A powerboat heads due northwest at 11 m/s relative to the water across a river that flows due north at 5.9 m/s. What is the velocity (both magnitude and direction) of the motorboat relative to the shore?
To find the velocity of the powerboat relative to the shore, we need to consider the vector addition of the velocity of the powerboat relative to the water and the velocity of the river.
Let's break down the velocities into their components:
Velocity of the powerboat relative to the water, VBw = 11 m/s
Velocity of the river, VR = 5.9 m/s
Since the powerboat is heading due northwest, we can decompose the velocity of the powerboat relative to the water into its northward and westward components.
The northward component of the powerboat's velocity will be the same as that of the river, since both are moving in the same direction. Therefore, the northward component of the velocity is 5.9 m/s.
The westward component of the powerboat's velocity will be the difference between the powerboat's velocity and the river's velocity. Since the powerboat is heading northwest, it is moving against the river's flow.
To find the westward component, we can use the Pythagorean theorem on the right triangle formed by the westward, northward, and total velocity components.
The magnitude of the powerboat's westward velocity component is given by:
VBw-west = √((VBw)^2 - (VR)^2)
VBw-west = √((11 m/s)^2 - (5.9 m/s)^2)
VBw-west ≈ √(121 m^2/s^2 - 34.81 m^2/s^2)
VBw-west ≈ √(86.19 m^2/s^2)
VBw-west ≈ 9.29 m/s
Now, to determine the magnitude and direction of the velocity of the motorboat relative to the shore, we can use the Pythagorean theorem again to find the magnitude of the total velocity:
V = √((VBw-west)^2 + (VBw-north)^2)
V = √((9.29 m/s)^2 + (5.9 m/s)^2)
V ≈ √(86.44 m^2/s^2 + 34.81 m^2/s^2)
V ≈ √(121.25 m^2/s^2)
V ≈ 11.01 m/s
The magnitude of the motorboat's velocity relative to the shore is approximately 11.01 m/s.
To find the direction, we can use trigonometry. The angle θ can be found by taking the inverse tangent of the northward component divided by the westward component:
θ = arctan((VBw-north) / (VBw-west))
θ = arctan(5.9 m/s / 9.29 m/s)
θ ≈ arctan(0.634)
θ ≈ 32.07°
Therefore, the direction of the motorboat's velocity relative to the shore is approximately 32.07° west of north.
To determine the velocity of the powerboat relative to the shore, we need to use vector addition.
1. Identify the given velocities:
- Velocity of the powerboat relative to the water = 11 m/s due northwest (let's call this velocity A)
- Velocity of the river relative to the shore = 5.9 m/s due north (let's call this velocity B)
2. Draw a diagram:
- Draw a line to represent the direction and magnitude of velocity A (11 m/s due northwest).
- Draw another line to represent the direction and magnitude of velocity B (5.9 m/s due north).
3. Add the vectors geometrically:
- Place the tail of B at the head of A (since we are adding B to A).
- Connect the tail of A to the head of B with a straight line. This line represents the resultant vector, which is the velocity of the motorboat relative to the shore.
4. Measure the magnitude and direction of the resultant vector:
- Use a ruler to determine the length of the resultant vector line.
- Measure the angle between the resultant vector line and the north direction.
5. Calculate the magnitude and direction of the resultant vector:
- Use the ruler measurement to determine the magnitude of the resultant vector.
- Use the protractor or trigonometric functions to determine the direction (angle) of the resultant vector relative to the north direction.
By following these steps, you should be able to determine the velocity (both magnitude and direction) of the motorboat relative to the shore.