Given the velocity function \(v(t) = 1.2t - 0.12t^2\), we need to find the displacement when the acceleration is 0.6 m s^-2.

First, find the acceleration by differentiating the velocity function:

\(a(t) = \frac{dv}{dt} = 1.2 - 0.24t\)

Set the acceleration equal to 0.6:

\(1.2 - 0.24t = 0.6\)

Solve for \(t\):

\(0.24t = 0.6\)

\(t = 2.5\)

Now, find the displacement by integrating the velocity function from 0 to 2.5:

\(s = \int_0^{2.5} (1.2t - 0.12t^2) \, dt\)

\(s = \left[0.6t^2 - 0.04t^3\right]_0^{2.5}\)

\(s = (0.6 \times 2.5^2 - 0.04 \times 2.5^3) - (0 - 0)\)

\(s = 3.125\)

Therefore, the displacement is 3.125 m.