a ship is headed towards east at a thrust speed of 7.00 knots. A strong wind pressure causes the ship to deviate to the north at 1.00 knots. The sea current is flowing to the southwest at 4.00 knots. Determine the velocity of the ship relative to the earth's surface.
V = 1[90o] + 4[225o].
V = (1*Cos90+4*Cos225)+(1*sin90+4*sin225)I = (0+2.83) + (1-2.83)I = 2.83 - 1.83i = 3.37n[-32.9o] = 3.37n[32.9o S. of E].
Is there other solution for this question? Thanks
To determine the velocity of the ship relative to the earth's surface, we need to calculate the resultant vector of the ship's thrust speed, wind pressure, and sea current.
Let's break down the given information into vectors:
1. The thrust speed of the ship towards the east has a magnitude of 7.00 knots.
This vector can be represented as 7.00 knots towards the east (positive x-direction).
2. The wind pressure causes the ship to deviate towards the north at 1.00 knots.
This vector can be represented as 1.00 knots towards the north (positive y-direction).
3. The sea current is flowing to the southwest at 4.00 knots.
We can break down this vector into x and y components. The x-component (westward) would be -4.00 knots, and the y-component (northward) would be 4.00 knots.
Now, let's calculate the x-component and y-component of the resultant vector by summing up the corresponding components of the individual vectors:
x-component (east/west):
Ship's thrust = 7.00 knots
Sea current = -4.00 knots
Total x-component = 7.00 knots - 4.00 knots = 3.00 knots (eastward)
y-component (north/south):
Wind pressure = 1.00 knots
Sea current = 4.00 knots
Total y-component = 1.00 knots + 4.00 knots = 5.00 knots (northward)
Finally, we can use the Pythagorean theorem to find the magnitude (velocity) of the resultant vector:
Magnitude of the resultant vector = √(x-component^2 + y-component^2)
= √(3.00^2 + 5.00^2)
= √(9.00 + 25.00)
= √34.00
≈ 5.83 knots (rounded to two decimal places)
Therefore, the velocity of the ship relative to the earth's surface is approximately 5.83 knots.
To determine the velocity of the ship relative to the Earth's surface, we will need to consider the vector components of the ship's thrust speed, wind pressure, and sea current.
The thrust speed of the ship towards the east is given as 7.00 knots. We can represent this velocity as a vector with a magnitude of 7.00 knots pointing towards the east (positive x-direction).
The wind pressure causes the ship to deviate to the north at 1.00 knot. We can represent this velocity as a vector with a magnitude of 1.00 knot pointing towards the north (positive y-direction).
The sea current is flowing to the southwest at 4.00 knots. We can represent this velocity as a vector with a magnitude of 4.00 knots pointing towards the southwest at a 45-degree angle counter-clockwise from the positive x-axis.
To determine the velocity of the ship relative to the Earth's surface, we need to add up the vector components of the ship's thrust speed, wind pressure, and sea current.
First, break down the sea current vector into its x and y-components. Since the vector makes a 45-degree angle with the positive x-axis and has a magnitude of 4.00 knots, the x-component and y-component can be determined using trigonometry:
x-component = 4.00 knots * cos(45°)
y-component = 4.00 knots * sin(45°)
Next, add up the x-components and y-components of all the vectors:
x-component_total = 7.00 knots + x-component_sea_current
y-component_total = 1.00 knots + y-component_sea_current
Finally, calculate the magnitude and direction of the resulting vector:
magnitude = sqrt(x-component_total^2 + y-component_total^2)
direction = atan2(y-component_total, x-component_total)
The magnitude and direction of the resulting vector represent the velocity of the ship relative to the Earth's surface.