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° .

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.