A passenger walks from one side of a ferry to the other as it approaches a dock.Passenger's velocity is 1.60 due north relative to the ferry, and 4.50 at an angle of 30.0 west of north relative to the water.
What is the magnitude of the ferry's velocity relative to the water?
What is the direction of the ferry's velocity relative to the water?
How exactly do I do this? I've tried tan^-1(2400/1400) but I get it wrong.... =-(
To solve this problem, you can use vector addition to determine the velocity of the ferry relative to the water. The magnitude and direction of the ferry's velocity can be calculated using trigonometry.
Step 1: Draw a diagram illustrating the vectors involved. Let's label the velocity of the passenger relative to the ferry as vector A and the velocity of the passenger relative to the water as vector B.
Step 2: Resolve vector A into its northward and westward components. The northward component is given as 1.60, and the westward component can be calculated as 4.50 * cos(30°).
Step 3: Now, find the sum of the northward components and the sum of the westward components. This will give you the total northward and westward components of the ferry's velocity relative to the water.
Step 4: Use the Pythagorean theorem to calculate the magnitude of the ferry's velocity relative to the water by taking the square root of the sum of the squares of the northward and westward components.
Step 5: Use trigonometry to find the direction of the ferry's velocity relative to the water. You can do this by finding the arctangent of the northward component divided by the westward component.
Now, let's calculate the magnitude and direction of the ferry's velocity relative to the water:
Step 1: Draw a diagram.
-> B (water)
/
A (ferry)->
Step 2: Resolving vector A into components:
- The northward component = 1.60
- The westward component = 4.50 * cos(30°) = 3.897
Step 3: Summing the components:
- The total northward component = 1.60
- The total westward component = 3.897
Step 4: Calculating the magnitude:
- Magnitude of the ferry's velocity = sqrt((total northward component)^2 + (total westward component)^2)
= sqrt((1.60)^2 + (3.897)^2)
= sqrt(2.56 + 15.187209)
= sqrt(17.747209)
≈ 4.207
Step 5: Calculating the direction:
- Direction of the ferry's velocity = arctan( (total northward component) / (total westward component) )
≈ arctan(1.60 / 3.897)
≈ 23.045°
So, the magnitude of the ferry's velocity relative to the water is approximately 4.207, and the direction is approximately 23.045° west of north.
Keep in mind that your previous attempt using the tangent function likely resulted in an incorrect answer because you were missing the step of resolving the vectors into their components.
To find the magnitude and direction of the ferry's velocity relative to the water, we will use vector addition.
First, let's break down the passenger's velocity into its horizontal (x) and vertical (y) components:
Given:
Passenger's velocity due north (Vy) = 1.60 m/s
Passenger's velocity at an angle of 30.0° west of north (Vw) = 4.50 m/s
Vertical component of passenger's velocity (Vy) = 1.60 m/s
Horizontal component of passenger's velocity (Vx) = Vw * cos(30°) = 4.50 m/s * cos(30°) = 3.897 m/s
Now, let's consider the ferry's velocity relative to the water:
The vertical component of the ferry's velocity (Vfy) will be the same as the passenger's vertical component (Vy) because they are moving together in the same direction.
The horizontal component of the ferry's velocity (Vfx) will be the sum of the passenger's horizontal component (Vx) and the velocity of the ferry relative to the land (Vfl):
Vfx = Vx + Vfl
The magnitude of the ferry's velocity relative to the water (Vf) can be found using the Pythagorean theorem:
Vf = √(Vfx^2 + Vfy^2)
Using the given values and calculations:
Vf = √(3.897^2 + 1.60^2) = √(15.1849 + 2.56) = √17.7449 = 4.21 m/s
The magnitude of the ferry's velocity relative to the water is approximately 4.21 m/s.
To find the direction of the ferry's velocity relative to the water, we can use trigonometry. The angle (θ) between the vertical component of the ferry's velocity and the positive x-axis can be calculated using the inverse tangent function:
θ = tan^(-1)(Vfy / Vfx)
θ = tan^(-1)(1.60 / 3.897) = tan^(-1)(0.41) ≈ 22.7°
Therefore, the direction of the ferry's velocity relative to the water is approximately 22.7° north of west.
So, to recap:
Magnitude of the ferry's velocity relative to the water = 4.21 m/s
Direction of the ferry's velocity relative to the water = approximately 22.7° north of west
If you previously tried tan^-1(2400/1400), it appears there was a misunderstanding or mistake in the calculations. Make sure to use the correct values and follow the steps provided above for an accurate answer.
Both velocities are vectors.
You could resolve each velocity into it's x- and y- components, which usually coincide with the East and North directions.
The velocity relative to water is the resultant, and the velocity of the ferry is yet unknown.
Velocity x-component y-component
1.6 0 1.6
? ? ?
------------------------------------
resultant =
4.5 4.5*cos(120) 4.5*sin(120)
From the table, it should be possible to find the individual components of the ferry by subtraction, and the velocity by vectorially summing the two components.
Post your answer for a check if necessary.