# Constant acceleration kinematics doubt

1. Sep 24, 2011

### zuppi

Hi,
I hope it is a very simple mistake and easy to spot.
Suppose you start at rest so initial velocity is 0 and you accelerate constantly over a certain distance.
An acceleration vs. distance graph (not time) would be like a rectangle. The area under the graph would be A=as=(s/t2)*s = v2.
According to v2=u2+2as, so that v2=2as and not as. I cant work out if the math or the general idea is wrong.

Hopefully I was clear enough. Thanks for any help!
zuppi

2. Sep 24, 2011

### Pengwuino

What you did was basically find $\int a ds$ which is not going to equal the velocity squared. Why should it? Just because the units are the same doesn't mean the actual quantities are the same.

3. Sep 24, 2011

### AlephZero

To get the velocity from the accleration, you have to find the area under the graph of acceleration against time, not acceleration against distance.

Actually, the area under the graph of acceleration against distance does mean something physical, but if you are just starting to study kinematics, you probably don't know enough physics yet to give a simple explanation of what it means.

4. Sep 24, 2011

### Pengwuino

Shhh, don't blab the secret out like that.

5. Sep 25, 2011

### zuppi

Thanks!
How can I get any useful information out of an acceleration vs distance graph if the acceleration is not constant? The area is not v2 or 1/2 v2 is it?

6. Sep 25, 2011

### Ken G

Now it seems you want to do calculus! Good for you, this is more or less the question Newton asked himself. Formulas like v2=2as are to make it so you don't have to do calculus, but they require constant a. The calculus says that you can use a=v*dv/ds and the area as is more generally an integral that results in v2/2, if v starts out zero. What's interesting is that this will be the area under the a vs. s graph for any function a(s), not just constant a. But until you get calculus, you will only treat constant a, and then the area is rectangular, just as you said. The kinematic formula tells you how to connect that area to the resulting final v, but when you notice that v2 comes out twice the area under the a(s) curve for constant a, it turns out that is also true for any a(s), which is I think just what you were speculating.

Last edited: Sep 25, 2011
7. Sep 25, 2011

### zuppi

Thats good so I can integrate a(s) to get half v2. How can I use calculus to transform that into a velocity or acceleration versus time graph so I can calculate the total time taken?

8. Sep 25, 2011

### Ken G

Ah, so you are taking physics at the calculus level. This kind of thing gets easier once you've seen it a few times. If you have a(s), and we know this = v*dv/ds for getting v(s), now you just need v(t). Sometimes it is easier to start with what you want to know, which here is what is going on with t, so then you are looking for t(v) rather than v(t), but you can see that it is basically the same thing. Here's how you use calculus to get t(v):
t = integral over dt = integral over ds of dt/ds = integral over ds of 1/v(s).
You see how I got that by basically saying dt = ds*dt/ds, which is just like multiplying and dividing by ds? The idea there is if s is your independent variable, it means you have a lot of information about how things depend on s. That means you want to convert your integrals into integrals over ds, so you don't want to do an integral like integral over dt of a(t), since t is not your independent variable here and you don't know anything about how things depend on t.