On a spacecraft two engines fire for a time of 565 s. One gives the craft an acceleration in the x direction of `a_(x)=5.10m//s^(2)`, while the other produces an acceleration in the y direction of `a_(y)=7.30m//s^(2)`. At the end of the firing period, the craft has velocity components of `v_(x)=3775m//sandv_(y)=4816m//s`. Find the magnitude and direction of the initial velocity. Express the direction as an angle with respect to the +x axis.

Options

A. `1130m//s,52.3^(@)`
B. `793m//s,37.7^(@)`
C. `1130m//s,37.7^(@)`
D. `793m//s,52.3^(@)`

Answer

To solve the problem, we need to find the initial velocity components \( u_x \) and \( u_y \) of the spacecraft before the engines fired. We can use the equations of motion for both the x and y directions.

### Step-by-Step Solution:

1. **Identify the given data:**
- Time of firing, \( t = 565 \, \text{s} \)
- Acceleration in the x direction, \( a_x = 5.10 \, \text{m/s}^2 \)
- Acceleration in the y direction, \( a_y = 7.30 \, \text{m/s}^2 \)
- Final velocity in the x direction, \( v_x = 3775 \, \text{m/s} \)
- Final velocity in the y direction, \( v_y = 4816 \, \text{m/s} \)

2. **Use the equation of motion for the x direction:**
\[ v_x = u_x + a_x \cdot t \]
Rearranging gives:
\[ u_x = v_x - a_x \cdot t \]
Substituting the values:
\[ u_x = 3775 \, \text{m/s} - (5.10 \, \text{m/s}^2 \cdot 565 \, \text{s}) \]
\[ u_x = 3775 \, \text{m/s} - 2881.5 \, \text{m/s} = 893.5 \, \text{m/s} \]

3. **Use the equation of motion for the y direction:**
\[ v_y = u_y + a_y \cdot t \]
Rearranging gives:
\[ u_y = v_y - a_y \cdot t \]
Substituting the values:
\[ u_y = 4816 \, \text{m/s} - (7.30 \, \text{m/s}^2 \cdot 565 \, \text{s}) \]
\[ u_y = 4816 \, \text{m/s} - 4124.5 \, \text{m/s} = 691.5 \, \text{m/s} \]

4. **Calculate the magnitude of the initial velocity:**
The magnitude \( u \) of the initial velocity can be found using the Pythagorean theorem:
\[ u = \sqrt{u_x^2 + u_y^2} \]
Substituting the values:
\[ u = \sqrt{(893.5 \, \text{m/s})^2 + (691.5 \, \text{m/s})^2} \]
\[ u = \sqrt{797,286.25 + 479,197.25} = \sqrt{1,276,483.5} \approx 1129.4 \, \text{m/s} \]

5. **Calculate the direction of the initial velocity:**
The angle \( \theta \) with respect to the positive x-axis can be calculated using:
\[ \tan(\theta) = \frac{u_y}{u_x} \]
\[ \theta = \tan^{-1}\left(\frac{691.5}{893.5}\right) \]
\[ \theta \approx \tan^{-1}(0.774) \approx 37.59^\circ \]

### Final Answers:
- Magnitude of initial velocity: \( \approx 1129.4 \, \text{m/s} \)
- Direction of initial velocity: \( \approx 37.59^\circ \) with respect to the +x axis.