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

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

What is the average speed of an object that is still at t=0t=0 and accelerates at a rate of a(t) =12-2ta(t) =12-2t from t in [3, 4]t in [3, 4]?