2} \to 0

, so

t = d / v_{b w}^{2} - v_{w}^{2} \to \infty

. The crossing takes unboundedly long as the current approaches the boat's speed — the route smoothly becomes impossible.

If the boat aims straight upstream (

θ = 90°

) with

v_{b w} > v_{w}

, what is its ground motion?

It moves upstream at ground speed

v_{b w} - v_{w}

and makes no progress across the river. All engine power fights the current with none left for crossing — a degenerate, useless heading.

Recall One-line self-test

Cover the answers above. If you can justify every True/False and name every error without peeking, you own this topic. Weakest link most students have ::: "current doesn't change min-time crossing" — because it feels like the river should slow you.

Connections

Relative motion — 1D and 2D; river-boat problems — the parent note these traps stress-test.
Vectors — addition, components, unit vectors — every "spot the error" here is really a vector-vs-scalar slip.
Frames of reference & Galilean transformation — the "why with respect to a frame" questions live here.
Newton's laws — inertial frames — the co-moving-frames edge case ( $v_{A B} = 0$ ) is why inertial frames are indistinguishable.
Rain-man umbrella problem — same subtract-the-observer logic; test yourself on its traps too.
Kinematics in 2D — projectile motion — the "perpendicular components are independent" idea reused.

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

True or false — justify