Question

Uugghhh help please.

The velocity of a 4.6 kg particle is given by vector v=(4t{i}+7t^2{j}), where v is in m/s and t is in seconds. At the instant when the net force on the particle has a magnitude of 40 N, what is the angle between the particle's acceleration and the particle's direction of motion? Answer in degrees.

I can't see relationship what the question asking for, so lost!

Answer 1

a = dv/dt = 4 i + 14 t

F = m a
40 = 4.6 sqrt (4^2 + 14^2 t^2)
(40/4.6)^2 = 16 + 196 t^2
solve for t
use that t to get v and a

Then
v dot a = |v||a| cos theta