A particle that is moving in an xy plane has a position vector given by `vecr=(3.00t^(3)-6.00t)hati+(7.00-8.00t^(4))hatj`, where `vecr` is measued in meters and t is measured in seconds. For t = 3.00 s, in unitvector notation, find (a) `vecr`, (b) `vecv`, and ( c) `veca`. (d) Find the angle between the positive direction of the x axis and a line that is tangent to the path of the particle at t = 3.00 s.

Explanation

To solve the problem step by step, we will find the position vector \(\vec{r}\), the velocity vector \(\vec{v}\), the acceleration vector \(\vec{a}\), and the angle between the positive x-axis and the tangent to the path of the particle at \(t = 3.00\) seconds.

### Step 1: Find the Position Vector \(\vec{r}\)

The position vector is given by:
\[ \vec{r} = (3.00t^3 - 6.00t) \hat{i} + (7.00 - 8.00t^4) \hat{j} \]
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