Question

A canoe has a velocity of 0.40 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

Short answer

The velocity of the canoe relative to the river is approximately 0.356 m/s, 51.3° south of west.

Step-by-step solution

Understand the Problem

We need to find the velocity of the canoe relative to the river. We have two given velocities: the velocity of the canoe relative to the earth, and the velocity of the river relative to the earth. We will use vector subtraction to find the velocity of the canoe relative to the river.

Break Down Velocities into Components

First, break down the velocity of the canoe relative to the earth (0.40 m/s southeast) into its east and south components. Since it's moving southeast, it contributes equally to both east and south directions. Thus, the components are:East component of the canoe = \(0.40 \times \cos(45^\circ) = 0.28\, \text{m/s}\).South component of the canoe = \(0.40 \times \sin(45^\circ) = 0.28\, \text{m/s}\).The velocity of the river is entirely in the east direction: 0.50 m/s east.

Calculate Relative Velocity Components

To find the velocity of the canoe relative to the river, perform vector subtraction on the east components:East component (canoe to river) = East component of canoe - East component of river = \(0.28 - 0.50 = -0.22\, \text{m/s}\).The south component remains the same as it's not affected by the river's flow:South component (canoe to river) = 0.28 m/s.

Calculate Magnitude of Relative Velocity

Use the Pythagorean theorem to find the magnitude of the canoe's velocity relative to the river:\(\text{Magnitude} = \sqrt{(-0.22)^2 + 0.28^2} = \sqrt{0.0484 + 0.0784} = \sqrt{0.1268} \approx 0.356\, \text{m/s}\).

Determine Direction of Relative Velocity

Calculate the direction using the tangent function:\(\theta = \tan^{-1}\left(\frac{0.28}{-0.22}\right)\).\(\theta \approx \tan^{-1}(-1.27) \approx -51.3^\circ \).Since the angle is measured from the east axis towards the south, the direction is approximately \(51.3^\circ\) south of west.

Key concepts

Vector Subtraction

In physics, particularly in the context of relative velocity, vector subtraction is a crucial method for comparing motion. To find the velocity of one object relative to another, you subtract the velocity vectors. Imagine velocities as arrows pointing in certain directions. The length tells you the speed, and the direction shows where it's headed. For example, if a canoe moves southeast, and a river flows east, you can visualize these movements as two arrows. To find out how fast and in which direction the canoe moves relative to the river, the arrows are subtracted. This process finds the **relative velocity** by eliminating the common motion with respect to a fixed point, like Earth. This subtraction gives us a new vector, showing us how the canoe would move if the river were still.

Velocity Components

Breaking down velocities into components simplifies complex vector calculations. When dealing with diagonal or angled motion, such as southeast, it's essential to think in terms of basic directions like east and south. This involves using trigonometry to split the velocity into parts:

• The **east component**, found using cosine, represents rightward movement.
• The **south component**, given by sine, indicates downward movement.

For example, the canoe traveling 0.40 m/s southeast has equal parts in both east and south directions. Calculating these, you'd have both components as 0.28 m/s. Understanding these components makes it easier to add or subtract velocities, as seen in our river example.

Pythagorean Theorem

The Pythagorean theorem is your go-to tool for finding the magnitude of a resultant vector in two dimensions. Once you've determined the individual east and south components of a velocity, you need to find how fast the object is moving in total.The theorem works by treating the east and south components as two sides of a right triangle, where the hypotenuse is the actual velocity we're interested in. Using the formula:\[\text{Magnitude} = \sqrt{(\text{East Component})^2 + (\text{South Component})^2}\]Plug in the values, and for our canoe example, the magnitude comes to approximately 0.356 m/s. This equation gives a precise calculation of the overall speed, regardless of the direction.

Direction Calculation

Determining direction is crucial once you know the magnitude of velocity. Direction tells you where an object is heading relative to a reference point, which is often as critical as knowing the speed.To calculate direction, trigonometry comes into play once more, particularly the tangent function. Here, you use the ratio between the south and east components of velocity to find the angle relative to a reference direction (usually east, or zero degrees):\[\theta = \tan^{-1}\left(\frac{\text{South Component}}{\text{East Component}}\right)\]For example, in the canoe scenario, this calculation reveals the direction is approximately \(-51.3^\circ\). Interpreted in a standard navigational context, it's measured *from* the east, heading towards west, indicating an angle south of west. Clear calculation and understanding of direction help in navigation and aligning with real-world conditions.