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

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Answer 1 · 2016-04-29T09:15:00

$v_a=5,57$

$v_a=1/(t_2-t_1)int _(t_1)^(t_2) a(t)*d t$

$v_a=1/(2-0)int _0^2(2t^2-t+4)*d t$

$v_a=1/2 [|2*t^³/3-t^2/2+4t|_0^2]$

$"assume t=0; v=0"$

$v_a=1/2[(2*2^3/3-2^2/2+4*2)-(0)]$

$v_a=1/2[16/3-2+8]$

$v_a=1/2[16/3+6]$

$v_a=1/2[(16+18)/3]$

$v_a=1/2[34/3]$

$v_a=34/6$

$v_a=5,57$

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