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

Answer:

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:

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: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
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: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$