Question

A boy is attempting to swim directly across a river; he is able to swim at a speed of \(0.500 \mathrm{m} / \mathrm{s}\) relative to the water. The river is \(25.0 \mathrm{m}\) wide and the boy ends up at \(50.0 \mathrm{m}\) downstream from his starting point. (a) How fast is the current flowing in the river? (b) What is the speed of the boy relative to a friend standing on the riverbank?

Short answer

Answer: The speed of the river's current is 1.00 m/s, and the boy's speed relative to the riverbank is 1.12 m/s.

Step-by-step solution

(Step 1: Set up the given information)

First, let's represent the horizontal motion with vector Vx, and the vertical motion with vector Vy. The boy's swimming speed across the river is Vy, which is 0.500 m/s. He ends up across the river (25.0 m) and downstream (50.0 m) from his starting point.

(Step 2: Calculate the time it takes for the boy to cross the river)

The time it takes for the boy to cross the river is the same in both directions. We can find this time by using the equation for distance: Distance = Speed × Time Rearranging for time, we get: Time = Distance / Speed We know the distance and speed of the boy in the vertical direction, so we'll plug them in to find the time: Time = 25.0 m / 0.500 m/s = 50.0 s

(Step 3: Calculate the speed of the river's current)

Now we'll use the horizontal distance the boy swam to find the speed of the river's current. Let Vx be the speed of the current, then: Horizontal Distance = Vx × Time Rearranging for Vx, we get: Vx = Horizontal Distance / Time Plugging in the values, we have: Vx = 50.0 m / 50.0 s = 1.00 m/s The speed of the river's current is 1.00 m/s.

(Step 4: Calculate the boy's speed relative to the riverbank)

To find the boy's speed relative to the riverbank, we need to combine the horizontal speed (Vx) and the vertical speed (Vy) as vectors. To do this, we'll use the Pythagorean theorem: VB_boy = √(Vx^2 + Vy^2) Plugging in the values, we get: VB_boy = √((1.00 m/s)^2 + (0.500 m/s)^2) = √(1.00 + 0.250) = 1.12 m/s (rounded to 2 decimal places) The boy's speed relative to the riverbank is 1.12 m/s.