Answer to Question #158605 in Physics for Brooke C

Question

A space vehicle is coasting at a constant velocity of 24.0 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 0.298 m/s² in the +x direction. After 52.0 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 (in degrees) measured from the +y direction.

Answer 1

(a) Find the magnitude of the vehicle's velocity relative to the space station. First, find what speed the vehicle gained in 52 s along +x direction:

$$v_x=a_xt=0.298\cdot52=15.5\text{ m/s}.$$

The magnitude is a vector sum of y- and x-components:

$$v=\sqrt{v_x^2+v_y^2}=28.6\text{ m/s}.$$

(b) The direction of the vehicle's velocity relative to the space station measured from the +y direction:

$$\theta=\text{atan}\frac{v_x}{v_y}=\text{atan}\frac{15.5}{24}=32.9°.$$