22. A space vehicle is coasting at a constant velocity of 21.0 m/s in the +y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.320 m/s2 in the +x direction. After 45.0 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle’s velocity relative to the space station. Express the direction as an angle measured from the +y direction.

The space vehicle is initially traveling in the +y direction and has no velocity in the +x direction. The thruster will create a velocity in the +x direction and will the y component of velocity unchanged. So the final velocity will be the vector sum of the x and y components which since they are perpendicular can be added using Pythagorean theorem.

Y component v_y = 21.0 m/s, a_y = 0. So final v_y = 21.0 m/s

X component v_0x = 0, a_x = 0.320 m/s², for a time of t = 45.0 s, we need to find final v_x

So the final magnitude of velocity is

To find the direction with respect to the +y direction consider the figure below

The tangent of the angle desired is found from

	PH 201 Homework Chapter 3 Solutions	Dr. Donovan's PH 201 Homework Page
Dr. Donovan's Classes Page	NMU Physics Department Web Page	NMU Main Page

Please send any comments or questions about this page to ddonovan@nmu.edu

This page last updated on January 11, 2020