# What is the average speed, on t in [0,5], of an object that is moving at 7 m/s at t=0 and accelerates at a rate of a(t) =2t^2+2 on t in [0,2]?

Jun 10, 2017

$32.8 \text{m"/"s}$

#### Explanation:

The average speed of the object is represented by the equation

overbrace(v_"av-x")^"speed" = "total distance traveled"/"time interval"

Since the acceleration is always positive, the total distance traveled is equal to the total displacement, so the average speed is the same as the average velocity, so we can use the equation

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

We must therefore find the total displacement of the object after $t = 5$ $\text{s}$.

To find the position equation from the acceleration equation, we need to do an integration. We need to integrate the acceleration equation twice because position is the second integral of acceleration (velocity lies in between, as the first integral). We can use the equation

${v}_{x} = {v}_{0 x} + {\int}_{0}^{t} {a}_{x} \mathrm{dt}$

to find the velocity equation, and then integrate the velocity equation to find the position equation. Knowing that

$\int {t}^{n} = \frac{1}{n + 1} {t}^{n + 1}$

the integral of color(red)(2)t^(color(blue)(2) is

$\left(\frac{1}{\textcolor{b l u e}{2} + 1}\right) \left(\textcolor{red}{2}\right) {t}^{\textcolor{b l u e}{2} + 1} = \frac{2}{3} {t}^{3}$

and the integral of color(red)(2, which is 2t^(color(blue)(0), is

$\left(\frac{1}{\textcolor{b l u e}{0} + 1}\right) \left(\textcolor{red}{2}\right) {t}^{\textcolor{b l u e}{0} + 1} = 2 t$

The initial velocity is $7 \text{m"/"s}$. Plugging these values into the integral equation, we have

${v}_{x} = 7 + \frac{2}{3} {t}^{3} + 2 t$

(excluding units)

The equation for position of time from velocity is

$x = {x}_{0} + {\int}_{0}^{t} {v}_{x} \mathrm{dt}$

Its starting position is assumed to be $x = 0$. Using the same technique as before, we have

$x = 0 + 7 t + \frac{1}{6} {t}^{4} + {t}^{2}$

(excluding units)

Now, all we have to do is find the position at $t = 5$ $\text{s}$. Plugging in $5$ for $t$ to this equation, we have

x = 0 + 7(5) + 1/6(5)^4 + (5)^2 = color(purple)(164.2 color(purple)("m"

Finally, let's plug this back into our average velocity equation:

${v}_{\text{av-x" = (color(purple)(164.2"m"))/(5"s") = color(green)(32.8"m"/"s}}$