Question
A drone is flying north at 4.0 m/s in still air, but there's a 5.0 m/s wind blowing east. Can you find the solution to this problem, specifically the drone's velocity relative to the ground and the time it takes to cross a 120m wide field?
Answers
GPT-4o mini
To find the drone's velocity relative to the ground, we can use vector addition. The drone is flying north at 4.0 m/s, and there's a wind blowing east at 5.0 m/s. We can represent these velocities as vectors:
- \( \vec{V}_{\text{drone}} = 4.0 \, \text{m/s} \, \text{north} = (0, 4.0) \, \text{m/s} \)
- \( \vec{V}_{\text{wind}} = 5.0 \, \text{m/s} \, \text{east} = (5.0, 0) \, \text{m/s} \)
Now we can add these two vectors together to find the resultant velocity of the drone relative to the ground:
\[
\vec{V}_{\text{ground}} = \vec{V}_{\text{drone}} + \vec{V}_{\text{wind}} = (5.0, 0) + (0, 4.0) = (5.0, 4.0) \, \text{m/s}
\]
Now, we need to calculate the magnitude of this resultant velocity vector:
\[
|\vec{V}_{\text{ground}}| = \sqrt{(5.0)^2 + (4.0)^2}
\]
\[
|\vec{V}_{\text{ground}}| = \sqrt{25 + 16} = \sqrt{41} \approx 6.4 \, \text{m/s}
\]
Next, we can calculate the direction of the velocity vector using the arctangent function:
\[
\theta = \tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right) = \tan^{-1}\left(\frac{4.0}{5.0}\right)
\]
\[
\theta \approx 0.6747 \, \text{radians} \, \text{or} \, 38.7^\circ \, \text{north of east}
\]
Now, to find the time it takes to cross a 120 m wide field, we first find the component of the drone's ground velocity that points directly across the field (to the north). This is simply the northward component of the velocity, which is 4.0 m/s.
Using the formula for time (distance = speed × time), we can rearrange to find the time:
\[
\text{time} = \frac{\text{distance}}{\text{speed}} = \frac{120 \, \text{m}}{4.0 \, \text{m/s}} = 30 \, \text{s}
\]
### Summary:
- The drone's velocity relative to the ground is approximately \( 6.4 \, \text{m/s} \) at an angle of \( 38.7^\circ \) north of east.
- The time it takes to cross a 120 m wide field is \( 30 \) seconds.
- \( \vec{V}_{\text{drone}} = 4.0 \, \text{m/s} \, \text{north} = (0, 4.0) \, \text{m/s} \)
- \( \vec{V}_{\text{wind}} = 5.0 \, \text{m/s} \, \text{east} = (5.0, 0) \, \text{m/s} \)
Now we can add these two vectors together to find the resultant velocity of the drone relative to the ground:
\[
\vec{V}_{\text{ground}} = \vec{V}_{\text{drone}} + \vec{V}_{\text{wind}} = (5.0, 0) + (0, 4.0) = (5.0, 4.0) \, \text{m/s}
\]
Now, we need to calculate the magnitude of this resultant velocity vector:
\[
|\vec{V}_{\text{ground}}| = \sqrt{(5.0)^2 + (4.0)^2}
\]
\[
|\vec{V}_{\text{ground}}| = \sqrt{25 + 16} = \sqrt{41} \approx 6.4 \, \text{m/s}
\]
Next, we can calculate the direction of the velocity vector using the arctangent function:
\[
\theta = \tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right) = \tan^{-1}\left(\frac{4.0}{5.0}\right)
\]
\[
\theta \approx 0.6747 \, \text{radians} \, \text{or} \, 38.7^\circ \, \text{north of east}
\]
Now, to find the time it takes to cross a 120 m wide field, we first find the component of the drone's ground velocity that points directly across the field (to the north). This is simply the northward component of the velocity, which is 4.0 m/s.
Using the formula for time (distance = speed × time), we can rearrange to find the time:
\[
\text{time} = \frac{\text{distance}}{\text{speed}} = \frac{120 \, \text{m}}{4.0 \, \text{m/s}} = 30 \, \text{s}
\]
### Summary:
- The drone's velocity relative to the ground is approximately \( 6.4 \, \text{m/s} \) at an angle of \( 38.7^\circ \) north of east.
- The time it takes to cross a 120 m wide field is \( 30 \) seconds.