A space vehicle is coasting at a constant velocity of 22.4 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 0.206 m/s2 in the x direction. After 40.5 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.

a. 23.9 m/s b. 69.58°

Explanation:

a. Since the space vehicle is moving at a constant velocity of 22.4 m/s in the y direction relative to the space station, its initial vertical velocity v = 22.4 m/s.

Also, since the space vehicle moves in the y direction, its initial horizontal velocity u = 0 m/s.

Since its acceleration a = 0.206 m/s² for time, t = 40.5 s, we find the final horizontal velocity u' from u' = u + at.

Substituting the values of the variables into the equation, we have

u' = u + at

u' = 0 m/s  + 0.206 m/s² × 40.5 s

= 8.343 m/s

≅ 8.34 m/s

The resultant velocity relative to the space station V = √(v² + u'²)

= √[(22.4m/s)² + (8.34 m/s)²]

= √[501.76 m²/s² + (69.56 m²/s²]

= √[571.32 m²/s²]

= 23.9 m/s

b. The direction of the vehicle's velocity relative to the space station is thus θ = tan⁻¹(v/u')

= tan⁻¹(22.4 m/s/8.34 m/s)

= tan⁻¹(2.6959)

= 69.58°

the answer would be a