If displacement S at time t is `S=t^(3)-3t^(2)-15t+12`, then acceleration at time `t=1 sec` is

Options

A. `6 "units" // sec ^(2)`
B. `-6 "units" // sec ^(2)`
C. 0
D. ` "units" // sec ^(2)`

Answer

To find the acceleration at time \( t = 1 \) second given the displacement function \( S(t) = t^3 - 3t^2 - 15t + 12 \), we will follow these steps:

### Step 1: Find the First Derivative (Velocity)
The first derivative of the displacement function \( S(t) \) with respect to time \( t \) gives us the velocity \( V(t) \).

\[ V(t) = \frac{dS}{dt} = \frac{d}{dt}(t^3 - 3t^2 - 15t + 12) \]

Using the power rule for differentiation:

\[ V(t) = 3t^2 - 6t - 15 \]

### Step 2: Find the Second Derivative (Acceleration)
The second derivative of the displacement function \( S(t) \) gives us the acceleration \( A(t) \).

\[ A(t) = \frac{d^2S}{dt^2} = \frac{d}{dt}(3t^2 - 6t - 15) \]

Again, using the power rule for differentiation:

\[ A(t) = 6t - 6 \]

### Step 3: Evaluate the Acceleration at \( t = 1 \) second
Now, we will substitute \( t = 1 \) into the acceleration function \( A(t) \):

\[ A(1) = 6(1) - 6 = 6 - 6 = 0 \]

### Conclusion
The acceleration at time \( t = 1 \) second is \( 0 \).

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