A ferry is crossing a river. The ferry is headed
due north with a speed of 1.7 m/s relative to
the water and the river’s velocity is 2.4 m/s
to the east.
a) What is magnitude of the boat’s velocity
relative to Earth?
Answer in units of m/s
b) Find the direction in which the ferry is
moving (measured from due east, with counterclockwise positive).
Answer in units of
◦
X = Hor. = 2.4 m/s.
Y = Ver. = 1.7 m/s.
a. Vb^2 = X^2 + Y^2.
Vb^2 = 5.76 + 2.89 = 8.65
Vb = 2.94 m/s. = Velocity of boat.
b. tanA = Y/X = 1.7/2.4 = 0.70833
A = 35.3o,CCW.
a) The magnitude of the boat's velocity relative to Earth can be found using the Pythagorean theorem. We have a right triangle formed by the boat's velocity relative to the water (1.7 m/s north) and the river's velocity (2.4 m/s east).
The magnitude of the boat's velocity relative to Earth, v, can be calculated as follows:
v = √[(1.7 m/s)² + (2.4 m/s)²]
v ≈ √(2.89 m²/s² + 5.76 m²/s²)
v ≈ √8.65 m²/s²
v ≈ 2.94 m/s
Therefore, the magnitude of the boat's velocity relative to Earth is approximately 2.94 m/s.
b) To find the direction in which the ferry is moving, we can use inverse tangent (arctan) to calculate the angle formed by the northward velocity (1.7 m/s) and eastward velocity (2.4 m/s).
The direction, θ, can be calculated as follows:
θ = tan^(-1)[(1.7 m/s) / (2.4 m/s)]
θ ≈ tan^(-1)(0.708)
θ ≈ 34.99°
Therefore, the ferry is moving in a direction of 34.99° (measured counterclockwise from due east).
To solve this problem, we can use vector addition. The boat's velocity relative to Earth is the vector sum of its velocity relative to the water and the velocity of the water itself.
a) To find the magnitude of the boat's velocity relative to Earth, we can use the Pythagorean theorem. The magnitude can be found using the formula:
|v| = sqrt((v1)^2 + (v2)^2)
where v1 is the magnitude of the boat's velocity relative to the water, and v2 is the magnitude of the river's velocity.
Given:
v1 = 1.7 m/s (magnitude of the boat's velocity relative to the water)
v2 = 2.4 m/s (magnitude of the river's velocity)
Using the formula, we have:
|v| = sqrt((1.7)^2 + (2.4)^2)
|v| = sqrt(2.89 + 5.76)
|v| = sqrt(8.65)
|v| ≈ 2.94 m/s
Therefore, the magnitude of the boat's velocity relative to Earth is approximately 2.94 m/s.
b) To find the direction in which the ferry is moving relative to Earth (measured from due east), we can use trigonometry. We can find the angle between the boat's velocity and the horizontal by using the equation:
θ = arctan(v2/v1)
Given:
v1 = 1.7 m/s (magnitude of the boat's velocity relative to the water)
v2 = 2.4 m/s (magnitude of the river's velocity)
Using the equation, we have:
θ = arctan(2.4/1.7)
θ ≈ 55.8°
Therefore, the direction in which the ferry is moving relative to Earth (measured from due east, with counterclockwise positive) is approximately 55.8°.
To solve the question, we can use vector addition to find the boat's velocity relative to Earth.
a) The boat's velocity relative to Earth can be found by adding the velocity of the boat relative to the water and the velocity of the river.
Given:
Velocity of the boat relative to the water = 1.7 m/s (due north)
Velocity of the river = 2.4 m/s (to the east)
To find the magnitude of the boat's velocity relative to Earth, we use the Pythagorean theorem:
Magnitude of the boat's velocity = √(velocity of the boat relative to the water)^2 + (velocity of the river)^2
Magnitude of the boat's velocity = √(1.7^2 + 2.4^2) = √(2.89 + 5.76) = √8.65 ≈ 2.94 m/s
Therefore, the magnitude of the boat's velocity relative to Earth is approximately 2.94 m/s.
b) To find the direction in which the ferry is moving, we use trigonometry. Using the given velocities, we can form a right triangle with the boat's velocity relative to the water as the vertical side and the velocity of the river as the horizontal side.
The angle θ that the boat's velocity makes with the east direction can be found using the tangent function:
tan(θ) = (velocity of the boat relative to the water) / (velocity of the river)
tan(θ) = 1.7 / 2.4
Taking the inverse tangent (arctan) of both sides, we find:
θ ≈ 36.87°
Therefore, the direction in which the ferry is moving, measured from due east with counterclockwise positive, is approximately 36.87°.