As an aid in working this problem, consult Interactive Solution 3.62. A ferryboat is traveling in a direction 32.0 ° north of east with a speed of 5.38 m/s relative to the water. A passenger is walking with a velocity of 1.50 m/s due east relative to the boat. What is (a) the magnitude and (b) the direction of the velocity of the passenger with respect to the water? Give the directional angle relative to due east.

To find the magnitude and direction of the velocity of the passenger with respect to the water, we can apply vector addition.

Step 1: Identify the given information.
- The velocity of the ferryboat relative to the water is traveling at 5.38 m/s in a direction 32.0° north of east.
- The velocity of the passenger relative to the boat is 1.50 m/s due east.

Step 2: Draw a diagram.
Draw a diagram to visualize the problem. Place the velocity vector of the ferryboat pointing 32.0° north of east. Then draw the velocity vector of the passenger pointing due east. The vector we are interested in finding is the resultant vector, which represents the velocity of the passenger with respect to the water.

Step 3: Resolve the vectors into components.
Convert the given information into the components using trigonometry.

For the velocity of the ferryboat:
- Velocity_x = 5.38 m/s * cos(32.0°)
- Velocity_y = 5.38 m/s * sin(32.0°)

For the velocity of the passenger:
- Velocity_x = 1.50 m/s
- Velocity_y = 0 m/s (since the passenger is walking due east)

Step 4: Add the respective components.
Add the x-components and y-components separately to obtain the resultant vector.

Resultant_x = Velocity_x(ferryboat) + Velocity_x(passenger)
Resultant_y = Velocity_y(ferryboat) + Velocity_y(passenger)

Step 5: Find the magnitude and direction of the resultant vector.
Use the Pythagorean theorem to find the magnitude:
Magnitude = sqrt(Resultant_x^2 + Resultant_y^2)

Use inverse tangent to find the direction (angle):
Direction_angle = arctan(Resultant_y / Resultant_x)

Step 6: Substitute the known values and solve.
Substitute the known values into the equations and calculate the magnitude and direction.

Magnitude = sqrt((5.38 m/s * cos(32.0°) + 1.50 m/s)^2 + (5.38 m/s * sin(32.0°) + 0 m/s)^2)

Direction_angle = arctan((5.38 m/s * sin(32.0°) + 0 m/s) / (5.38 m/s * cos(32.0°) + 1.50 m/s))

Step 7: Calculate the answers.
Using a calculator, compute the values for magnitude and direction_angle.

Magnitude ≈ 6.32 m/s
Direction_angle ≈ 39.7°

So, (a) the magnitude of the velocity of the passenger with respect to the water is approximately 6.32 m/s, and (b) the direction of the velocity is approximately 39.7° relative to due east.

I feel like you are in Dr. W class at ABU