A spacecraft is traveling with a velocity of

v0x = 5690 m/s
along the +x direction. Two engines are turned on for a time of 845 s. One engine gives the spacecraft an acceleration in the +x direction of
ax = 1.35 m/s2,
while the other gives it an acceleration in the +y direction of
ay = 8.35 m/s2.
At the end of the firing, find
vx and vy.

To find the final velocities (vx and vy) of the spacecraft, we can use the equations of motion.

The initial velocity in the x-direction (v0x) is given as 5690 m/s, and the acceleration in the x-direction (ax) is given as 1.35 m/s^2. The time for which the engine is turned on is 845 s.

Using the equation of motion in the x-direction:
vx = v0x + ax * t

Substituting the given values:
vx = 5690 m/s + (1.35 m/s^2) * (845 s)
vx = 5690 m/s + 1140.75 m/s
vx = 6830.75 m/s

Therefore, the final velocity in the x-direction (vx) is 6830.75 m/s.

Similarly, the initial velocity in the y-direction (v0y) is not given, but we assume it to be 0 m/s since it's not specified. The acceleration in the y-direction (ay) is given as 8.35 m/s^2.

Using the equation of motion in the y-direction:
vy = v0y + ay * t

Substituting the given values:
vy = 0 m/s + (8.35 m/s^2) * (845 s)
vy = 0 m/s + 7038.75 m/s
vy = 7038.75 m/s

Therefore, the final velocity in the y-direction (vy) is 7038.75 m/s.

Hence, the final velocities of the spacecraft are vx = 6830.75 m/s (in the x-direction) and vy = 7038.75 m/s (in the y-direction).