# Homework Help: Describing the motion of a graph.

1. Feb 18, 2012

### koolkidx45

Hi i have a graph given in kinematics. I need to describe the motion and calculate the average velocity during the entire duration. I have attached a picture. How would i calculate the average velocity

No equations give

I said that the velocity is not constant. I dont know if the average velocity woul dbe negative?

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2. Feb 18, 2012

### genericusrnme

How would you define average velocity?

$\Delta v = \frac{\Delta x}{\Delta t}$

Something like that, yes?
Use that definition, what comes out?

3. Feb 18, 2012

### koolkidx45

Thanks for your reply. But would the average velocity be negative looking at the trend of the graph? And how would i describe this motion?

4. Feb 18, 2012

### HallsofIvy

From your graph, it looks like x is 25 when t is about 2 and x is about 2 when t= 30. So what are $\Delta x$, $\Delta t$, and $\Delta x/\Delta t$?

5. Feb 18, 2012

### genericusrnme

If you apply the definition of the average velocity you will see that it is negative since the change in position is definitely negative (it seems to start at some high x and ends up at x=0)
So yes, the average velocity of would be negative.

To describe the motion, we could look at the average velocity between different times and see how they relate.

Between t=10 and t=20 the position has gone from x=15 to x=5, so the average velocity during this time is given as'

$\Delta v = \frac{\Delta x}{\Delta t} = \frac{-10}{10} = -1$

If we look at the average velocity at some later time, say t=20 and t=30 the position has gone from x=5 to perhaps x~2 and so

$\Delta v = \frac{\Delta x}{\Delta t} = \frac{-3}{10} = -0.3$

So we can see that the average velocity is decreasing as time increases.
We can also see that the position is finally going to reach x=0 as t gets to be quite big.

We could describe this by saying that the position 'asymptotes' to x=0 for large t and that the velocity also 'asymptotes' to 0 for large t.

Do you understand my reasoning here?

6. Feb 18, 2012

### koolkidx45

Yes but what is the general description of this motion. Like is the object moving in an westward direction?

7. Feb 18, 2012

### genericusrnme

That depends on what direction you want to set as the positive x direction.
If you set west as your positive x direction then you're moving westward, if you set north as your positive x direction then you're moving northward.
It is completely up to you which direction you set positive x to be facing

8. Feb 18, 2012

### technician

I would say that approximatey 25m was covered in approximately 30 to 40 secs.
So the only sensible answer for me is that the average velocity is between 25/30 and 25/40 m/s
ie 0.83m/s and 0.63m/s