Question #e7978

1 Answer
Sep 19, 2016

Answer:

#"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, #Deltav#, that occurs in a given time interval, #Deltat#.

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.

#color(purple)(bar(ul(|color(white)(a/a)color(black)(a_"avg" = "change in velocity"/"change in time" = (Deltav)/(Deltat))color(white)(a/a)|)))#

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

#Deltav = v_"final" - v_"initial"#

In this case, the treadmill started from rest, which means that #v_"initial" = 0#. You thus have

#Deltav = v_"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

#Deltat = "5 min"#

Now, notice that the average acceleration is given in meters per second per second, #"m/s/s"#, which is another way of expressing meters per square second, #"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_"avg" = v_"final"/(Deltat)#

Rearrange to solve for #v_"final"#

#v_"final" = a_"avg" * Deltat#

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.