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

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Answer 1 · 2017-07-16T15:13:19

The average speed is $= - 0.5 m s^{- 1}$ $=-0.5ms^-1$

The speed is the integral of the acceleration

$a (t) = 3 t - 4$ $a(t)=3t-4$

$v (t) = \int (3 t - 4) d t = \frac{3}{2} t^{2} - 4 t + C$ $v(t)=int(3t-4)dt=3/2t^2-4t+C$

Plugging in the initial conditions at $t = 0$ $t=0$

$v (0) = 0 - 0 + C = 0$ $v(0)=0-0+C=0$ , $\Rightarrow$ $=>$ , $C = 0$ $C=0$

The average velocity is

$(3 - 2) ¯ v = \int_{2}^{3} (\frac{3}{2} t^{2} - 4 t) d t$ $(3-2)barv=int_2^3(3/2t^2-4t)dt$

$= {[\frac{1}{2} t^{3} - 2 t^{2}]}_{2}^{3}$ $=[1/2t^3-2t^2]_2^3$

$= (\frac{27}{2} - 18) - (4 - 8)$ $=(27/2-18)-(4-8)$

$¯ v = - 0.5$ $barv=-0.5$

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