The position of a particle is given by the equation `f(t)=t^(3)-6t^(2)+9t` where t is measured in second and s in meter. Find the acceleration at time t. What is the acceleration at 4 s?

Explanation

To find the acceleration of the particle given the position function \( f(t) = t^3 - 6t^2 + 9t \), we will follow these steps:

### Step 1: Find the velocity function
The velocity \( v(t) \) is the first derivative of the position function \( f(t) \) with respect to time \( t \).

\[ v(t) = \frac{df(t)}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 9t) \]

Using the power rule of differentiation:

- The derivative of \( t^3 \) is \( 3t^2 \).
- The derivative of \( -6t^2 \) is \( -12t \).
- The derivative of \( 9t \) is \( 9 \).

Thus, the velocity function is:

\[ v(t) = 3t^2 - 12t + 9 \]

### Step 2: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \) with respect to time \( t \).

\[ a(t) = \frac{dv(t)}{dt} = \frac{d}{dt}(3t^2 - 12t + 9) \]

Again, applying the power rule:

- The derivative of \( 3t^2 \) is \( 6t \).
- The derivative of \( -12t \) is \( -12 \).
- The derivative of \( 9 \) is \( 0 \).

Thus, the acceleration function is:

\[ a(t) = 6t - 12 \]

### Step 3: Find the acceleration at \( t = 4 \) seconds
Now we will substitute \( t = 4 \) into the acceleration function:

\[ a(4) = 6(4) - 12 \]

Calculating this gives:

\[ a(4) = 24 - 12 = 12 \, \text{m/s}^2 \]

### Final Answer
The acceleration at \( t = 4 \) seconds is \( 12 \, \text{m/s}^2 \).

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