A space vehicle is coasting at a constant velocity of 22.3 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 0.203 m/s2 in the x direction. After 56.7 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.

Explanation:

Initial velocity in y direction Vy = 22.3 m /s

initial acceleration in x direction ax = .203 m /s ²

time of acceleration t = 56.7 s

final velocity in x direction

v  = u + a t

Vx = 0 + .203 x 56.7 = 11.51 m /s

Final velocity in y direction will remain same as initial velocity in y direction = 22.3 m /s because there is no acceleration in y direction .

Magnitude of final velocity

= √ ( Vx² + Vy²)

= √ (22.3² + 11.51² )

= √ ( 497.29 + 132.48)

= 25.1 m /s

Direction of final velocity  from y direction be Ф

TanФ = Vx / Vy = 11.51 / 22.3 = .516

Ф = 27.3° .

the answer is a 3.8cm

explanation: