posted by Taylor .
A space vehicle is coasting at a constant velocity of 20.7 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 0.260 m/s2 in the +x direction. After 42.0 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find the following quantities.
(a) the magnitude of the vehicle's velocity
(b) the direction of the vehicle's velocity relative to the space station Express the direction as an angle measured from the +y direction.
° to the right of the +y direction
X = hor. = at = 0.260m/s^2 * 42s = 10.92m/s.
Y = ver. = 20.7m/s.
a. V = sqrt(X^2 + Y^2),
V = sqrt((10.92)^2 + (20.7)^2) = 23.4m/s.
b. tanA = Y / X = 20.7 / 10.92=1.8956.
A = 62.2 deg.,N of E.
A = 90 - 62.2 = 27.8 deg.,E of N.
Where are you getting 62.2 deg
you take the inverse tan. So tan-1(1.8956)= 62.2