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

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Answer 1 · 2015-12-27T06:05:51

$15.32 m/s$ $15.32"m/s"$

Acceleration is a function of time here so the average velocity will be the area under the acceleration - time graph between the limits given.

graph{2x^2-3x-2 [-10, 10, -5, 5]}
$:.v_(ave)=int_2^4a.dt$

$v_(ave)=int_2^4(2t^2-3t-2).dt$

$v_(ave)=[(2t^3)/3-(3t^2)/2-2t]_(2)^(4)$

$v_(ave)=[(2xx4^(3))/(3)-(3xx4^(2))/2-(2xx4)]-[(2xx2^(3)-(3xx2^(2))/2-(2xx2)]$

$v_(ave)=[128/3-24-8]-[16/3-6-4]$

$v_(ave)=10.66-(-4.66)$

$v_(ave)=15.32"m/s"$

What is the average speed of an object that is still at t=0t=0t=0 and accelerates at a rate of a(t) =2t^2-3t-2a(t)=2t2−3t−2a(t) =2t^2-3t-2 from t in [2, 4]t∈[2,4]t in [2, 4]?