Question

A spacecraft 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?

Answer 1

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

Explanation:

1. For a constant acceleration:
   `v_(f)=v_(0)+at`, where
   `v_(f)` is the final velocity in a direction after the acceleration is applied,
   `v_(0)` 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.
2. Then for the x direction it is known that the initial velocity is
   `v_(0x) =` 5320 m/s, the acceleration (the applied by the engine) in x direction is
   `a_(x)` 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:
   `v_(fx)=v_(0x)+a_(x)t=5320(m)/(s) +1.79(m)/(s^(2) )*739s=<strong>6642.81(m)/(s)</strong>`
3. In the same fashion, for the y direction, the initial velocity is
   `v_(0y) =` 0 m/s, the acceleration in y direction is
   `a_(y)` 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:
   `v_(fy)=v_(0y)+a_(y)t=0(m)/(s) +7.18(m)/(s^(2) )*739s=<strong>5306.02(m)/(s)</strong>`