# Question e7978

Sep 19, 2016

$\text{1.4 m/s}$

#### Explanation:

The idea here is that the average acceleration of an object can be described as the average change in velocity, $\Delta v$, that occurs in a given time interval, $\Delta t$.

In other words, the average acceleration tells you the rate at which the velocity of the object changed from an initial value to a final value in a period of time.

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}_{\text{avg" = "change in velocity"/"change in time}} = \frac{\Delta v}{\Delta t}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The change in velocity can be calculated by subtracting the initial velocity from the final velocity

$\Delta v = {v}_{\text{final" - v_"initial}}$

In this case, the treadmill started from rest, which means that ${v}_{\text{initial}} = 0$. You thus have

$\Delta v = {v}_{\text{final}}$

The change in time is said to be equal to $5$ minutes. In this case, it doesn't matter when the treadmill started moving, all that matters is the total time it spent moving. You thus have

$\Delta t = \text{5 min}$

Now, notice that the average acceleration is given in meters per second per second, $\text{m/s/s}$, which is another way of expressing meters per square second, ${\text{m/s}}^{2}$.

Convert the change in time from minutes to seconds

5 color(red)(cancel(color(black)("min"))) * "60 s"/(1color(red)(cancel(color(black)("min")))) = "300 s"

You now have what you need to use the equation for average acceleration

${a}_{\text{avg" = v_"final}} / \left(\Delta t\right)$

Rearrange to solve for ${v}_{\text{final}}$

${v}_{\text{final" = a_"avg}} \cdot \Delta t$

Plug in your values to find

v_"final" = 4.7 * 10^(-3)"m"/("s" * color(red)(cancel(color(black)("s")))) * 300color(red)(cancel(color(black)("s")))= color(green)(bar(ul(|color(white)(a/a)color(black)("1.4 m/s")color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only provided one sig fig for the time interval.