We know that for variable acceleration case

Average velocity #v_(text{average})=(x(t)-x_0)/t# .......(1)

Given acceleration #a(t)=36-t^2# and at #t=0# object is still.

#=># #v_0=0#, time #t in [0,5]#

Now general expressions for

velocity #v(t)=dx/dt=v_0+int_0^ta(t)cdot dt# .......(2)

and Position #x(t)=x_0+int_0^tv(t)cdotdt# ......(3)

where #x_0 and v_0# are initial position and initial velocity respectively.

From equation (2)

#v(t)=int_0^t (36-t^2)cdot dt#, on integration yields

#v(t)=[36t-t^3/3]_0^t+C#, #C# being constant of integration.

We obtain #v(t)=(36t-t^3/3+C)-(36xx0-0^3/3+C)#

or #v(t)=36t-t^3/3#

Inserting this value in equation (3), we obtain

#x(t)=x_0+int_0^t(36t-t^3/3)cdotdt#, on integration over the given range of time #t# yields

#x(5)=x_0+int_0^5(36t-t^3/3)cdotdt#

#x(5)=x_0+[36t^2/2-1/3cdott^4/4+C_1]_0^5#, where #C_1# is constant of integration

#x(5)=x_0+{36 5^2/2-1/3cdot5^4/4+C_1}-{36 0^2/2-1/3cdot0^4/4+C_1}#

or #x(5)=x_0+450-52.08dot3#

or #x(5)=x_0+397.91dot6#

Inserting this value in equation (1), we obtain

#v_(text{average})=(cancel (x_0)+397.91dot6-cancel(x_0))/5#

#v_(text{average})=79.58dot3 text{ units}#