Remember how one calculates average value of any quantity:
\langle f \rangle = \frac{1}{t_2-t_1}\int_{t_1}^{t_2} f(t)dt
We need to calculate \langle v \rangle, where v is the velocity, for that we will need to first find the velocity it self, but we do know that
\frac{dv}{dt}=a, where a is the acceleration or in other words:
v(t_2)-v(t_1)= \int_{t_1}^{t_2}a(t)dt
We have been given:
a(t)=4-t and v(0)=0, lets us plug these in the above formula, with t_1=0, t_2=t
We get,
v(t)=\int_{0}^{t}(4-t)dt
v(t)= (4t-\frac{t^2}{2})
Now for its average value we use the first formula used
\langle v \rangle = \frac{1}{3-2}\int_{2}^{3}(4t-\frac{t^2}{2}) dt
\langle v \rangle = \frac{1}{3-2}(2t^2-\frac{t^3}{6}) with limits 2 and 3.
Inserting limits
\langle v \rangle = (50-19/6)
\langle v \rangle = (281/6)=46.8 units/sec
