A velocity is always measured relative to something. Walking at $1.5\ \text{m/s}$ on a moving walkway, you move faster relative to the floor than relative to the walkway. When frames move relative to each other, velocities combine by Adding Vectors by Components.
Key idea
$$\vec{v}_{\text{object, ground}} = \vec{v}_{\text{object, medium}} + \vec{v}_{\text{medium, ground}}.$$ For a boat: its velocity relative to the shore is its velocity relative to the water plus the water's velocity relative to the shore.
Why it works
In one second, the water carries the boat some distance downstream while the boat moves through the water in the direction it points. Its total displacement relative to the shore is the sum of the two, so its velocity is too. Because the two motions add as vectors, a current at right angles to the boat's heading doesn't help or hinder the crossing; it only moves the boat sideways.
Worked examples
Worked example: Along a line
A plane flies at $220\ \text{km/h}$ relative to the air into a $40\ \text{km/h}$ headwind. What is its ground speed?
- Headwind means opposite directions: $220 + (-40)$.
- Ground speed $= 180\ \text{km/h}$.
Answer: $180\ \text{km/h}$
Worked example: Straight across a current
A boat that moves at $4\ \text{m/s}$ in still water points straight across a $60\ \text{m}$ river flowing at $3\ \text{m/s}$.
- Crossing time uses only the across part: $t = \dfrac{60}{4} = 15\ \text{s}$.
- Drift downstream: $3 \times 15 = 45\ \text{m}$.
- Speed relative to the shore: $\sqrt{4^2 + 3^2} = 5\ \text{m/s}$.
Answer: $15\ \text{s}$; lands $45\ \text{m}$ downstream; $5\ \text{m/s}$
Worked example: Aiming upstream
Same boat and river. At what angle from straight across must the boat point to land directly opposite?
- The upstream part of its velocity must cancel the current: $4\sin\theta = 3$.
- $\theta = \sin^{-1}(0.75) \approx 48.6^\circ$ upstream of straight across.
- Its across speed is now $4\cos 48.6^\circ \approx 2.65\ \text{m/s}$, so the crossing takes about $22.7\ \text{s}$, longer than before.
Answer: about $48.6^\circ$ upstream
Common mistake
Adding the river's speed to the boat's speed when the boat points straight across ("$4 + 3 = 7\ \text{m/s}$, so it crosses faster"). The current is perpendicular to the heading, so it adds as a vector: it changes where the boat lands, not how long it takes to cross. Check: crossing time = width ÷ (the across part of the velocity).