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 #
