Posted by Taylor on Thursday, October 13, 2011 at 11:25am.
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
m/s
(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

physics  Henry, Friday, October 14, 2011 at 9:51pm
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.

physics  GG, Monday, January 16, 2012 at 9:53am
Where are you getting 62.2 deg

physics  Kyleigh, Thursday, September 18, 2014 at 12:55pm
you take the inverse tan. So tan1(1.8956)= 62.2
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