To find \(\frac{ds}{dt}\) when \(t = \frac{4}{5}\) seconds, we need to differentiate the expression \(s = (2 + 3t)(5t - 4)\) with respect to \(t\).

Let's differentiate each term using the product rule:

\(\frac{d}{dt}[(2 + 3t)(5t - 4)] = (5t - 4) \cdot \frac{d}{dt}(2 + 3t) + (2 + 3t) \cdot \frac{d}{dt}(5t - 4)\)

\(\frac{d}{dt}[(2 + 3t)(5t - 4)] = (5t - 4) \cdot 3 + (2 + 3t) \cdot 5\)

Now, substitute \(t = \frac{4}{5}\) seconds:

\(\frac{ds}{dt} = (5 \cdot \frac{4}{5} - 4) \cdot 3 + (2 + 3 \cdot \frac{4}{5}) \cdot 5\)

Simplify the expression:

\(\frac{ds}{dt} = (4 - 4) \cdot 3 + (2 + \frac{12}{5}) \cdot 5\)

\(\frac{ds}{dt} = 0 \cdot 3 + (\frac{10 + 12}{5}) \cdot 5\)

\(\frac{ds}{dt} = \frac{22}{5} \cdot 5\)

\(\frac{ds}{dt} = 22\) units per second

So, the correct answer is 22 units per second.