At the entrance channel of a harbor, the tidal current has a velocity of 4.22 km/hr in a direction 17.4o south of east. Suppose a ship caught in this current has a speed of 17.5 km/hr relative to the water. If the helmsman keeps the bow of the ship aimed north, what will be the speed of the ship relative to the ground?
speed N = 17.5 - 4.22 sin 17.4
speed E = 4.22 cos 17.4
magnitude of speed = sqrt(N^2+E^2)
direction = tan^-1 (E/N) in true compass degrees (east of north)
To find the speed of the ship relative to the ground, we need to determine the resulting velocity vector by considering the velocity of the current and the velocity of the ship.
Given:
- Current velocity = 4.22 km/hr
- Direction: 17.4 degrees south of east
- Ship's speed relative to the water = 17.5 km/hr
- Ship's direction: North
Step 1: Convert all velocities to vector form.
The current velocity is given in vector form as:
Vc = 4.22 km/hr at an angle of 180 degrees - 17.4 degrees (in the southeastern direction).
The ship's velocity relative to the water is given in vector form as:
Vs = 17.5 km/hr vertically upward (northward).
Step 2: Add the vector velocities.
To find the resulting velocity, we add the current velocity vector and the ship's velocity vector:
Vr = Vc + Vs
Step 3: Calculate the magnitude and direction of the resulting velocity.
To find the magnitude of the resulting velocity, we can use the Pythagorean theorem:
Magnitude of Vr = √(Vrx^2 + Vry^2)
To find the direction of the resulting velocity, we can use the inverse tangent function:
Direction of Vr = arctan(Vry/Vrx)
Step 4: Substitute the values and calculate.
Vrx = Vc * cos(180° - 17.4°)
Vry = Vc * sin(180° - 17.4°)
Magnitude of Vr = √(Vrx^2 + Vry^2)
Direction of Vr = arctan(Vry/Vrx)
Substituting the values and calculating:
Vrx = 4.22 km/hr * cos(180° - 17.4°) = 4.22 km/hr * cos(162.6°)
Vry = 4.22 km/hr * sin(180° - 17.4°) = 4.22 km/hr * sin(162.6°)
Magnitude of Vr = √(Vrx^2 + Vry^2)
Direction of Vr = arctan(Vry/Vrx)
Finally, we will have the resulting speed of the ship relative to the ground.
To find the speed of the ship relative to the ground, we can use vector addition.
First, let's break down the given velocities into x and y components.
The tidal current velocity has a magnitude of 4.22 km/hr and is directed 17.4° south of east. To calculate the x and y components, we use trigonometry:
x-component of tidal current velocity = 4.22 km/hr * cos(17.4°) = 3.97 km/hr
y-component of tidal current velocity = 4.22 km/hr * sin(17.4°) = 1.28 km/hr
Since the ship is traveling in the direction opposite to the current, the x-component of the ship's velocity relative to the water will be equal in magnitude but opposite in direction to the x-component of the tidal current velocity:
x-component of ship's velocity relative to water = -3.97 km/hr
The y-component of the ship's velocity relative to the water will be 0 because the ship is keeping its bow aimed north:
y-component of ship's velocity relative to water = 0 km/hr
Now, we can add the ship's velocity relative to the water and the tidal current velocity components to find the ship's velocity relative to the ground:
x-component of ship's velocity relative to ground = -3.97 km/hr
y-component of ship's velocity relative to ground = 0 km/hr + 1.28 km/hr (since the tidal current is only affecting the ship's y-component)
The magnitude of the ship's velocity relative to the ground is found using the Pythagorean theorem:
magnitude of ship's velocity relative to ground = √((-3.97 km/hr)^2 + (1.28 km/hr)^2) = 4.14 km/hr
Therefore, the speed of the ship relative to the ground is approximately 4.14 km/hr.