# An object travels North at 3 m/s for 7 s and then travels South at 4 m/s for  6 s. What are the object's average speed and velocity?

Oct 11, 2017

Speed = 3 m/s
Velocity = +0.2 m/s South (or -0.2 m/s)

#### Explanation:

Two formulas that will come in handy here are:

$\frac{\text{Speed" = "distance}}{t}$

$\frac{\text{Velocity" = "displacement}}{t}$

So firstly, let's calculate the amount of distance travelled over the course of the two time intervals:

For Interval 1:

$d = \text{speed} \cdot t$
$= 3 \left(\frac{m}{s}\right) \cdot 7 s$
$= 21 m$

For Interval 2:

$d = \text{speed} \cdot t$
$= 4 \left(\frac{m}{s}\right) \cdot 6 s$
$= 24 m$

Notice that since distance is a scalar quantity (i.e. it has no direction, just a magnitude), we do not need to worry about the North or South here.

Now that we have the total distance travelled, all we need to do is divide by the total time, and we'll get our answer:

"speed" = d_("total")/t_("total")

$= \frac{21 m + 24 m}{7 s + 6 s}$

$= 3.4615 \frac{m}{s} \implies$ 3 m/s
*rounded to one significant digit.

Now, velocity. The idea is similar, but this time we need to worry about a displacement over time, not a distance. Now, displacement is a vector quantity, meaning that both magnitude and direction matter. Hence, in this case we will have to take the North and South into account.

Firstly, I am going to define North as my positive direction here. This may seem redundant to do, but for the numbers to work out it is key to specify which direction you're setting as positive. You could set South to be your positive direction and still get the right answer, but you'd need to interpret your final answer differently, as you will see.

Now let's calculate displacement:

Interval 1 (Travelling North):

$\text{Displacement" (Deltax) = "Velocity} \cdot t$

$= \left(+ 3 \frac{m}{s}\right) \cdot 7 s$

$= + 21 m$

Note that I've added a plus sign up there. Again, it may seem redundant, but when working with vector quantities it is helpful to explicitly label the directions of the components you're working with.

Interval 2 (Travelling South):

$\Delta x = \text{Velocity} \cdot t$

$= \left(- 4 \frac{m}{s}\right) \cdot 6 s$

= $- 24 m$

To reiterate, because South is defined as my negative direction, I gave my Southward velocity a negative sign, which in turn led to my southward displacement to have a negative sign.

Now the final step:

"Velocity" = (Deltax_("total"))/t_("total")

$= \frac{+ 21 m - 24 m}{7 s + 6 s}$
$= - 0.23076 \frac{m}{s} \implies$ -0.2 m/s
*rounded to one significant digit.

What does this answer mean? Well, recall that since North is defined as our positive, this means that we're traveling -0.2 m/s relative to North, or in other words, +0.2 m/s South.

If you had set South to be your positive, you'd have got a positive answer, which would have the same interpretation.

Hope that helped :)