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

WHY do we need relative motion?

WHAT we want: the velocity of object A as seen by observer B (call it $\vec v_{AB}$).

WHY it matters: A boat's engine pushes it through water, not through the ground. The current carries the water. So the ground sees boat-velocity = (velocity in water) + (water velocity). Mixing these frames up is the #1 source of errors.

Deriving the relative-velocity rule from scratch

Why this step? "Position of A relative to B" literally means start at B, go to A — that arrow is $\vec r_A - \vec r_B$.

Now differentiate with respect to time (frames don't rotate, $t$ is universal in Newtonian physics):

$\frac{d\vec r_{AB}}{dt} = \frac{d\vec r_A}{dt} - \frac{d\vec r_B}{dt}$

Two consequences you must internalize:

• $\vec v_{AB} = -\vec v_{BA}$ (A sees B moving opposite to how B sees A).
• Chain rule of frames: $\vec v_{AC} = \vec v_{AB} + \vec v_{BC}$. The middle index cancels like fractions.

1D relative motion

Choose a sign convention (right = +). Then everything is just signed addition.

2D relative motion & the river-boat problem

Set up axes: $x$ = across the river (width $d$), $y$ = downstream. Let:

• $\vec v_{bw}$ = velocity of boat wrt water (the engine's doing — magnitude fixed by motor).
• $\vec v_w$ = velocity of water wrt ground (the current, along $y$).
• $\vec v_{bg}$ = velocity of boat wrt ground = what we actually see.

By the chain rule: $\vec v_{bg} = \vec v_{bw} + \vec v_w$

Why this step? The boat swims through the water; the whole water sheet drifts. Add them.

Case 1 — Shortest time to cross

WHAT: minimize time. HOW: point the boat straight across ($\vec v_{bw}$ entirely along $x$).

Crossing depends only on the across-component, and the full engine speed $v_{bw}$ is across: $t_{\min} = \frac{d}{v_{bw}}$ Drift downstream during this time: $x_{\text{drift}} = v_w\, t_{\min} = \frac{v_w\,d}{v_{bw}}$

Why this step? The current only adds a $y$-velocity; it never speeds or slows the crossing. So heading purely across spends 100% of engine power on crossing → minimum time. You still land downstream — that's the cost.

Case 2 — Shortest path (land directly opposite)

WHAT: zero downstream drift. HOW: aim upstream at angle $\theta$ so the upstream component of $\vec v_{bw}$ exactly cancels the current.

$v_{bw}\sin\theta = v_w \;\Rightarrow\; \boxed{\sin\theta = \frac{v_w}{v_{bw}}}$

Why this step? Net downstream velocity must be 0. Only the upstream component of the boat can fight the current.

The effective across-speed is the remaining component: $v_{\text{across}} = v_{bw}\cos\theta = \sqrt{v_{bw}^2 - v_w^2}$ $t = \frac{d}{\sqrt{v_{bw}^2 - v_w^2}}$

Forecast-then-Verify

Common mistakes (Steel-man)

Flashcards

Connections

• Vectors — addition, components, unit vectors — relative velocity is vector subtraction.
• Kinematics in 2D — projectile motion — same "independent perpendicular components" logic.
• Frames of reference & Galilean transformation — the formal basis of $\vec v_{AB}=\vec v_A-\vec v_B$.
• Newton's laws — inertial frames — relative velocity is constant between inertial frames.
• Rain-man umbrella problem — direct cousin of the river-boat problem.

Concept Map

Hinglish (regional understanding)