The velocity of a ball in an x-y coordinate system is (10, -5) where distance is measured in metres. A second coordinate system, p-q, uses units of feet (1 ft = 0.3048 m). The p-axis is oriented at alpha = 15 degrees relative to the x-axis. The origin of the p-q system is located at (10m, 2m) in the x-y system. What is the velocity vector of the ball in the p-q coordinate system?
Basic math, rotation matrix. I'm still not 100% on why, but the way we are taught, the rotation matrix is [cos sin; -sin, cos] for a CCW rotation.
The Attempt at a Solution
[p;q]=3.28ft/m*([cos15 sin15; -sin15 cos15] [10; -5]+[10;2])
I *think* I'm correct on the +[10;2] rather than a negative since from the x-y frame, it moves in the postive direction.
However, I can't seem to get the answer of "(27.5, -24.3) ft/s" even when I mess around with varying negatives and stuff.
There's a followup question which involves going back the other way, which I would have done as:
[x;y]=0.3048([cos(-15) sin(-15); -sin(-15) cos(-15)][p;q]-[10;2])
But seeing as the first one isn't giving the correct answer I doubt that this will work either....