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, 4]#?

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Answer 1 · 2016-10-20T09:30:49

Given acceleration of the object at t th sec

#a(t)=12-2t#

#=>(dv(t))/(dt)=12-2t#

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

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

#"Given at "t=0, v(0)=0" " So" " c=0#

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

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

So the distance traversed during #t in abs(3,4)#

#s=int_3^4(12t-t^2)dt#

#=[(12t^2)/2-t^3/3]_3^4#

#=6*4^2-4^3/3 -6*3^2+3^3/3=89/3#

So the average speed

#V_"average"="Distance covered"/"Time taken"=(89/3)/1=89/3#