To find \(\frac{ds}{dt}\), we take the derivative of \(S\) with respect to \(t\). Applying the product rule, we get:

\(\frac{ds}{dt} = \frac{d}{dt}(4t + 3) \cdot (t - 2) + (4t + 3) \cdot \frac{d}{dt}(t - 2)\)

Simplify the derivatives:

\(\frac{ds}{dt} = 4 \cdot (t - 2) + (4t + 3) \cdot 1\)

Expand and simplify further:

\(\frac{ds}{dt} = 4t - 8 + 4t + 3\)

Combine like terms:

\(\frac{ds}{dt} = 8t - 5\)

Now, to find \(\frac{ds}{dt}\) when \(t = 5\) secs:

\(\frac{ds}{dt} = 8 \cdot 5 - 5 = 40 - 5 = 35\)