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Irodov 1.7Trickykinematicsrelative-motionriver-crossing

River Crossing - Two Swimmers

Problem 1.7

Two swimmers leave point

A

on one bank of the river to reach point

B

lying right across on the other bank. One of them crosses the river along the straight line

AB

while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get from

A

B

. What was the velocity

u

of his walking if both swimmers reached the destination simultaneously? The stream velocity

v_0 = 2.0 \text{ km/h}

, the velocity of each swimmer with respect to water

v' = 2.5 \text{ km/h}

Answer:

u = 3.0 \text{ km/h}

Solution Path

Swimmer 1 crosses diagonally at a reduced speed of 1.5 km/h (fighting the current). Swimmer 2 crosses perpendicular at 2.5 km/h but drifts 0.8d downstream and must walk back. Setting total times equal and cancelling d gives a walking speed of u = 3.0 km/h.

01Problem Restatement

1/6

Two swimmers start from the same point

A

on one bank of a river and need to reach point

B

directly across on the other bank. They take different strategies.Swimmer 1 swims diagonally, aiming upstream so that the current and his swimming combine to carry him in a straight line from

A

B

.Swimmer 2 swims perpendicular to the banks (the shortest water path), but the current carries him downstream to some point

C

. He then walks along the bank from

C

back to

B

.Given:
-

v_0 = 2.0 \text{ km/h}

(stream velocity)
-

v' = 2.5 \text{ km/h}

(each swimmer's speed relative to water)
- Both swimmers arrive at

B

at the same timeWe need to find:

u

, the walking velocity of Swimmer 2 along the bank, in km/h.

Find

u

(walking velocity in km/h)

02Physical Picture and Strategy

2/6

Picture a river flowing from left to right. Point

A

is on the bottom bank, point

B

is directly across on the top bank. The river has width

d

(which will cancel out).Swimmer 1 (diagonal path): He aims his body upstream at an angle so that the downstream current exactly cancels his upstream component. His net velocity points straight from

A

B

. He crosses at a reduced effective speed because part of his swimming effort goes into fighting the current.Swimmer 2 (perpendicular + walk): He points his body straight across the river. He crosses faster (using his full swimming speed for the crossing), but the current pushes him downstream to point

C

. He then walks back along the bank from

C

B

at velocity

u

.Strategy: We will find the total time for each swimmer in terms of the river width

d

. Since both arrive at

B

simultaneously, we set

t_1 = t_2

and solve for

u

. The width

d

cancels.

Two strategies: diagonal (slower crossing, no drift) vs. perpendicular (faster crossing, must walk back)

03Step 1: Swimmer 1 (Diagonal Crossing)

3/6

Swimmer 1 aims upstream at an angle so that his net velocity is directed straight across the river from

A

B

. His velocity relative to water is

v'

, and the stream flows at

v_0

.Using the Pythagorean theorem for velocity components (the swimmer's velocity triangle):

v_{\perp} = \sqrt{v'^2 - v_0^2}

where

v_{\perp}

is the net cross-river speed. Substituting:

v_{\perp} = \sqrt{(2.5)^2 - (2.0)^2} = \sqrt{6.25 - 4.00} = \sqrt{2.25} = 1.5 \text{ km/h}

This means: Swimmer 1 crosses the river at only 1.5 km/h because a large portion of his effort (2.0 out of 2.5 km/h) goes into fighting the current.His total time to cross width

d

t_1 = \frac{d}{v_{\perp}} = \frac{d}{1.5}

t_1 = d / 1.5

(cross-river speed is only 1.5 km/h)

04Step 2: Swimmer 2 (Perpendicular + Walk)

4/6

Swimmer 2 points straight across the river, so his full swimming speed

v' = 2.5 \text{ km/h}

goes into crossing. He crosses faster than Swimmer 1.Using the definition of velocity for the crossing phase:

t_{\text{cross}} = \frac{d}{v'} = \frac{d}{2.5}

While crossing, the current carries him downstream. Using displacement = velocity

\times

time:

x = v_0 \cdot t_{\text{cross}} = v_0 \cdot \frac{d}{v'} = \frac{2.0 \cdot d}{2.5} = 0.8\,d

This means: Swimmer 2 lands at point

C

, which is

0.8d

downstream from

B

.He must walk this distance back at velocity

u

t_{\text{walk}} = \frac{x}{u} = \frac{0.8\,d}{u}

His total time:

t_2 = \frac{d}{2.5} + \frac{0.8\,d}{u}

t_2 = d/2.5 + 0.8d/u

(fast crossing, but must walk back the drift)

05Step 3: Set Equal and SolveKEY INSIGHT

5/6

Both swimmers reach

B

at the same time, so

t_1 = t_2

\frac{d}{1.5} = \frac{d}{2.5} + \frac{0.8\,d}{u}

The river width

d

appears in every term, so it cancels:

\frac{1}{1.5} = \frac{1}{2.5} + \frac{0.8}{u}

Computing the left side and first term on the right:

\frac{1}{1.5} = \frac{2}{3}, \qquad \frac{1}{2.5} = \frac{2}{5}

\frac{0.8}{u} = \frac{2}{3} - \frac{2}{5} = \frac{10 - 6}{15} = \frac{4}{15}

Solving for

u

u = \frac{0.8 \times 15}{4} = \frac{12}{4} = 3.0 \text{ km/h}

This means: the walking speed must be exactly 3.0 km/h for both swimmers to arrive simultaneously. The answer is independent of river width

d

u = 3.0 \text{ km/h}

06Verification and Summary

6/6

Verification: Let us check by computing both times with

d

as a parameter.

t_1 = \frac{d}{1.5} = 0.667d \text{ h}

t_2 = \frac{d}{2.5} + \frac{0.8d}{3.0} = 0.4d + 0.267d = 0.667d \text{ h}

t_1 = t_2 \; \checkmark

Dimensional check:

u

has units of km/h

\checkmark

Reasonableness: 3.0 km/h is a brisk walking pace, which is physically sensible for a swimmer who needs to walk along a riverbank.

\checkmark

Summary: Swimmer 1 crosses slower (1.5 km/h net) because he fights the current to go straight. Swimmer 2 crosses faster (2.5 km/h) but drifts 0.8d downstream and must walk back. Setting their total times equal and cancelling

d

gives a walking speed of

u = 3.0

km/h.

\boxed{u = 3.0 \text{ km/h}}

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