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1) Maybe this gives you a clue:
\[\frac{dv}{dt^2}=\frac{dv}{2tdt}=\frac{a}{2t}\]
2) Find the slope,

explain like i'm five? ^^"

so I have slope y = 4.8711x

how do you find a

Can you like give me what your problem that you have to solve is? :S

x here t^2?
If yes, then a=1/2*4.8711
:D

it's to compute the experimental value of g (gravity) from the slope value which is y = 4.8711x

You're giving me an equation, but I have no idea what this equation means. Can you clarify?

imron, how'd you know to substitute that way though?

so i have this graph where the slope of the distance vs time^2 is y=4.8711x

you could just draw the graph, but nvm I got it now.

i did

|dw:1347849013566:dw|
If this graph is what you get?

Kk. you have
\(\huge s(x)=\frac{1}{2}at^2\)
s(x) is the distance, a is acceleration, and t is time.

yezzir

assume we don't know a

or g rather

We don't know a, but it's a constant.

4.8711x = (1/2)at^2

do the x and t^2 cancel out somehow?

since x represents t^2 in the graph?

They are the same thing.

So there you go. Cancel them both out, multiply both sides by 2, and here you have a.

hmm

wtf who are you and tracks?

sorry. proceed

I finished already. ;)

|dw:1347849594808:dw|

it feels not right canceling x and t^2

Well, you defined x as t^2. So too bad.

The x axis IS t^2.

yeahhhh but

flutterit i'll see what happens

thanks for all your help, the derivations were cool

Yeah, I likes dem too :D

awwwwwww yeahhhh math.