Math Problem Statement

A swimmer wants to cross a river, from point A to point B, as shown in the figure. The distance d1 (from A to C) is 200 m , the distance d2 (from C to B) is 150 m , and the speed vr of the current in the river is 5 km/hour . Suppose that the swimmer's velocity relative to the water makes an angle of θ=45degrees with the line from A to C, as indicated in the figure.o swim directly from A to B, what speed us , relative to the water, should the swimmer have? Express the swimmer's speed numerically, to three significant figures, in kilometers per hour.

Solution

To solve this problem, we need to determine the swimmer's speed relative to the water (vsv_s) that will allow them to swim directly from point A to point B, compensating for the current.

Given Data:

  • d1=200md_1 = 200 \, \text{m} (distance from A to C)
  • d2=150md_2 = 150 \, \text{m} (distance from C to B)
  • vr=5km/hv_r = 5 \, \text{km/h} (speed of the river current)
  • θ=45\theta = 45^\circ (angle at which swimmer swims relative to the line from A to C)

Step 1: Convert speeds to consistent units.

The current speed is given in km/h. We need to convert this to m/s: vr=5km/h=5×10003600m/s=1.3889m/sv_r = 5 \, \text{km/h} = \frac{5 \times 1000}{3600} \, \text{m/s} = 1.3889 \, \text{m/s}

Step 2: Resolve swimmer's velocity components.

The swimmer's velocity relative to the water vsv_s has components:

  • A horizontal component vscos(θ)v_s \cos(\theta)
  • A vertical component vssin(θ)v_s \sin(\theta)

We want to ensure that the swimmer swims directly from A to B. The current will push the swimmer downstream, so the swimmer's horizontal component of velocity must cancel the effect of the river current.

Thus, the horizontal component of the swimmer's velocity relative to the water must equal the speed of the current: vscos(45)=vrv_s \cos(45^\circ) = v_r

Since cos(45)=22\cos(45^\circ) = \frac{\sqrt{2}}{2}, we have: vs22=1.3889m/sv_s \cdot \frac{\sqrt{2}}{2} = 1.3889 \, \text{m/s} vs=1.3889×22m/sv_s = \frac{1.3889 \times 2}{\sqrt{2}} \, \text{m/s} vs2.77781.4142m/sv_s \approx \frac{2.7778}{1.4142} \, \text{m/s} vs1.962m/sv_s \approx 1.962 \, \text{m/s}

Step 3: Convert the swimmer's speed to km/h.

To convert from m/s to km/h, multiply by 3.6: vs=1.962m/s×3.6km/h per m/sv_s = 1.962 \, \text{m/s} \times 3.6 \, \text{km/h per m/s} vs7.07km/hv_s \approx 7.07 \, \text{km/h}

Final Answer:

The swimmer's speed relative to the water should be approximately 7.07 km/h.


Would you like more details or have any questions?

Here are 5 related questions:

  1. How would the swimmer's speed change if the angle θ\theta were different?
  2. What would happen if the current speed were doubled?
  3. How do you determine the swimmer's velocity if they swim at an angle other than 45°?
  4. How could we find the time it would take the swimmer to cross the river?
  5. How would the distances d1d_1 and d2d_2 change if the swimmer were not compensating for the current?

Tip: When dealing with currents in river problems, always break the swimmer’s velocity into components, especially when dealing with angles. This makes it easier to handle the directionality of the current.

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Math Problem Analysis

Mathematical Concepts

Velocity
Trigonometry
Vector components
Units conversion

Formulas

v_r = 5 km/h
v_s cos(θ) = v_r
v_s = (v_r × 2) / √2
v_s = v_s (m/s) × 3.6

Theorems

Pythagoras' theorem
Trigonometric identity for cos(45°)

Suitable Grade Level

Grades 10-12