If the position of the block in a block-spring oscillator is given by
`x(t) = (3.00 xx 10^(-3)m) cos [(4.00 "rad/s") t + 0.120"rad"]`, what is the block's position, velocity and acceleration at time t= 2.30s?

Explanation

To solve the problem, we need to find the position, velocity, and acceleration of a block in a block-spring oscillator at a specific time \( t = 2.30 \, \text{s} \). The position of the block is given by the equation:

\[ x(t) = (3.00 \times 10^{-3} \, \text{m}) \cos \left( (4.00 \, \text{rad/s}) t + 0.120 \, \text{rad} \right) \]

### Step 1: Calculate the Position at \( t = 2.30 \, \text{s} \)

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