By definition, the speed after a time \(t\) of the pebble is \(v(t)=g\cdot t\) where \(g\approx 10m/s^2\) is the G-force's acceleration.
Therefore the average value over one second is \(\ \textless \ v\ \textgreater \ =\int_0^1v(t)dt=\left[\frac12gt^2\right]_0^1=\frac g2\)
Using \(g=10m/s^2\) we therefore have an average speed of \(\boxed{\ \textless \ v\ \textgreater \ =5m/s}\)
