A crippled ship is being assisted to port. The motor is stuck, and is moving the boat east with a speed of 15 km/h. The current is flowing a 9 km/h at 30 degrees east of south.
A. If the maximum speed of the tugboat which is trying to rescue the disabled ship is 35 km/h, calculate the direction the tug should pull in order to get the ship to a dock directly north of its position.
B. Calculate the resultant velocity of the boat.
If the tug pulls at speed v on a course of θ, then in x-y coordinates,
15 + 9cos60° + 35 sinθ = 0
19.5 + 35 sinθ = 0
sinθ = -0.5571
θ = N 33.9° W
Now you can figure part B.
A. To calculate the direction the tug should pull to get the ship to a dock directly north of its position, we need to find the resultant velocity of the ship and determine the angle at which the tugboat should apply its force.
First, let's break down the given information:
1. The speed of the ship moving east (with respect to the water) is 15 km/h.
2. The speed of the current flowing south is 9 km/h at an angle of 30 degrees east of south.
3. The maximum speed of the tugboat is 35 km/h.
To find the resultant velocity of the ship, we need to add the velocities of the ship and the current at the given angles. Convert the angles to standard Cartesian coordinates:
1. The ship's velocity vector is 15 km/h due east.
2. The current's velocity vector can be split into its north-south and east-west components. The southward component is calculated as 9 km/h * sin(30 degrees), which gives us approximately 4.5 km/h downward. The eastward component is calculated as 9 km/h * cos(30 degrees), which gives us approximately 7.8 km/h to the east.
Now, add the individual velocities:
1. The resultant velocity in the east direction is 15 km/h + 7.8 km/h = 22.8 km/h.
2. The resultant velocity in the south direction is -4.5 km/h (negative because it's against the desired north direction).
Now, we can find the magnitude and direction of the resultant velocity vector. Use the Pythagorean theorem:
Resultant velocity (magnitude) = √[(22.8 km/h)^2 + (-4.5 km/h)^2] ≈ 23.33 km/h
To find the direction, we can use the inverse tangent function:
Resultant velocity (direction) = atan(-4.5 km/h / 22.8 km/h) ≈ -11 degrees
The resultant velocity vector is approximately 23.33 km/h at an angle of -11 degrees (south of east). Therefore, the tugboat should pull the ship in the opposite direction, which is 180 degrees + (-11 degrees) = 169 degrees east of north.
B. The resultant velocity of the boat can be calculated using the same approach as in part A.
The ship's velocity vector is 15 km/h due east.
The current's velocity vector can be split into its north-south and east-west components. The southward component is calculated as 9 km/h * sin(30 degrees), which gives us approximately 4.5 km/h downward. The eastward component is calculated as 9 km/h * cos(30 degrees), which gives us approximately 7.8 km/h to the east.
Adding the individual velocities:
1. The resultant velocity in the east direction is 15 km/h + 7.8 km/h = 22.8 km/h.
2. The resultant velocity in the south direction is -4.5 km/h (negative because it's against the desired north direction).
The magnitude of the resultant velocity can be found using the Pythagorean theorem:
Resultant velocity (magnitude) = √[(22.8 km/h)^2 + (-4.5 km/h)^2] ≈ 23.33 km/h
The direction of the resultant velocity can be found using the inverse tangent function:
Resultant velocity (direction) = atan(-4.5 km/h / 22.8 km/h) ≈ -11 degrees
Therefore, the resultant velocity of the boat is approximately 23.33 km/h at an angle of -11 degrees (south of east).