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Magazine Chapter 3: Problem 35
A canoe has a velocity of \(0.40 \mathrm{~m} / \mathrm{s}\) southeast relative to the earth. The canoe is on a river that is flowing \(0.50 \mathrm{~m} / \mathrm{s}\) east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.
Short Answer
Expert verified
The canoe's velocity relative to the river is approximately 0.36 m/s at 52掳 southwest.
Step by step solution
01
Understand the Problem
We need to find the velocity of the canoe relative to the river. This requires us to subtract the velocity of the river from the velocity of the canoe.
02
Break Down Velocities to Components
The velocity of the canoe is 0.40 m/s southeast, which can be broken down into its east and south components using trigonometric functions. Assuming southeast is 45掳, the east and south components become 0.40 m/s * cos(45掳) = 0.40 m/s * 0.7071 = 0.2828 m/s each.
03
Expand River Velocity into Components
The river flows 0.50 m/s east relative to the earth. Hence, the river's velocity components are 0.50 m/s east and 0 m/s south.
04
Subtract the River's Velocity Components from Canoe's Velocity Components
Subtract the east component of the river's velocity from the east component of the canoe's velocity: 0.2828 m/s - 0.50 m/s = -0.2172 m/s (west direction). The south component remains unchanged at 0.2828 m/s.
05
Calculate the Magnitude of the Resultant Velocity
Use the Pythagorean theorem to calculate the resultant magnitude: \(v = \sqrt{(-0.2172)^2 + (0.2828)^2}\). This equals approximately 0.36 m/s.
06
Determine the Direction
The direction \( \theta \) can be found using \( \theta = \tan^{-1}\left(\frac{0.2828}{0.2172}\right)\), which gives approximately 52掳 southwest.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Components
When dealing with velocities in directions that aren't purely along the axes, it's important to break them down into vector components. This means we express the velocity in terms of two perpendicular directions, usually along the x-axis (east-west) and y-axis (north-south). This helps in simplifying calculations especially when dealing with angles.
- For our case, the canoe's velocity is given as 0.40 m/s towards the southeast.
- Imagine a vector pointing southeast; it equally divides between east and south.
- We use trigonometric functions, like cosine and sine, to break down this vector into components that lie along the east (x) and south (y) directions.
- Using \(\cos(45^\circ)\) and \(\sin(45^\circ)\), both components for the canoe's velocity are equal and can be calculated as \(0.40 \text{ m/s} \times 0.7071 = 0.2828 \text{ m/s}\).
Velocity Addition
To find how fast one thing is moving relative to another, velocity addition or subtraction is used. In this scenario, we want the canoe's velocity in relation to the river. This requires the concept of relative velocity, which is derived by vector addition and subtraction.
- First, recognize the velocity's existence in the real-world context: the river moves 0.50 m/s east, and the canoe 0.40 m/s southeast.
- Since the river affects the canoe's perceived motion, subtracting the river's components from the canoe's gives the velocity relative to the river.
- Subtract 0.50 m/s east from the canoe's east component 0.2828 m/s, resulting in \(-0.2172 \text{ m/s}\), which indicates a westward direction.
- The south component remains unaffected at 0.2828 m/s.
Trigonometric Functions
Trigonometric functions, like sine and cosine, are essential for resolving vector components when angles are involved. For directions like southeast, which are not purely east, west, north, or south, using these functions simplifies component calculations significantly.
- Consider an angle of 45掳 for southeast.
- For the canoe's velocity, \(\cos(45^\circ)\) and \(\sin(45^\circ)\) allow the conversion of the original velocity into east and south components respectively.
- Due to trigonometric symmetry, \(\cos(45^\circ) = \sin(45^\circ) = 0.7071\).
- Each vector component for the canoe鈥檚 direction can be calculated as \(0.40 \text{ m/s} \times 0.7071 = 0.2828 \text{ m/s}\).
Pythagorean Theorem
The Pythagorean theorem is an enduring concept in geometry, extending directly into the realm of physics. It provides a method for calculating the magnitude of a resultant vector from its components.
- Consider the subtracted components from the canoe's motion relative to the river: westward (-0.2172 m/s) and southward (0.2828 m/s).
- According to the Pythagorean theorem, the magnitude of the canoe's velocity vector relative to the river can be found using \[v = \sqrt{(-0.2172)^2 + (0.2828)^2}\].
- Solving this equation gives approximately 0.36 m/s.
- This provides not only the quantitative speed but also reaffirms the careful component breakdown and relative evaluation performed earlier.
