Aspacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. two engines are turned on for a time of 739 s. one engine gives the spacecraft an acceleration in the +x direction of ax = 1.79 m/s2, while the other gives it an acceleration in the +y direction of ay = 7.18 m/s2. at the end of the firing, what is a) vx and b) vy?

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: , where   is the final velocity in a direction after the acceleration is applied, is the initial velocity in that direction  before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.Then for the x direction it is known that the initial velocity is 5320 m/s, the acceleration (the applied by the engine) in x direction is 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the  is 739 s. Then: In the same fashion, for the y direction, the initial velocity is   0 m/s, the acceleration in y direction is 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction:

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