What is the average speed of an object that is still at #t=0# and accelerates at a rate of #a(t) =t/6# from #t in [0, 1]#?

1 Answer
Jan 4, 2016

Answer:

You also need the initial speed of the object #u_0#. The the answer is:

#u_(av)=0.042+u_0#

Explanation:

Definition of acceleration:

#a(t)=(du)/dt#

#a(t)*dt=du#

#int_0^ta(t)dt=int_(u_0)^udu#

#int_0^t(t/6)dt=int_(u_0)^udu#

#1/6int_0^t(t)dt=int_(u_0)^udu#

#1/6(t^2/2-0^2/2)=u-u_0#

#u(t)=t^2/12+u_0#

To find the average speed:

#u(0)=0^2/12+u_0=u_0#

#u(1)=1^2/12+u_0=1/12-u_0#

#u_(av)=(u_0+u_1)/2#

#u_(av)=(u_0+1/12+u_0)/2#

#u_(av)=(2u_o+1/12)/2#

#u_(av)=(2u_0)/2+(1/12)/2#

#u_(av)=u_0+1/24#

#u_(av)=0.042+u_0#