Question

A spacecraft is travelling with a velocity of Vox = 6910 m/s along the +x direction. 2 engines are turned on for a time of 782s. One engine gives the spacecraft and acceleration in the +X direction of ax = 1.02 m/s^2, while the other gives it an acceleration in the +Y direction of ay = 7.91 m/s^2. Add the end of the firing, what is a) Vx and b) Vy?

Answer 1

a)

$$v_x=v_{0x}+a_xt=6910+(1.02)(782)=7708\frac{m}{s}$$

b)

$$v_y=v_{0y}+a_yt=(7.91)(782)=6186\frac{m}{s}$$