Question #5a595

So, the velocity of your spacecraft forms a `30^@` 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

`{(v_(0x) = v_0 cos(30^@)), (v_(0y) = v_0 sin(30^@)) :}`

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

`{ (a_x = "6.3 m/s"^2), (a_y = "2.85 m/s"^2) :}`

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

`{(v_x = v_(0x) + a_x * t), (v_y = v_(0y) + a_y * t) :}`

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 * sqrt(3)/2 + 2992.5 = "5287.5 m/s"`

and

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

`v_y = 2650 * 1/2 + 1353.75 = "2678.8 m/s"`

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

`v_"final" = sqrt(v_x^2 + v_y^2)`

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