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 solve the problem, we need to find the acceleration of a particle whose position is given by the equation \( f(t) = t^3 - 6t^2 + 9t \). 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 \( t \).

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

Calculating the derivative:

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

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

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

Calculating the derivative:

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

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

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

Calculating the value:

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

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

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