# Question 5a595

Jun 23, 2015

The velocity of the spacecraft will be 5930 m/s.

#### Explanation:

So, the velocity of your spacecraft forms a ${30}^{\circ}$ angle with the horizontal. This means that the spacecraft's motion can be described as a combination of a vertical movement and of a horizontal movement.

The vertical and horizontal components of the initial velocity will be

$\left\{\begin{matrix}{v}_{0 x} = {v}_{0} \cos \left({30}^{\circ}\right) \\ {v}_{0 y} = {v}_{0} \sin \left({30}^{\circ}\right)\end{matrix}\right.$

At some point in time, the two engines of the ship fire for 475 s. The two accelerations induced by the two separate engines are directed east, or along the y direction, and north, or along the x direction.

This means that you can write

$\left\{\begin{matrix}{a}_{x} = {\text{6.3 m/s"^2 \\ a_y = "2.85 m/s}}^{2}\end{matrix}\right.$

The acceleration oriented along the y axis will influence the vertical component of the initial velocity, and the acceleration oriented along the x axis will influence the horizontal component of the initial velocity.

This means that you can write

$\left\{\begin{matrix}{v}_{x} = {v}_{0 x} + {a}_{x} \cdot t \\ {v}_{y} = {v}_{0 y} + {a}_{y} \cdot t\end{matrix}\right.$

Plug in your values and solve for the vertical and horizontal components of the final velocity.

v_x = 2650 "m"/"s" cos(30^@) + 6.3"m"/"s"^(cancel(2)) * 475cancel("s")

${v}_{x} = 2650 \cdot \frac{\sqrt{3}}{2} + 2992.5 = \text{5287.5 m/s}$

and

v_y = 2650 "m"/"s" sin(30^@) + 2.85"m"/"s"^(cancel(2)) * 475cancel("s")

${v}_{y} = 2650 \cdot \frac{1}{2} + 1353.75 = \text{2678.8 m/s}$

To get the final velocity of the spacecraft, use Pythagoras' Theorem

${v}_{\text{final}} = \sqrt{{v}_{x}^{2} + {v}_{y}^{2}}$

v_"final" = sqrt(5287.5""^2 + 2678.8""^2) = color(green)("5930 m/s")#