A ferry is crossing a river. The ferry is headed
due north with a speed of 2.3 m/s relative to
the water and the river’s velocity is 3.4 m/s
to the east.
a) What is magnitude of the boat’s velocity
relative to Earth?
Answer in units of m/s.
024 (part 2 of 2) 10.0 points
b) Find the direction in which the ferry is
moving (measured from due east, with counterclockwise positive).
Answer in units of ◦
.
a) The magnitude of the boat's velocity relative to Earth can be found using the Pythagorean theorem. We can consider the boat's velocity relative to the water as one vector and the river's velocity as another vector.
Let's call the boat's velocity relative to the water Vb and the river's velocity Vr.
Vb = 2.3 m/s (due north)
Vr = 3.4 m/s (to the east)
Using the Pythagorean theorem, the magnitude of the boat's velocity relative to Earth (Ve) is given by:
Ve = √(Vb^2 + Vr^2)
Plugging in the values:
Ve = √(2.3^2 + 3.4^2) ≈ √(5.29 + 11.56) ≈ √16.85 ≈ 4.10 m/s
Hence, the magnitude of the boat's velocity relative to Earth is approximately 4.10 m/s.
b) To find the direction in which the ferry is moving, we can use trigonometry. Since the river's velocity is to the east, it forms an angle of 90 degrees with the north direction.
Let's call the angle between the boat's velocity relative to Earth and the east direction θ.
Using trigonometry, we can find:
tan(θ) = Vr / Vb = 3.4 / 2.3 ≈ 1.48
Taking the inverse tangent (arctan) of both sides:
θ ≈ arctan(1.48) ≈ 55.6 degrees
Since the question asks for the direction measured from due east, we subtract this angle from 90 degrees:
Direction = 90 - 55.6 ≈ 34.4 degrees
Therefore, the direction in which the ferry is moving, measured from due east with counterclockwise positive, is approximately 34.4 degrees.
To solve this problem, we can use vector addition to find the velocity of the ferry relative to the Earth.
a) To find the magnitude of the boat's velocity relative to Earth, we can use the Pythagorean theorem:
Magnitude of boat's velocity = √((velocity of boat relative to water)² + (velocity of river)²)
First, let's calculate the magnitude of the boat's velocity:
velocity of boat relative to water = 2.3 m/s
velocity of river = 3.4 m/s
Magnitude of boat's velocity = √((2.3 m/s)² + (3.4 m/s)²)
Magnitude of boat's velocity = √(5.29 + 11.56)
Magnitude of boat's velocity = √16.85
Magnitude of boat's velocity ≈ 4.1 m/s
So, the magnitude of the boat's velocity relative to Earth is approximately 4.1 m/s.
b) To find the direction in which the ferry is moving, we can use trigonometry. We need to calculate the angle between the boat's velocity and the x-axis (east) in a counterclockwise direction.
Let's consider the boat's velocity relative to the Earth as the resultant vector. We can break it down into its x and y components.
velocity of boat relative to Earth (x-component) = velocity of boat relative to water = 2.3 m/s
velocity of boat relative to Earth (y-component) = velocity of river = 3.4 m/s
To find the angle, we can use the tangent function:
tanθ = (velocity of boat relative to Earth (y-component)) / (velocity of boat relative to Earth (x-component))
θ = arctan((velocity of boat relative to Earth (y-component)) / (velocity of boat relative to Earth (x-component)))
θ = arctan(3.4 m/s / 2.3 m/s)
Using a calculator, we find:
θ ≈ 55.3°
So, the direction in which the ferry is moving, measured from due east in a counterclockwise direction, is approximately 55.3°.