Final answer:
The velocity of the passenger with respect to the water is 6.84 m/s in a direction 37.1◦ east of south.
Step-by-step explanation:
To determine the velocity of the passenger with respect to the water, we need to consider the relative velocities of the ferryboat and the passenger.
The ferryboat is traveling in a direction 38.0◦ north of east with a speed of 5.50 m/s relative to the water.
The passenger is walking with a velocity of 2.50 m/s due east relative to the boat.
We can use vector addition to find the resultant velocity of the passenger with respect to the water.
Since the passenger is moving due east and the boat is traveling in a direction 38.0◦ north of east, we can break down the boat's velocity into its horizontal and vertical components:
Vx = Vboat * cos(38.0◦) = 5.50 m/s * cos(38.0◦)
= 4.34 m/s
Vy = Vboat * sin(38.0◦) = 5.50 m/s * sin(38.0◦)
= 3.36 m/s
The passenger's velocity relative to the water is then:
Vpassenger = Vx + Vpassenger = 4.34 m/s + 2.50 m/s
= 6.84 m/s
The magnitude of the passenger's velocity with respect to the water is 6.84 m/s. The direction can be found using trigonometry:
θ = tan¯¹(Vy/Vx) = tan¯¹(3.36 m/s / 4.34 m/s)
= 37.1◦
The passenger's velocity with respect to the water is 6.84 m/s in a direction 37.1◦ east of south.