On a spacecraft, two engines are turned on for 684 s at a moment when the velocity of the craft has x and y components of `v_(o x)=4370m//sandv_(oy)=6280m//s`. While the engines are firing, the craft undergoes a displacement that has components of `x=4.11xx10^(6)mandy=6.07xx10^(6)m`. Find the x and y components of the spacecraft's acceleration.

Options

A. `{:(a_(x),a_(y)),(9.58m//s^(2),5.06m//s^(2)):}`
B. `{:(a_(x),a_(y)),(6.39m//s^(2),10.1m//s^(2)):}`
C. `{:(a_(x),a_(y)),(4.79m//s^(2),7.59m//s^(2)):}`
D. `{:(a_(x),a_(y)),(5.06m//s^(2),9.58m//s^(2)):}`

Answer

To find the x and y components of the spacecraft's acceleration, we will use the second equation of motion, which is given by:

\[ s = ut + \frac{1}{2} a t^2 \]

where:
- \( s \) is the displacement,
- \( u \) is the initial velocity,
- \( t \) is the time,
- \( a \) is the acceleration.

### Step 1: Set up the equations for the x and y components

For the x-direction:
- Displacement \( s_x = 4.11 \times 10^6 \, \text{m} \)
- Initial velocity \( u_x = 4370 \, \text{m/s} \)
- Time \( t = 684 \, \text{s} \)

The equation becomes:
\[ 4.11 \times 10^6 = 4370 \times 684 + \frac{1}{2} a_x (684)^2 \]

For the y-direction:
- Displacement \( s_y = 6.07 \times 10^6 \, \text{m} \)
- Initial velocity \( u_y = 6280 \, \text{m/s} \)

The equation becomes:
\[ 6.07 \times 10^6 = 6280 \times 684 + \frac{1}{2} a_y (684)^2 \]

### Step 2: Calculate the x-component of acceleration \( a_x \)

1. Calculate \( 4370 \times 684 \):
\[ 4370 \times 684 = 2989080 \, \text{m} \]

2. Substitute this value into the x-direction equation:
\[ 4.11 \times 10^6 = 2989080 + \frac{1}{2} a_x (684)^2 \]

3. Rearranging gives:
\[ 4.11 \times 10^6 - 2989080 = \frac{1}{2} a_x (684)^2 \]

4. Calculate \( 4.11 \times 10^6 - 2989080 \):
\[ 4.11 \times 10^6 - 2989080 = 1121910 \, \text{m} \]

5. Now calculate \( (684)^2 \):
\[ (684)^2 = 467056 \]

6. Substitute this back into the equation:
\[ 1121910 = \frac{1}{2} a_x (467056) \]

7. Solve for \( a_x \):
\[ a_x = \frac{1121910 \times 2}{467056} \approx 4786.6 \, \text{m/s}^2 \]

### Step 3: Calculate the y-component of acceleration \( a_y \)

1. Calculate \( 6280 \times 684 \):
\[ 6280 \times 684 = 4285920 \, \text{m} \]

2. Substitute this value into the y-direction equation:
\[ 6.07 \times 10^6 = 4285920 + \frac{1}{2} a_y (684)^2 \]

3. Rearranging gives:
\[ 6.07 \times 10^6 - 4285920 = \frac{1}{2} a_y (684)^2 \]

4. Calculate \( 6.07 \times 10^6 - 4285920 \):
\[ 6.07 \times 10^6 - 4285920 = 1784080 \, \text{m} \]

5. Substitute \( (684)^2 \) again:
\[ 1784080 = \frac{1}{2} a_y (467056) \]

6. Solve for \( a_y \):
\[ a_y = \frac{1784080 \times 2}{467056} \approx 7630.6 \, \text{m/s}^2 \]

### Final Results:
- \( a_x \approx 4786.6 \, \text{m/s}^2 \)
- \( a_y \approx 7630.6 \, \text{m/s}^2 \)