First, see that in the time interval 3.44 to 9.49, the average velocity is [tex]2.9\text{m} / (9.49\text{s} - 3.44\text{s})=2.9\text{m}/6.05\text{s} \approx 0.479\text{m/s}[/tex]. So, as we have uniform acceleration, the velocity must be linearly increasing over this entire interval, so for the average to be 0.479 m/s over this interval, the velocity must be 0.479 m/s in the exact middle of this interval, or at 5.465s.
We now note that the object starts from rest, which means that at 0s, the velocity is 0 m/s. So, in 5.465 seconds, the velocity increases by 0.479 m/s. We again have that the object undergoes uniform acceleration, meaning that the acceleration over this interval is a constant [tex]\frac{0.479\text{m/s}}{5.465\text{s}} \approx 0.0876 \text{m/s}^2[/tex].
Finally, note again that as we are looking at uniform acceleration, by the same principle at the beginning, the average velocity of the object during the time interval from 18.6s to 22.92s is the same as the velocity at the exact middle of this interval, or at 20.76s. We have that acceleration is constant and 0.0876 m/s^2, and initial velocity is 0 at 0s. So, in 20.76 seconds, the object will have accelerated [tex]0.0876\text{m/s}^2 \cdot 20.76\text{s} \approx 1.82 \text{m/s}[/tex].
So, average velocity will be 1.82 m/s over the time interval 18.6s to 22.92s.
