Optimizing swimming direction in a moving river current

Key takeaway: This is a relative velocity optimization problem: the swimmer’s velocity must combine with the river current to produce the shortest crossing time.

Part (a): Minimize the time to reach the point directly opposite

To minimize the time, Lucky should maximize the component of his swimming velocity straight across the river. The river current is horizontal, so the across-river component comes from Lucky’s swimming direction.

Let $\theta$ be the angle between Lucky’s swimming direction and the river bank. Since Lucky swims at $1\,\text{m s}^{-1}$, his across-river speed is

$1\sin\theta.$

To reach a point directly opposite, the downstream drift can be canceled by choosing the angle so that the horizontal component matches the river speed:

$1\cos\theta=0.5.$

Thus

$\cos\theta=0.5 \implies \theta=60^\circ.$

So Lucky should swim at $\boxed{60^\circ}$ to the river bank.

Part (b)(i): New angle to reach the downstream point faster

The owner moves 50 m downstream to $M'$. Let the river width be $w$ meters. From the original setup, Lucky’s across-river speed remains determined by his swimming angle, but now he must also make up the 50 m downstream shift.

The shortest-time strategy is to align Lucky’s ground velocity directly toward $M'$. Using vector components, the swimmer’s velocity relative to the bank must satisfy

$\text{downstream component of swimming} - 0.5 = \frac{50}{T},$

while the cross-river component covers the river width.

Since the prompt asks for the angle to the nearest degree, the standard setup gives a triangle with horizontal displacement 50 m and vertical displacement equal to the river width. The resulting angle is found from

$\tan\theta = \frac{\text{across distance}}{50}.$

Because the river width is not explicitly stated in the text provided here, the angle must be determined from the full original figure or data associated with the problem statement.

Part (b)(ii): Travel time

The travel time is

$T=\frac{\text{distance traveled across the ground path}}{1}$

because Lucky swims at $1\,\text{m s}^{-1}$. Once the geometric dimensions are known, the time comes from the resultant vector path length divided by 1. The current changes the direction of the path, but the swimmer’s own speed remains fixed.

Key principle

For river-current problems, the key is to break velocity into components:

• component parallel to the bank: affected by the current
• component perpendicular to the bank: determines crossing time

If the downstream target is shifted, the optimal direction changes so that Lucky’s resultant motion points toward the new destination. Without the missing river width, a numeric answer for parts (b)(i) and (b)(ii) cannot be completed from the text alone.

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Pitfalls the pros know 👇 A common mistake is to treat the swimmer’s speed of 1 m/s as the ground speed. It is not. That speed is relative to the water, so the current must be added as a separate velocity vector. Another error is choosing the angle from the river current instead of from the river bank, which changes the trigonometric ratios. For part (a), students also often maximize the downstream motion, but the goal is to minimize time, so the crucial piece is the cross-river component, not the drift. For part (b), the shifted destination changes the geometry, so you need the actual river width and displacement triangle before plugging into trig.

What if the problem changes? If the river current were stronger, say $0.8\,\text{m s}^{-1}$, then the angle from the river bank would need to be steeper to keep the same crossing strategy, because Lucky would need a larger horizontal swimming component to offset the current. If the owner moved upstream instead of downstream, the optimal direction would tilt the other way, and the downstream component would be reduced rather than increased. These variants all use the same vector decomposition method, but the sign of the horizontal component changes the setup.

Tags: relative velocity, vector components, river current