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jaersyn
Can someone explain to me why the slope on a distance vs. time^2 graph represents 1/2 the acceleration of that object? And also how to figure out the spring constant given a force vs. displacement graph? Thanks
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? ^^"
y = 1/2 * gt^2 the slope of the line is y = ax^2 so my thinking is x^2 corresponds to t^2 while a corresponds to 1/2*gt but i don't know how to express that algebraically
The slope of distance vs time^2: \[\frac{dy}{dt^2}= \frac{dy}{2tdt}=\frac{v}{2t}=\frac{a}{2} \] Is this clear enough?
\(\Large\frac{d^2s}{dt^2}=a\) This is acceleration by definition. \(\Large\frac{ds}{dt}=at+C_1=v\) This is velocity by definition. \(\Large s=\frac{1}{2}at^2+C_1t+C_2\) This is distance. Now we figure out what the constants are. C1 here is the initial velocity, while C2 here is the starting distance. Thus we have \(\Large\frac{ds}{dt}=at+v_0=v\) \(\Large s=\frac{1}{2}at^2+v_0t+s_0\)
So, where'd the 1/2 come from? well take the derivative of at^2. You'll get 2at, which is twice off of what we want, which is at. In fact, we fix this issue by dividing at^2 by 2. Try the derivative of (1/2)at^2, you should get at.
so I have slope y = 4.8711x
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.
|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.
From the derivation before, you can see that the slope is 1/2*a. a=2*4.8711 (I was mistaken back then :) )
Well you have a 1/2 in that equation that comes from integrating that I just explained earlier. a is assumed to be constant (gravity), and there you go it's the equation.
We don't know a, but it's a constant.
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.
So, where did the magical formula come from? Well I explained it way back there. \(\Huge s(x)=\frac{1}{2}at^2+v_0t+s_0\) But, we know that the initial velocity and the initial distance are both 0, we we effectively get \(\Huge s(x)=\frac{1}{2}at^2\)
wtf who are you and tracks?
I finished already. ;)
it feels not right canceling x and t^2
Well, you defined x as t^2. So too bad.
flutterit i'll see what happens
thanks for all your help, the derivations were cool
Yeah, I likes dem too :D