To find the unit vector in the direction of the plane's velocity and the magnitude of its average velocity, we can break down the problem into a few logical steps. Let's start by determining the change in position of the plane over the given time interval.
Step 1: Calculate the Change in Position
The initial position of the plane at time \( t = 0 \) is given as \( (100 \, \text{m}, 200 \, \text{m}, 1000 \, \text{m}) \), and the position after 130 seconds is \( (2500 \, \text{m}, 1200 \, \text{m}, 1000 \, \text{m}) \).
To find the change in position, we subtract the initial position from the final position:
- Change in x-coordinate: \( 2500 \, \text{m} - 100 \, \text{m} = 2400 \, \text{m} \)
- Change in y-coordinate: \( 1200 \, \text{m} - 200 \, \text{m} = 1000 \, \text{m} \)
- Change in z-coordinate: \( 1000 \, \text{m} - 1000 \, \text{m} = 0 \, \text{m} \)
Thus, the change in position vector \( \Delta \mathbf{r} \) is:
\( \Delta \mathbf{r} = (2400 \, \text{m}, 1000 \, \text{m}, 0 \, \text{m}) \)
Step 2: Calculate the Average Velocity
The average velocity \( \mathbf{v}_{\text{avg}} \) can be calculated using the formula:
\( \mathbf{v}_{\text{avg}} = \frac{\Delta \mathbf{r}}{\Delta t} \)
Where \( \Delta t = 130 \, \text{s} \). Plugging in the values:
- Average velocity in x-direction: \( \frac{2400 \, \text{m}}{130 \, \text{s}} \approx 18.46 \, \text{m/s} \)
- Average velocity in y-direction: \( \frac{1000 \, \text{m}}{130 \, \text{s}} \approx 7.69 \, \text{m/s} \)
- Average velocity in z-direction: \( \frac{0 \, \text{m}}{130 \, \text{s}} = 0 \, \text{m/s} \)
Thus, the average velocity vector is:
\( \mathbf{v}_{\text{avg}} = (18.46 \, \text{m/s}, 7.69 \, \text{m/s}, 0 \, \text{m/s}) \)
Step 3: Calculate the Magnitude of Average Velocity
The magnitude of the average velocity \( |\mathbf{v}_{\text{avg}}| \) can be found using the formula:
\( |\mathbf{v}_{\text{avg}}| = \sqrt{(v_x^2 + v_y^2 + v_z^2)} \)
Substituting the values we calculated:
\( |\mathbf{v}_{\text{avg}}| = \sqrt{(18.46^2 + 7.69^2 + 0^2)} \)
Calculating this gives:
- \( 18.46^2 \approx 340.51 \)
- \( 7.69^2 \approx 59.24 \)
So, \( |\mathbf{v}_{\text{avg}}| = \sqrt{(340.51 + 59.24)} \approx \sqrt{399.75} \approx 19.99 \, \text{m/s} \)
Step 4: Determine the Unit Vector in the Direction of Velocity
The unit vector \( \mathbf{u} \) in the direction of the average velocity can be calculated as:
\( \mathbf{u} = \frac{\mathbf{v}_{\text{avg}}}{|\mathbf{v}_{\text{avg}}|} \)
Substituting the average velocity and its magnitude:
\( \mathbf{u} = \left( \frac{18.46}{19.99}, \frac{7.69}{19.99}, 0 \right) \)
Calculating each component gives:
- \( u_x \approx 0.923 \)
- \( u_y \approx 0.385 \)
- \( u_z = 0 \)
Thus, the unit vector in the direction of the plane's velocity is approximately:
\( \mathbf{u} \approx (0.923, 0.385, 0) \)
Summary of Results
In summary, we have determined that:
- The magnitude of the average velocity is approximately \( 19.99 \, \text{m/s} \).
- The unit vector in the direction of the plane's velocity is approximately \( (0.923, 0.385, 0) \).
