a) 
b) 5843 N
Explanation:
a)
The position of the UFO at time t is given by the vector:

Therefore it has 3 components:

We start by finding the velocity of the UFO, which is given by the derivative of the position:

And then, by differentiating again, we find the acceleration:

The weight of the UFO is W = 12,500 N, so its mass is:

Therefore, the components of the force on the UFO are given by Newton's second law:

So, Substituting t = 2 s, we find:

So the net force on the UFO at t = 2 s is

b)
The magnitude of a 3-dimensional vector is given by

where
are the three components of the vector
In this problem, the three components of the net force are:

Therefore, substituting into the equation, we find the magnitude of the net force:
