A nuclear submarine leaves its base and travels at 23.5 mph. For 2.00 hrs it travels along a course 32.1 degrees north of west. It then turns an additional 21.5 degrees north of west and travels for another 1.00 hrs. How far from its base is it?

d1 = 23.5mi/h * 2h = 47 Mi[180o-32.1o] =

47Mi[147.9o]

d2 = 23.5mi/h * 1h = 23.5mi[180-53.6] =
23.5mi[126.4o]

d1 + d2 = 47mi[147.9] + 23.5mi[126.4o]
X = 47*cos147.9 + 23.5*cos126.4=-53.8 mi
Y = 47*sin147.9 + 23.5*sin126.4=43.9 mi

D^2=X^2 + Y^2=-53.8^2 + 43.9^2=4821.65
D = 69.4 miles.

To find the distance from the base, we need to break down the submarine's motion into vertical and horizontal components.

Step 1: Find the horizontal component of the distance covered in the first part of the journey.
The distance covered along the initial course is given by:
Distance = Speed × Time

So, the horizontal distance can be calculated as:
Horizontal distance = 23.5 mph × 2.00 hrs

Step 2: Find the vertical component of the distance covered in the first part of the journey.
Since the course is 32.1 degrees north of west, we can use trigonometry to find the vertical component.

Vertical distance = Horizontal distance × tan(32.1°)

Step 3: Find the horizontal component of the distance covered in the second part of the journey.
The speed and time remain the same, but now the course is further north of west by 21.5 degrees.

Horizontal distance = 23.5 mph × 1.00 hr

Step 4: Find the vertical component of the distance covered in the second part of the journey.
Similar to Step 2, we need to find the vertical distance corresponding to a course 21.5 degrees north of west.

Vertical distance = Horizontal distance × tan(21.5°)

Step 5: Calculate the total horizontal distance covered.
Total horizontal distance = Horizontal distance (Step 1) + Horizontal distance (Step 3)

Step 6: Calculate the total vertical distance covered.
Total vertical distance = Vertical distance (Step 2) + Vertical distance (Step 4)

Step 7: Find the distance from the base using the Pythagorean theorem.
Distance from the base = sqrt((Total horizontal distance)^2 + (Total vertical distance)^2)

To find the distance from the submarine's base, we can use trigonometry and vector addition. Here's how to calculate it step by step:

1. Convert the speeds to meters per second:
- 23.5 mph = (23.5 * 1609.34 m) / (1 hr * 3600 s) = 10.514 m/s

2. Convert the time intervals to seconds:
- 2 hours = 2 hr * 3600 s = 7200 s
- 1 hour = 1 hr * 3600 s = 3600 s

3. Calculate the displacement for the first leg of the submarine's journey:
- The submarine travels at 32.1 degrees north of west.
- Convert this angle to radians: 32.1 degrees = (32.1 * π) / 180 = 0.5591 radians.
- To find the displacement, we can multiply the velocity by the time and the cosine of the angle: Displacement = (velocity) * (time) * cos(angle)
- Displacement = (10.514 m/s) * (7200 s) * cos(0.5591 radians) = 65,317.79 m

4. Calculate the displacement for the second leg of the submarine's journey:
- The submarine turns an additional 21.5 degrees north of west.
- Convert this angle to radians: 21.5 degrees = (21.5 * π) / 180 = 0.3745 radians.
- Displacement = (velocity) * (time) * cos(angle)
- Displacement = (10.514 m/s) * (3600 s) * cos(0.3745 radians) = 13,363.90 m

5. Add the displacements vectorially:
- To combine the displacements as vectors, we need to break them into their x and y components. The x-component is given by Displacement * cos(angle), and the y-component is given by Displacement * sin(angle).
- For the first displacement: x-component = (65,317.79 m) * cos(0.5591 radians) = 65,317.79 m * 0.8296 = 54,139.40 m
y-component = (65,317.79 m) * sin(0.5591 radians) = 65,317.79 m * 0.5586 = 36,449.70 m
- For the second displacement: x-component = (13,363.90 m) * cos(0.3745 radians) = 13,363.90 m * 0.9299 = 12,414.29 m
y-component = (13,363.90 m) * sin(0.3745 radians) = 13,363.90 m * 0.3685 = 4,921.46 m

6. Add the x and y components to find the total displacement:
- Total x-component = 54,139.40 m + 12,414.29 m = 66,553.69 m
- Total y-component = 36,449.70 m + 4,921.46 m = 41,371.16 m

7. Use the Pythagorean theorem to find the magnitude of the total displacement:
- Magnitude = √(Total x-component ^ 2 + Total y-component ^ 2)
- Magnitude = √(66,553.69 m^2 + 41,371.16 m^2) = √(4,420,730,676.34 m^2) = 66,445.32 m

Therefore, the submarine is approximately 66,445.32 meters (or about 41.27 miles) from its base.