Question

An object travels North at `5 m/s` for `3 s` and then travels South at `4 m/s` for `6 s`. What are the object's average speed and velocity?

Answer 1

Average speed: `4.33"m"/"s"`

Average velocity: `-1"m"/"s"`

Explanation:

We're asked to calculate both the average speed and the average velocity for a given displacement.

Let's calculate the average velocity first.

The average velocity of an object moving in one dimension (which we'll call `y`, as it's moving north and south) is defined as

`v_(av-y) = (Deltay)/(Deltat)`

where

• `Deltay` is the total change in the object's position (what we must find), and

• `Deltat` is the total change in time, which is

`overbrace(3"s")^"first displacement" + overbrace(6"s")^"second displacement" = overbrace(9"s")^"total"`

To find the total net change in the position of the object, we need to calculate how far it is from the origin after both displacements. We'll label north "positive" and south "negative". The two displacements are

`1`:

`5"m"/(cancel("s"))(3cancel("s")) = 15` `"m"`

`2`:

`-4"m"/(cancel("s"))(6cancel("s")) = -24` `"m"`

The total displacement is the sum of the constituent displacements:

`15` `"m" + (-24` `"m") = -9` `"m"`

Therefore, the average velocity of the object is

`v_(av-y) = (Deltay)/(Deltat) = (-9"m")/(9"s") = color(red)(-1)` `color(red)("m"/"s")`

Now, let's calculate the average speed of the object, which is slightly different than the average velocity. The average speed is the total distance traveled divided by the time interval (still `9` `"s"`):

`overbrace(v_(av-y))^"speed" = ("total distance traveled")/(Deltat)`

To find the total distance traveled, we neglect all signs when it comes to the object's position. Thus, traveling `15` `"m"` north and then `24` `"m"` south is a total of

`15` `"m" + 24` `"m" = 39` `"m"`

The average speed of the object is thus

`overbrace(v_(av-y))^"speed" = ("total distance traveled")/(Deltat) = (39"m")/(9"s") = color(blue)(4.33"m"/"s"`