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

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

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, 6]t in [3, 6]?