Question

A satellite in outer space is moving at a constant velocity of 21.2 m/s in the +y direction when one of its onboard thrusters turns on, causing an acceleration of 0.400 m/s^2 in the +x direction. The acceleration lasts for 48.0 s, at which point the thruster turns off. What is the final velocity of the satellite in the +x direction after the thruster turns off?

a. 19.6 m/s
b. 39.6 m/s
c. 20.0 m/s
d. 0.0 m/s

Answer 1

The final velocity of the satellite in the +x direction after the thruster turns off is 19.2 m/s, which is closest to option a, 19.6 m/s, given the choices in the question.

Step-by-step explanation:

The question pertains to the final velocity of a satellite in the +x direction after acceleration from an onboard thruster. To find the final velocity (vf) in the +x direction, we use the kinematic equation vf = vi + at, where vi is the initial velocity, a is the acceleration, and t is the time. The satellite starts with an initial velocity of 0 m/s in the +x direction since it is initially traveling only in the +y direction. It accelerates at 0.400 m/s2 in the +x direction for 48.0 s. Therefore, the final velocity in the +x direction is vf = 0 m/s + (0.400 m/s2 × 48.0 s), which equals 19.2 m/s.