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}
