A 0.150 kg particle moves along an x axis according to `x(t)=-13.00+2.00t+4.00t^(2)-3.00t^(3)`, with x in meters and t in seconds. In unit-vector notation, what is the net force acting on the particle at `t=2.60" s"` ?

Explanation

To solve the problem, we need to find the net force acting on a particle given its position as a function of time. The steps to solve this are as follows:

### Step 1: Write down the position function
The position of the particle is given by the equation:
\[ x(t) = -13.00 + 2.00t + 4.00t^2 - 3.00t^3 \]

### Step 2: Differentiate the position function to find velocity
To find the velocity \( v(t) \), we differentiate the position function \( x(t) \) with respect to time \( t \):
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