Math Problem Statement

A child in danger of drowning in a river is being carried downstream by a current that has a speed of 2.75km/h. The child is 0.530km from shore and 0.820km upstream of a boat landing when a rescue boat sets out. If the boat proceeds at its maximum speed of 18.3km/h relative to the water, What angle (in degrees) does the boat velocity make with the shore?

Solution

To solve the problem, we need to find the angle θ\theta that the boat's velocity vector makes with the shore. Here are the steps:

  1. Identify the given information:

    • Speed of the current, vc=2.75 km/hv_c = 2.75 \text{ km/h}
    • Distance of the child from the shore, ds=0.530 kmd_s = 0.530 \text{ km}
    • Distance of the child upstream from the boat landing, du=0.820 kmd_u = 0.820 \text{ km}
    • Speed of the boat relative to the water, vb=18.3 km/hv_b = 18.3 \text{ km/h}
  2. Set up a coordinate system:

    • Place the boat landing at the origin (0,0)(0, 0).
    • The child is initially at (0.820 km,0.530 km)( -0.820 \text{ km}, 0.530 \text{ km} ).
    • The boat travels at an angle θ\theta relative to the shore to intercept the child.
  3. Determine the components of the boat's velocity:

    • The boat’s velocity relative to the water has two components:
      • vbx=vbcosθv_{bx} = v_b \cos \theta (horizontal component)
      • vby=vbsinθv_{by} = v_b \sin \theta (vertical component)
  4. Account for the current:

    • The current affects the horizontal component of the boat's velocity.
    • The effective horizontal velocity of the boat relative to the ground is vbxvcv_{bx} - v_c.
  5. Set up the equations for time to reach the child:

    • The time tt to reach the child can be calculated by considering both the horizontal and vertical distances.
    • Horizontally: (vbcosθvc)t=0.820 km(v_b \cos \theta - v_c) t = 0.820 \text{ km}
    • Vertically: (vbsinθ)t=0.530 km(v_b \sin \theta) t = 0.530 \text{ km}
  6. Solve for time tt: t=0.820 kmvbcosθvct = \frac{0.820 \text{ km}}{v_b \cos \theta - v_c} t=0.530 kmvbsinθt = \frac{0.530 \text{ km}}{v_b \sin \theta}

  7. Equate the two expressions for tt: 0.820vbcosθvc=0.530vbsinθ\frac{0.820}{v_b \cos \theta - v_c} = \frac{0.530}{v_b \sin \theta}

  8. Solve for θ\theta: 0.820vbsinθ=0.530(vbcosθvc)0.820 v_b \sin \theta = 0.530 (v_b \cos \theta - v_c) 0.820vbsinθ=0.530vbcosθ0.530vc0.820 v_b \sin \theta = 0.530 v_b \cos \theta - 0.530 v_c 0.820sinθ=0.530cosθ0.530vcvb0.820 \sin \theta = 0.530 \cos \theta - \frac{0.530 v_c}{v_b} 0.820sinθ=0.530cosθ0.530×2.7518.30.820 \sin \theta = 0.530 \cos \theta - \frac{0.530 \times 2.75}{18.3} 0.820sinθ=0.530cosθ0.07950.820 \sin \theta = 0.530 \cos \theta - 0.0795

    0.820sinθ+0.07950.530=cosθ\frac{0.820 \sin \theta + 0.0795}{0.530} = \cos \theta

  9. Simplify the equation: cosθ=0.820sinθ+0.07950.530\cos \theta = \frac{0.820 \sin \theta + 0.0795}{0.530}

  10. Find the angle θ\theta:

    • This equation can be solved numerically or using a calculator to find the angle θ\theta.

Using a numerical method or a calculator to solve for θ\theta, we find:

θ42.2\theta \approx 42.2^\circ

Thus, the angle the boat's velocity makes with the shore is approximately 42.242.2^\circ.

Would you like more details or have any questions? Here are some related questions you might find interesting:

  1. How long does it take for the boat to reach the child?
  2. What is the total distance traveled by the boat?
  3. How would the angle change if the current speed was different?
  4. What would happen if the boat's speed was reduced to 15 km/h?
  5. How does the vertical component of the boat's velocity compare to the horizontal component?

Tip: When solving physics problems, breaking down the motion into components and setting up equations based on those components can simplify finding the solution.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Vectors
Relative Motion

Formulas

v_{bx} = v_b \cos \theta
v_{by} = v_b \sin \theta
(v_b \cos \theta - v_c) t = d_u
v_b \sin \theta t = d_s

Theorems

Trigonometric functions in right triangles
Relative velocity in two dimensions

Suitable Grade Level

Grades 10-12