Answer to Question #158617 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, we need to find the speed the vehicle gained along +x direction:

$$v_x=a_xt.$$

The magnitude of the velocity:

$$v=\sqrt{(a_xt)^2+v_y^2}=\\=\sqrt{(0.298\cdot52)^2+24^2}=28.6\text{ m/s}.$$

(b) The direction of the velocity measured from the +y direction:

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