A ferry is crossing a river. The ferry is headed
due north with a speed of 2.8 m/s relative to
the water and the river’s velocity is 4.9 m/s
to the east.
a) What is magnitude of the boat’s velocity
relative to Earth
Add the two vectors. The two vectors (ferry rel. to water and water rel. to land) are at right angles, so the magnitude is the hypotenuse.
|V| = 5.64 m/s
To find the magnitude of the boat's velocity relative to Earth, we can use the concept of vector addition. The boat's velocity relative to Earth is the sum of its velocity relative to the water and the velocity of the river.
Given:
Velocity of the boat relative to water = 2.8 m/s (due north)
Velocity of the river = 4.9 m/s (to the east)
To find the magnitude of the boat's velocity relative to Earth, we can use the Pythagorean theorem. The magnitude of a vector is calculated by finding the square root of the sum of the squares of its components.
Let's break down the given velocities into their x and y components:
Velocity of the boat relative to water:
- x-component = 0 m/s (since it is moving north)
- y-component = 2.8 m/s (magnitude)
Velocity of the river:
- x-component = 4.9 m/s (magnitude)
- y-component = 0 m/s (since it is moving east)
Now, add the x-components and y-components separately:
x-component = 0 m/s + 4.9 m/s = 4.9 m/s
y-component = 2.8 m/s + 0 m/s = 2.8 m/s
Using the Pythagorean theorem:
Magnitude of the boat's velocity relative to Earth = √(x-component^2 + y-component^2)
Magnitude = √(4.9^2 + 2.8^2)
Magnitude = √(24.01 + 7.84)
Magnitude = √31.85
Therefore, the magnitude of the boat's velocity relative to Earth is approximately 5.65 m/s.