A ferry is crossing a river.The ferry is headed due north with a speed of 2.5 m/s relative to the water and the river's velocity is 2.0 m/s to the east. What is the magnitude and direction of the ferry's velocity relative to Earth.

( Give the angle from east 0 degrees )

Ferry velocity with respect to earth =

Ferry velocity with respect to water + water velocity with respect to earth

magnitude = sqrt[2.5^2 + 2.0^2]
= 3.20 m/s

direction = arctan(2.5/2)
= 51.3 degrees north of east.

To find the magnitude and direction of the ferry's velocity relative to Earth, we can use vector addition.

Let's consider the ferry's velocity relative to the water as vector v1, which is headed due north at a speed of 2.5 m/s. The river's velocity, vector v2, is 2.0 m/s to the east.

To find the resulting velocity vector, we need to add these two vectors.

First, we need to find the x-component and y-component of each vector.

v1 has zero x-component and a positive y-component of 2.5 m/s (since it is headed due north).

v2 has a positive x-component of 2.0 m/s (since it is to the east) and zero y-component.

Now, let's add the x-components and y-components separately.

The resulting x-component is 2.0 m/s (the sum of the x-components of v1 and v2).

The resulting y-component is 2.5 m/s (the sum of the y-components of v1 and v2).

Using these components, we can calculate the magnitude and direction of the resulting velocity vector.

The magnitude (speed) is given by the formula:
magnitude = sqrt((x-component)^2 + (y-component)^2)

magnitude = sqrt((2.0 m/s)^2 + (2.5 m/s)^2)
magnitude = sqrt(4.0 m^2/s^2 + 6.25 m^2/s^2)
magnitude = sqrt(10.25 m^2/s^2)
magnitude ≈ 3.2 m/s

The direction can be found using trigonometry.

The direction (angle) θ can be calculated using the formula:
θ = atan(y-component / x-component)

θ = atan(2.5 m/s / 2.0 m/s)
θ = atan(1.25)
θ ≈ 51.34 degrees

Therefore, the magnitude of the ferry's velocity relative to Earth is approximately 3.2 m/s, and its direction is approximately 51.34 degrees east of north.

To find the magnitude and direction of the ferry's velocity relative to Earth, we can use vector addition.

First, let's break down the velocities into their x and y components. The ferry's velocity relative to water is purely in the north direction, so its x-component is 0 m/s (since it is moving purely in the y-direction).

The river's velocity is purely in the east direction, so its y-component is 0 m/s (since it is moving purely in the x-direction).

Now, we add these components to find the ferry's velocity relative to Earth:

x-component of ferry's velocity relative to Earth = x-component of ferry's velocity relative to water + x-component of river's velocity
y-component of ferry's velocity relative to Earth = y-component of ferry's velocity relative to water + y-component of river's velocity

Since the x-component of the ferry's velocity relative to water is 0 m/s and the y-component of the river's velocity is 0 m/s, the x-component of the ferry's velocity relative to Earth is also 0 m/s.

The y-component of the ferry's velocity relative to Earth is the sum of the y-component of the ferry's velocity relative to water (2.5 m/s) and the y-component of the river's velocity (0 m/s), which is 2.5 m/s.

So, the magnitude of the ferry's velocity relative to Earth is the square root of the sum of the squares of its x and y components:

magnitude = sqrt((x-component)^2 + (y-component)^2)
= sqrt((0 m/s)^2 + (2.5 m/s)^2)
= sqrt(0 + 6.25)
= sqrt(6.25)
≈ 2.5 m/s

Since the x-component is 0 m/s, the angle from east (0 degrees) is the same as the angle from the y-axis. We can find this angle using trigonometry:

angle = arctan(y-component/x-component)
= arctan(2.5 m/s/0 m/s)
= arctan(undefined)

Here, the angle is undefined because the x-component is zero. This means the ferry's velocity relative to Earth is purely in the y-direction, or due north.

Therefore, the magnitude of the ferry's velocity relative to Earth is approximately 2.5 m/s, and its direction is due north.