In an xy plane, a 0.450 kg object moves is such a way that `x(t)=-16.0+3.00t-5.00t^(3)` and `y(t)=26.0+8.00t-10.0t^(2)`, where x and y are measured in meters and t in seconds. At t = 0.800 s, find (a) the magnitude and (b) the angle, relative to the positive direction of the x axis, of the net force on the object, and ( c ) the angle of the object's travel direction.

Explanation

To solve the problem, we need to find the net force acting on the object at \( t = 0.800 \, \text{s} \) using the given position functions \( x(t) \) and \( y(t) \). We will follow these steps:

### Step 1: Find the velocity components
The velocity components \( v_x \) and \( v_y \) can be obtained by differentiating the position functions with respect to time.

1. **Differentiate \( x(t) \)**:
\[
x(t) = -16.0 + 3.00t - 5.00t^3
...