Newton's second law problem?

Question

A 0.37 kg particle moves in an xy plane according to x(t) = - 16 + 2 t - 2 t3 and y(t) = 19 + 5 t - 4 t2, with x and y in meters and t in seconds. At t = 0.6 s, what are (a) the magnitude and (b) the angle (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?

Answer 1

Looks like to be calculus based physics.
x(t) = - 16 + 2t - 2t^3
y(t) = 19 + 5t - 4t^2
v_x = dx/dt = 2 - 6t^2
v_y = dy/dt = 5 - 8t
a_x = d(dx/dt)/dt = - 12t
a_y = d(dy/dt)/dt = - 8
At t = 0.6 s, v_x = -0.16, v_y = 0.2, a_x = -7.2, and a_y = -8
(a) 0.37*sqrt(7.2^2 + 8^2) = 3.98 (N)
(b) 180(degree) + tan^(-1) (8/7.2) = 228(degree)
(c) 180(degree) + tan^(-1) (-0.2/0.16) = 129(degree)