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Sagot :
To determine the magnitude of the boat's resultant vector, we need to consider both the boat's velocity heading north and the current's impact which pushes it 65 degrees west of north. Here is a step-by-step solution for this problem:
1. Decompose the velocities into their respective components:
- Boat's velocity components:
The boat is heading directly north at 10 mph. This means the entire velocity is in the north (y) direction, and there is no west (x) component.
[tex]\[ \text{boat}_x = 0 \quad (\text{west-east component}) \][/tex]
[tex]\[ \text{boat}_y = 10 \quad (\text{north-south component}) \][/tex]
- Current's velocity components:
The current has a speed of 5 mph and is pushing 65 degrees west of north. To find the components:
[tex]\[ \text{current}_x = 5 \cos(65^\circ) \][/tex]
[tex]\[ \text{current}_y = 5 \sin(65^\circ) \][/tex]
Since it’s west of north, the cosine component (x direction) should be negative (westward).
[tex]\[ \text{current}_x = -5 \cos(65^\circ) \approx -2.113 \][/tex]
[tex]\[ \text{current}_y = 5 \sin(65^\circ) \approx 4.532 \][/tex]
2. Sum the components to get the resultant components:
- Resultant x-component:
[tex]\[ \text{resultant}_x = \text{boat}_x + \text{current}_x = 0 + (-2.113) = -2.113 \][/tex]
- Resultant y-component:
[tex]\[ \text{resultant}_y = \text{boat}_y + \text{current}_y = 10 + 4.532 = 14.532 \][/tex]
3. Calculate the magnitude of the resultant vector:
The magnitude of the resultant vector can be found using the Pythagorean theorem:
[tex]\[ |\overrightarrow{R}| = \sqrt{(\text{resultant}_x)^2 + (\text{resultant}_y)^2} \][/tex]
Substituting in the values:
[tex]\[ |\overrightarrow{R}| = \sqrt{(-2.113)^2 + (14.532)^2} \][/tex]
[tex]\[ |\overrightarrow{R}| = \sqrt{4.465 + 211.168} \][/tex]
[tex]\[ |\overrightarrow{R}| = \sqrt{215.633} \approx 14.68 \text{ mph} \][/tex]
Therefore, the magnitude of the boat's resultant vector is [tex]\(\boxed{14.68 \text{ mph}}\)[/tex].
1. Decompose the velocities into their respective components:
- Boat's velocity components:
The boat is heading directly north at 10 mph. This means the entire velocity is in the north (y) direction, and there is no west (x) component.
[tex]\[ \text{boat}_x = 0 \quad (\text{west-east component}) \][/tex]
[tex]\[ \text{boat}_y = 10 \quad (\text{north-south component}) \][/tex]
- Current's velocity components:
The current has a speed of 5 mph and is pushing 65 degrees west of north. To find the components:
[tex]\[ \text{current}_x = 5 \cos(65^\circ) \][/tex]
[tex]\[ \text{current}_y = 5 \sin(65^\circ) \][/tex]
Since it’s west of north, the cosine component (x direction) should be negative (westward).
[tex]\[ \text{current}_x = -5 \cos(65^\circ) \approx -2.113 \][/tex]
[tex]\[ \text{current}_y = 5 \sin(65^\circ) \approx 4.532 \][/tex]
2. Sum the components to get the resultant components:
- Resultant x-component:
[tex]\[ \text{resultant}_x = \text{boat}_x + \text{current}_x = 0 + (-2.113) = -2.113 \][/tex]
- Resultant y-component:
[tex]\[ \text{resultant}_y = \text{boat}_y + \text{current}_y = 10 + 4.532 = 14.532 \][/tex]
3. Calculate the magnitude of the resultant vector:
The magnitude of the resultant vector can be found using the Pythagorean theorem:
[tex]\[ |\overrightarrow{R}| = \sqrt{(\text{resultant}_x)^2 + (\text{resultant}_y)^2} \][/tex]
Substituting in the values:
[tex]\[ |\overrightarrow{R}| = \sqrt{(-2.113)^2 + (14.532)^2} \][/tex]
[tex]\[ |\overrightarrow{R}| = \sqrt{4.465 + 211.168} \][/tex]
[tex]\[ |\overrightarrow{R}| = \sqrt{215.633} \approx 14.68 \text{ mph} \][/tex]
Therefore, the magnitude of the boat's resultant vector is [tex]\(\boxed{14.68 \text{ mph}}\)[/tex].
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