# What is the average speed of an object that is moving at 22 m/s at t=0 and accelerates at a rate of a(t) =t-2 on t in [0,2]?

May 26, 2017

$21 \text{m"/"s}$

#### Explanation:

Since average speed is given by the equation

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

we need to find the position of the object at times $t = 0$ and $t = 2$.

Since no initial position is given, we'll assume its position at $t = 0$ is $0$.

We're given an equation for how the object's acceleration varies with time, and what we need to find is how its position varies with time, so we can find its position at time $t = 2$. We can do this by integrating the acceleration equation to find its velocity equation, then integrating that to find its position equation. If you're not familiar with integrals, just know that the integral of ${t}^{n}$ is denoted $\int {t}^{n} \mathrm{dt}$ and is equal to $\frac{1}{n + 1} {t}^{n + 1}$ (except for $n = - 1$).

The velocity equation for the object is

v_x = v_(0x) + int_0^t a_xdt = 22"m"/"s" + 1/2(1 "m"/("s"^3))t^2 - (2"m"/("s"^2))t

And the position equation with respect to time is

$x = {x}_{0} + {\int}_{0}^{t} {v}_{x} \mathrm{dt} = 0 + \frac{1}{6} \left(1 \text{m"/("s"^3))t^3 - 1/2(2"m"/("s"^2))t^2 + (22"m"/"s}\right) t$

Which, after simplifying, yields

$x = \frac{1}{6} \left(1 \text{m"/("s"^3))t^3 - (1"m"/("s"^2))t^2 + (22"m"/"s}\right) t$

Plugging in $t = 2$ yields a position of

x = 1/6(1"m"/(cancel("s"^3)))(2cancel("s"))^3 - (1"m"/(cancel("s"^2)))(2cancel("s"))^2 + (22"m"/cancel("s"))(2cancel("s"))

= color(red)(41"m" ($2$ significant digits)

Lastly, let's calculate the average velocity, using our first equation:

${v}_{\text{av" = ((color(red)(41"m")) - (0"m"))/(2"s") = color(blue)(21"m"/"s}}$

Rounded to two significant figures.