If `s=(t^(3))/(3)-(1)/(2)t^(2)-6t+5`, Find the value of `(d^(2)s)/(dt^(2))` at `t=1`

Explanation

To find the value of \(\frac{d^2s}{dt^2}\) at \(t=1\) for the given function \(s = \frac{t^3}{3} - \frac{1}{2}t^2 - 6t + 5\), we will follow these steps:

### Step 1: Differentiate \(s\) to find \(\frac{ds}{dt}\)

The function is:
\[ s = \frac{t^3}{3} - \frac{1}{2}t^2 - 6t + 5 \]

Now we differentiate \(s\) with respect to \(t\):
\[ \frac{ds}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) - \frac{d}{dt}\left(\frac{1}{2}t^2\right) - \frac{d}{dt}(6t) + \frac{d}{dt}(5) \]

Calculating each term:
- \(\frac{d}{dt}\left(\frac{t^3}{3}\right) = t^2\)
- \(\frac{d}{dt}\left(-\frac{1}{2}t^2\right) = -t\)
- \(\frac{d}{dt}(-6t) = -6\)
- \(\frac{d}{dt}(5) = 0\)

Combining these results:
\[ \frac{ds}{dt} = t^2 - t - 6 \]

### Step 2: Differentiate \(\frac{ds}{dt}\) to find \(\frac{d^2s}{dt^2}\)

Now we differentiate \(\frac{ds}{dt}\):
\[ \frac{d^2s}{dt^2} = \frac{d}{dt}(t^2 - t - 6) \]

Calculating each term:
- \(\frac{d}{dt}(t^2) = 2t\)
- \(\frac{d}{dt}(-t) = -1\)
- \(\frac{d}{dt}(-6) = 0\)

Combining these results:
\[ \frac{d^2s}{dt^2} = 2t - 1 \]

### Step 3: Evaluate \(\frac{d^2s}{dt^2}\) at \(t=1\)

Now we substitute \(t = 1\) into \(\frac{d^2s}{dt^2}\):
\[ \frac{d^2s}{dt^2} \bigg|_{t=1} = 2(1) - 1 = 2 - 1 = 1 \]

### Final Answer:
The value of \(\frac{d^2s}{dt^2}\) at \(t=1\) is:
\[ \boxed{1} \]