To solve the problem and find the magnitude and direction of the spacecraft's initial velocity before the engines were turned on, follow these steps:
1. Extract Given Data:
- Acceleration in the [tex]\( x \)[/tex]-direction, [tex]\( a_x = 5.10 \, \text{m/s}^2 \)[/tex].
- Acceleration in the [tex]\( y \)[/tex]-direction, [tex]\( a_y = 7.30 \, \text{m/s}^2 \)[/tex].
- Time the engines were on, [tex]\( t = 630 \, \text{s} \)[/tex].
- Final velocity in the [tex]\( x \)[/tex]-direction, [tex]\( v_{x_{\text{final}}} = 3740 \, \text{m/s} \)[/tex].
- Final velocity in the [tex]\( y \)[/tex]-direction, [tex]\( v_{y_{\text{final}}} = 4472 \, \text{m/s} \)[/tex].
2. Calculate Initial Velocities:
- For the [tex]\( x \)[/tex]-direction:
[tex]\[
v_{x_{\text{initial}}} = v_{x_{\text{final}}} - a_x \times t
\][/tex]
Substituting the given values:
[tex]\[
v_{x_{\text{initial}}} = 3740 \, \text{m/s} - (5.10 \, \text{m/s}^2 \times 630 \, \text{s}) = 527 \, \text{m/s}
\][/tex]
- For the [tex]\( y \)[/tex]-direction:
[tex]\[
v_{y_{\text{initial}}} = v_{y_{\text{final}}} - a_y \times t
\][/tex]
Substituting the given values:
[tex]\[
v_{y_{\text{initial}}} = 4472 \, \text{m/s} - (7.30 \, \text{m/s}^2 \times 630 \, \text{s}) = -127 \, \text{m/s}
\][/tex]
3. Calculate the Magnitude of the Initial Velocity:
The magnitude of the initial velocity [tex]\( v_{\text{initial}} \)[/tex] can be found using the Pythagorean theorem:
[tex]\[
v_{\text{initial}} = \sqrt{(v_{x_{\text{initial}}})^2 + (v_{y_{\text{initial}}})^2}
\][/tex]
Substituting the values:
[tex]\[
v_{\text{initial}} = \sqrt{(527)^2 + (-127)^2} \approx 542.09 \, \text{m/s}
\][/tex]
4. Calculate the Direction of the Initial Velocity:
The direction of the initial velocity θ, measured counterclockwise from the [tex]\( +x \)[/tex]-axis, can be found using the inverse tangent function:
[tex]\[
\theta = \tan^{-2} \left( \frac{v_{y_{\text{initial}}}}{v_{x_{\text{initial}}}} \right)
\][/tex]
Substituting the values:
[tex]\[
\theta = \tan^{-2} \left( \frac{-127}{527} \right) \approx -13.55^\circ
\][/tex]
Since the arctangent function gives an angle with respect to the [tex]\( x \)[/tex]-axis, and the result is negative, this angle indicates a direction below the positive [tex]\( x \)[/tex]-axis (clockwise). However, when describing direction as counterclockwise from the [tex]\( +x \)[/tex]-axis, use the negative sign as is to maintain convention.
Thus, the magnitude of the spacecraft's initial velocity is approximately [tex]\( 542.09 \, \text{m/s} \)[/tex] and the direction is [tex]\( -13.55^\circ \)[/tex] counterclockwise from the [tex]\( +x \)[/tex]-axis.
