What is the average speed of an object that is still at #t=0# and accelerates at a rate of #a(t) = 36-t^2# from #t in [0, 5]#?

1 Answer
Mar 5, 2016

Answer:

Average speed #=79.58dot3 text{ units}#

Explanation:

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}#