Problem 85 - Algebra and Trigonometry

Question

A river has a constant current of \(3 \mathrm{~km} / \mathrm{h}\). At what angle to a boat dock should a motorboat capable of maintaining a constant speed of \(20 \mathrm{~km} / \mathrm{h}\) be headed in order to reach a point directly opposite the dock? If the river is \(\frac{1}{2}\) kilometer wide, how long will it take to cross?

Step-by-Step Solution

Answer

The angle \(\theta\) is \(\cos^{-1}\left(\frac{3}{20}\right)\) and it takes about 0.025 hours to cross the river.

Step 1: Analyze given information

The velocity of the river current is given as \(3 \mathrm{~km/h}\). The motorboat's speed is \(20 \mathrm{~km/h}\), and the width of the river is \(\frac{1}{2}\) kilometer. The boat needs to head at an angle to counteract the river's current to reach the point directly opposite the dock.

Step 2: Establish the coordinate system

Consider the river flow along the x-axis with the dock at the origin. The boat should be headed at an angle \(\theta\) to the shore, making an angle with the y-axis.

Step 3: Break down the boat's velocity

Let the boat's velocity be \(\vec{v}_b = 20 \mathrm{~km/h}\). Decompose the velocity into components. The x-component of the boat's velocity due to heading against the current will be \(20 \cos \theta\), and the y-component will be \(20 \sin \theta\).

Step 4: Set up the equation to counteract the current

To reach the point directly opposite, the x-component of the boat’s velocity must counteract the river current. Therefore, \(20 \cos \theta = 3\).

Step 5: Solve for \(\theta\)

Solve for the angle \(\theta\): \[ \cos \theta = \frac{3}{20} \]Thus, \[\theta = \cos^{-1}\left(\frac{3}{20}\right)\].

Step 6: Calculate the time to cross the river

The y-component of the boat’s velocity is responsible for crossing the river width. Use \[\frac{1}{2} = 20 \sin(\theta) \times t\] where \(t\) is the time to cross. First, compute \(\sin(\theta)\) using \(\sin^2 \theta + \cos^2 \theta = 1\).

Step 7: Find \(\sin(\theta)\)

Calculate \[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{3}{20}\right)^2} = \sqrt{1 - \frac{9}{400}} = \sqrt{\frac{391}{400}} = \frac{\sqrt{391}}{20} \].

Step 8: Solve for time

Substitute \(\sin(\theta)\) back into the equation: \[ \frac{1}{2} = 20 \cdot \frac{\sqrt{391}}{20} \cdot t \]This simplifies to: \[ \frac{1}{2} = \sqrt{391} \cdot t \]Therefore, \[ t = \frac{1}{2 \sqrt{391}} \approx 0.025 \text{ hours} \].

Key Concepts

Vector Decomposition

In physics, vector decomposition is a method used to break down a vector into its component parts. In this problem, the boat's velocity is decomposed into horizontal and vertical components. The boat's total velocity is given as 20 km/h. We focus on two parts:

• The horizontal (x-component) which counters the river current.
  
• The vertical (y-component) which helps the boat cross the river's width.
  

These components can be derived using trigonometric functions. Decomposing the vector helps in solving the problem efficiently.

Trigonometric Functions

Trigonometric functions are mathematical functions of angles that relate the angles to the lengths of sides of a right triangle. In our scenario, the boat must counter the river's current by heading at an angle \(\theta\) to the shore. We use cosine (cos) and sine (sin) to find the x and y components of the boat's velocity:

• The x-component: \(20 \cos \theta\)
  
• The y-component: \(20 \sin \theta\)
  

The cosine function helps determine the x-component, which must balance the river current, while the sine function helps us find the y-component needed to traverse the river.

Relative Velocity

Relative velocity refers to the velocity of an object as observed from a particular frame of reference. The problem involves the boat's velocity relative to the riverbank.

• The boat's speed in still water is 20 km/h.
  
• The river current is moving at 3 km/h.
  

To reach directly opposite the starting point, the boat's relative velocity in the x-direction must be zero. This requires countering the river's velocity of 3 km/h by controlling the direction and speed of the boat. Hence, the equation \(20 \cos \theta = 3\) helps achieve this balance.

Time Calculation

Calculating the time to cross the river involves the y-component of the boat's velocity. We know the river's width is 0.5 km. Using the formula for distance:
\[ d = v \times t \] Here,

• \(d\) is the river's width: 0.5 km,
  
• \(v\) is the boat's y-component of velocity: \(20 \sin \theta\)
  

From earlier steps, we calculated \(\sin \theta = \frac{\sqrt{391}}{20}\). Substituting this back:
\[0.5 = 20 \times \frac{\sqrt{391}}{20} \times t \] Simplifying, we get \(t = 0.025\) hours. This is the time the boat takes to cross the river.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to find the angle when trigonometric values are known. In this case, we use the inverse cosine function to calculate \(\theta\). Given:
\[-\cos \theta = \frac{3}{20} \] We solve for \(\theta \) using the inverse cosine function:
\[\theta = \cos^{-1}\left(\frac{3}{20}\right)\] This angle ensures that the boat is headed appropriately to counteract the river's current. Using \(\theta\), we find the required components of velocity and time to cross the river. It allows us to determine the precise heading required for the boat to efficiently accomplish its task.