A 0.26 kg particle moves in an xy plane according to x(t) = - 10 + 2 t - 3 t3(t cubed) and y(t) = 29 + 3 t - 4 t2 (t squared), with x and y in meters and t in seconds. At t = 2.0 s, what are (a) the magnitude and (b) the angle (within (-180°, 180°] interval relative to the positive direction of the x axis) of the net force on the particle, and (c) what is the angle of the particle's direction of travel? To solve this question I differentiated the two equations given and also found the accceleration.I then subbed in my value for t (t=2). After some more calculations I got 31.04 as my answer for part (a)146.309 degrees for part (b) and 123.69 degrees for part (c) These answers are wrong and i cant understand where I am going wrong. Any help with this problem would be much appreciated!