River crossing canoe velocity and landing point

Key takeaway: Resultant velocity for a canoe crossing a river depends on combining the canoe's still-water speed with the river current as perpendicular vectors.

Vector components in the canoe crossing

The speeds 5 km/h and 2 km/h act at right angles in this river problem, so the canoe’s motion must be treated as a vector sum. One velocity points straight across the river, and the other points downstream with the current. That means the resultant velocity is not 5 km/h or 2 km/h by itself, but the diagonal formed by combining both components.

Because the canoeist steers perpendicular to the current, the canoe has a cross-river component of 5 km/h and a downstream component of 2 km/h. These values stay constant as long as the paddling direction is unchanged and the current is uniform.

Resultant velocity and direction

For the resultant velocity, use the Pythagorean relationship because the two components are perpendicular. The speed is

$v_R = \sqrt{5^2 + 2^2} = \sqrt{29} \approx 5.39\text{ km/h}$

To find the direction, measure the angle downstream from the straight-across direction. Using tangent,

$\tan \theta = \frac{2}{5}$

so

$\theta = \tan^{-1}(0.4) \approx 21.8^\circ$

This means the canoe moves at about 5.39 km/h, angled 21.8° downstream from the direction perpendicular to the bank.

Landing point on the opposite bank

The river is 400 m wide, which is 0.4 km. The time to cross depends on the across-river component only, because that is the part carrying the canoe from one bank to the other. Since the perpendicular speed is 5 km/h,

$t = \frac{0.4}{5} = 0.08\text{ h}$

Convert that to minutes if needed:

$0.08 \times 60 = 4.8\text{ minutes}$

During that time, the current moves the canoe downstream at 2 km/h, so the downstream drift is

$d = 2 \times 0.08 = 0.16\text{ km} = 160\text{ m}$

So the canoe lands 160 m downstream from the point directly opposite the start.

Common mistakes with river motion

A frequent error with perpendicular velocity problems is using the resultant speed to find the crossing time. That gives the wrong answer because only the across-river component determines how long the canoe takes to reach the far bank. Another mistake is mixing units, especially forgetting to convert 400 m into 0.4 km before using km/h.

It also helps to draw a vector diagram before calculating. The triangle makes it clear which side is the across-river speed, which side is the current, and which side is the actual path. That visual prevents confusion between direction and distance.

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Pitfalls the pros know 👇 The hardest part of this river crossing vector problem is deciding which speed belongs in the crossing-time calculation. Many students accidentally use the resultant speed of 5.39 km/h to find how long the canoe takes to cross, but that is not correct because the canoe only gets across the river from the 5 km/h perpendicular component. If you use the diagonal speed, your landing point will also come out wrong.

Another place people lose marks is with units. The river width is given as 400 m, while the speeds are in km/h, so the width should be converted to 0.4 km before finding time. After that, the downstream drift can be found using the 2 km/h current. A clean sketch with labeled components usually makes the method much easier to follow and reduces sign or direction mistakes.

What if the problem changes? If the canoeist paddles at 6 km/h instead of 5 km/h while the current stays at 2 km/h, the same method still works but the answers change. The resultant speed becomes

$\sqrt{6^2 + 2^2} = \sqrt{40} \approx 6.32\text{ km/h}$

and the angle downstream is

$\tan^{-1}(2/6) \approx 18.4^\circ.$

For the same 400 m wide river, the crossing time would be based on 6 km/h across the river, so the time is shorter and the downstream drift is smaller.

A second variation is to keep the 5 km/h canoe speed but change the river width to 600 m. The direction stays the same, but the canoe spends more time in the current, so it lands farther downstream. That version is useful for checking whether you understand that width affects drift distance, while the across-river speed controls time.

Tags: resultant velocity, river current vectors, vector displacement