This is a classic relative velocity problem, where an athlete is swimming in a river with a current. We'll solve the problem step-by-step.

Part (a) – Finding the velocity of the water ($v_w$) with respect to the ground:

Given dataThis is a classic relative velocity problem, where an athlete is swimming in a river with a current. We'll solve the problem step-by-step.

Part (a) – Finding the velocity of the water ($v_w$) with respect to the ground:

Given

• $w = 28$ m (width of the river)
• $d = 38$ m (distance downstream)
• $v_s = 0.25$ m/s (swimming speed relative to the water)

The swimmer crosses the river perpendicularly, so the velocity of the swimmer relative to the water is entirely in the perpendicular direction.

The time $t$ it takes to cross the river is determined by the width of the river and the swimmer's speed perpendicular to the current:

$t = \frac{w}{v_s} = \frac{28}{0.25} = 112 \text{ seconds}$

During this time, the water current carries the swimmer downstream by a distance $d = 38$ meters. Since the swimmer moves downstream due to the water's flow, we can calculate the velocity of the water ($v_w$) using:

$v_w = \frac{d}{t} = \frac{38}{112} \approx 0.339 \text{ m/s}$

Thus, the velocity of the water with respect to the ground is:

$v_w \approx 0.339 \text{ m/s}$

---

Part (b) – Finding the velocity of the swimmer with respect to the ground ($v_{s,g}$):

To calculate the swimmer's speed relative to a friend at rest on the ground, we need to combine the swimmer's velocity in the water and the water's velocity relative to the ground. Since the swimmer's velocity is perpendicular to the water flow and the water's velocity is along the river, the two velocities form a right-angled triangle.

Using the Pythagorean theorem:

$v_{s,g} = \sqrt{v_s^2 + v_w^2} = \sqrt{(0.25)^2 + (0.339)^2}$

Let's calculate that:

$v_{s,g} = \sqrt{0.0625 + 0.114921} = \sqrt{0.177421} \approx 0.421 \text{ m/s}$

Thus, the swimmer's speed with respect to the ground is approximately:

$v_{s,g} \approx 0.421 \text{ m/s}$

---

Final Answers:

• Part (a): The velocity of the water with respect to the ground is $v_w \approx 0.339 \, \text{m/s}$.
• Part (b): The speed of the swimmer with respect to a friend on the ground is $v_{s,g} \approx 0.421 \, \text{m/s}$.

Would you like more details on any part of the solution?

---

Here are 5 related questions to explore further:

1. How would the swimmer's time to cross change if the current speed doubled?
2. What would happen if the swimmer aimed upstream at a certain angle? How would that affect his trajectory?
3. If the swimmer's speed relative to water increased, how would that affect the downstream drift?
4. How does this problem relate to vector addition in physics?
5. What happens to the swimmer's ground speed if the water is still?

Tip: Always break down motion problems into perpendicular components to make the math easier!