Question

A satellite in outer space is moving at a constant velocity of 20.7 m/s in the +y direction when one of its onboard thruster turns on, causing an acceleration of 0.330 m/s2 in the +x direction. The acceleration lasts for 49.0 s, at which point the thruster turns off. What is the magnitude of the satellite's velocity when the thruster turns off?

Answer 1

Answer:

26.267 m/s

Explanation:

$v_y$ = Velocity in y direction = 20.7 m/s

Velocity in x direction

$v_x=u_x+at\\\Rightarrow v_x=0+0.33\times 49\\\Rightarrow v_x=16.17\ m/s$

Resultant velocity is given by

$v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{16.17^2+20.7^2}\\\Rightarrow v=26.267\ m/s$

The magnitude of the satellite's velocity when the thruster turns off is 26.267 m/s