A Coast Guard ship is traveling at a constant velocity of 4.13 m/s, due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of 3900 m with respect to the ship, in a direction 23.0 ° south of east. Six minutes later, he notes that the object's position relative to the ship has changed to 1360 m, 61.0 ° south of west. What are (a) the magnitude and (b) direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.

Question

A Coast Guard ship is traveling at a constant velocity of 4.13 m/s, due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of 3900 m with respect to the ship, in a direction 23.0 ° south of east. Six minutes later, he notes that the object's position relative to the ship has changed to 1360 m, 61.0 ° south of west. What are (a) the magnitude and (b) direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.

Answer 1

To determine the magnitude and direction of the velocity of the object relative to the water, we can break down the problem into two parts:

1. Find the velocity of the ship relative to the ground.
2. Use the information about the object's position relative to the ship to find its velocity relative to the ground.
3. Subtract the velocity of the ship from the velocity of the object to find its velocity relative to the water.

Let's start with the first step:

1. Find the velocity of the ship relative to the ground:
Since the ship is traveling at a constant velocity due east, we know that the velocity of the ship relative to the ground is also 4.13 m/s, due east.

Now, let's move on to the second step:

2. Find the velocity of the object relative to the ground:
The object's initial position is 3900 m, 23.0° south of east, relative to the ship. After six minutes, the object's position changes to 1360 m, 61.0° south of west, relative to the ship.
We need to convert these position changes into velocity changes by dividing by the time interval, which is 6 minutes or 360 seconds.

The change in position for the object in the east-west direction is:
Δx = 1360 m - 3900 m = -2540 m
The change in position for the object in the north-south direction is:
Δy = 0 m - 3900 m = -3900 m

Now, calculate the velocity of the object in the east-west direction:
vx = Δx / Δt = -2540 m / 360 s = -7.06 m/s (negative because it's in the west direction)

Calculate the velocity of the object in the north-south direction:
vy = Δy / Δt = -3900 m / 360 s = -10.83 m/s (negative because it's in the south direction)

Now we can move on to the third step:

3. Find the velocity of the object relative to the water:
To find the velocity of the object relative to the water, we subtract the velocity of the ship from the velocity of the object.

vx_relative = vx_object - vx_ship = -7.06 m/s - 4.13 m/s = -11.19 m/s (negative because it's in the west direction)

vy_relative = vy_object - vy_ship = -10.83 m/s - 0 m/s = -10.83 m/s (negative because it's in the south direction)

Finally, we can find the magnitude and direction of the velocity of the object relative to the water:

Magnitude of the velocity (v_relative):
|v_relative| = √(vx_relative^2 + vy_relative^2)
|v_relative| = √((-11.19 m/s)^2 + (-10.83 m/s)^2) = √(125.11 + 117.49) = √242.6 = 15.57 m/s

Direction of the velocity relative to due west:
θ = arctan(vy_relative / vx_relative)
θ = arctan(-10.83 m/s / -11.19 m/s) = arctan(0.97) ≈ 45.4°

So, the magnitude of the velocity of the object relative to the water is approximately 15.57 m/s, and the direction is about 45.4° to the south of due west.