What is the average speed of an object that is still at #t=0# and accelerates at a rate of #a(t) =12-2t# from #t in [3, 6]#?

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Answer 1 · 2016-11-08T02:56:42

Given
Acceleration #a(t)=12-2t#

#=>(d(v(t)))/(dt)=12-2t#

#=>int(d(v(t)))=int(12-2t)dt#

#=>v(t)=12t-2xxt^2/2+c#

#=>v(t)=12t-t^2+c#

where t = integration constant

Now by the given condition

at t=0, v(0)=0

So #v(0)=0=12xx0-0^2+c=>c=0#

Hence

#=>v(t)=12t-t^2#

#=>(d(s(t)))/(dt)=12t-t^2#

#=>int_3^6(d(s(t)))=int_3^6(12t-t^2)#

#=>s(6)-s(3)=[6t^2-t^3/3]_3^6#

#=6*6^2-6^3/3-6*3^2+3^3/3#

#=216-72-54+9=99#

Average speed in time [3,6] is

#=(s(6)-s(3))/(6-3)=99/3=33 unit#