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

May 26, 2017

$- 0.5 \text{m"/"s}$

#### Explanation:

Since average speed is given by the equation

${v}_{\text{av-x}} = \frac{\Delta x}{\Delta t}$

We need to find the positions of the car at times $t = 2$ and $t = 3$.

We can derive first the velocity equation from the acceleration, and then the position from that velocity equation, via integration, and we'll assume the initial position and velocity are $0$:

${v}_{x} \left(t\right) = {\int}_{0}^{t} {a}_{x} \mathrm{dt} = \frac{2}{3} {t}^{3} - {t}^{2} - 2 t$

$x \left(t\right) = {\int}_{0}^{t} {v}_{x} \mathrm{dt} = \frac{1}{6} {t}^{4} - \frac{1}{3} {t}^{3} - {t}^{2}$

Now we have our position equation, $x \left(t\right)$, so let's plug in the values $2$ and $3$ for $t$ to find the position at those times:

$x \left(2\right) = \frac{1}{6} {\left(2\right)}^{4} - \frac{1}{3} {\left(2\right)}^{3} - {\left(2\right)}^{2} = - 4.0 \text{m}$

$x \left(3\right) = \frac{1}{6} {\left(3\right)}^{4} - \frac{1}{3} {\left(3\right)}^{3} - {\left(3\right)}^{2} = - 4.5 \text{m}$

So, the average velocity of the car on the interval $t \in \left[2 , 3\right]$ is

v_("av-x") = ((-4.5 "m") - (-4.0 "m"))/(3"s" - 2"s") = color(red)(-0.5"m"/"s"