Question
A ship is sailing through the water in the English Channel with a velocity of 22 knots along a bearing of 157° (knots being a unit used to measure the speed of aircrafts and boats). The current has a velocity of 5 knots along a bearing of 213°. Find the resultant velocity and direction of the ship. (Remember that bearing is measured clockwise from the north axis).
(1 point)
Responses
25 knots at 166.5°
25 knots at 166.5°
27 knots at 350°
27 knots at 350°
166.5 knots at 25°
166.5 knots at 25°
350 knots at 27°
(1 point)
Responses
25 knots at 166.5°
25 knots at 166.5°
27 knots at 350°
27 knots at 350°
166.5 knots at 25°
166.5 knots at 25°
350 knots at 27°
Answers
GPT-4o mini
To find the resultant velocity of the ship taking into account the current, we can break down the velocities into their Cartesian components and then combine them.
1. **Convert the velocities to Cartesian components:**
For the ship (22 knots at a bearing of 157°):
- The angle from the positive x-axis (East) is given by \(90° - 157° = -67°\), hence it translates to \(360° - 67° = 293°\).
- The components for the ship are:
\[
V_{sx} = 22 \cos(293°) = 22 \cos(-67°) \approx 22 \times 0.3907 \approx 8.60 \text{ knots (to the West)}
\]
\[
V_{sy} = 22 \sin(293°) = 22 \sin(-67°) \approx 22 \times -0.9200 \approx -20.24 \text{ knots (to the South)}
\]
For the current (5 knots at a bearing of 213°):
- The angle from the positive x-axis (East) is given by \(90° - 213° = -123°\), thus it translates to \(360° - 123° = 237°\).
- The components for the current are:
\[
V_{cx} = 5 \cos(237°) = 5 \cos(-123°) \approx 5 \times -0.8090 \approx -4.05 \text{ knots (to the West)}
\]
\[
V_{cy} = 5 \sin(237°) = 5 \sin(-123°) \approx 5 \times -0.5878 \approx -2.94 \text{ knots (to the South)}
\]
2. **Combine the components:**
Total x-component (West-East direction):
\[
V_x = V_{sx} + V_{cx} = -8.60 - 4.05 = -12.65 \text{ knots}
\]
Total y-component (South-North direction):
\[
V_y = V_{sy} + V_{cy} = -20.24 - 2.94 = -23.18 \text{ knots}
\]
3. **Find the magnitude of the resultant velocity:**
\[
V_r = \sqrt{(-12.65)^2 + (-23.18)^2} \approx \sqrt{160.92 + 537.84} \approx \sqrt{698.76} \approx 26.45 \text{ knots}
\]
4. **Find the direction of the resultant velocity:**
\[
\theta = \tan^{-1}\left(\frac{V_y}{V_x}\right) = \tan^{-1}\left(\frac{-23.18}{-12.65}\right) \approx \tan^{-1}(1.83) \approx 61.18°
\]
This angle is in the third quadrant (because both components are negative), thus the bearing is:
\[
180° + 61.18° \approx 241.18°
\]
Therefore, this means:
\[
\text{Bearing} \approx 241.18°
\]
Compared to the provided options, the closest answer based on our calculations needs to be compared again, as it might not perfectly match.
Thus, the resultant velocity is approximately **26.45 knots at a bearing of 241°.** (None of the provided options matched this result).
Checking the values, the ship is indeed moving against the current effectively.
If we were to find the closest option based on potential misinterpretation or rounding, we would see that:
- **25 knots at 166.5°** is misrepresented as the resultant is significantly heading more southerly towards southwest.
The calculated result seems to indicate misalignment with exercise intent despite the cleanliness of the component breakdown concluding.
1. **Convert the velocities to Cartesian components:**
For the ship (22 knots at a bearing of 157°):
- The angle from the positive x-axis (East) is given by \(90° - 157° = -67°\), hence it translates to \(360° - 67° = 293°\).
- The components for the ship are:
\[
V_{sx} = 22 \cos(293°) = 22 \cos(-67°) \approx 22 \times 0.3907 \approx 8.60 \text{ knots (to the West)}
\]
\[
V_{sy} = 22 \sin(293°) = 22 \sin(-67°) \approx 22 \times -0.9200 \approx -20.24 \text{ knots (to the South)}
\]
For the current (5 knots at a bearing of 213°):
- The angle from the positive x-axis (East) is given by \(90° - 213° = -123°\), thus it translates to \(360° - 123° = 237°\).
- The components for the current are:
\[
V_{cx} = 5 \cos(237°) = 5 \cos(-123°) \approx 5 \times -0.8090 \approx -4.05 \text{ knots (to the West)}
\]
\[
V_{cy} = 5 \sin(237°) = 5 \sin(-123°) \approx 5 \times -0.5878 \approx -2.94 \text{ knots (to the South)}
\]
2. **Combine the components:**
Total x-component (West-East direction):
\[
V_x = V_{sx} + V_{cx} = -8.60 - 4.05 = -12.65 \text{ knots}
\]
Total y-component (South-North direction):
\[
V_y = V_{sy} + V_{cy} = -20.24 - 2.94 = -23.18 \text{ knots}
\]
3. **Find the magnitude of the resultant velocity:**
\[
V_r = \sqrt{(-12.65)^2 + (-23.18)^2} \approx \sqrt{160.92 + 537.84} \approx \sqrt{698.76} \approx 26.45 \text{ knots}
\]
4. **Find the direction of the resultant velocity:**
\[
\theta = \tan^{-1}\left(\frac{V_y}{V_x}\right) = \tan^{-1}\left(\frac{-23.18}{-12.65}\right) \approx \tan^{-1}(1.83) \approx 61.18°
\]
This angle is in the third quadrant (because both components are negative), thus the bearing is:
\[
180° + 61.18° \approx 241.18°
\]
Therefore, this means:
\[
\text{Bearing} \approx 241.18°
\]
Compared to the provided options, the closest answer based on our calculations needs to be compared again, as it might not perfectly match.
Thus, the resultant velocity is approximately **26.45 knots at a bearing of 241°.** (None of the provided options matched this result).
Checking the values, the ship is indeed moving against the current effectively.
If we were to find the closest option based on potential misinterpretation or rounding, we would see that:
- **25 knots at 166.5°** is misrepresented as the resultant is significantly heading more southerly towards southwest.
The calculated result seems to indicate misalignment with exercise intent despite the cleanliness of the component breakdown concluding.