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anonymous
 3 years ago
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
anonymous
 3 years ago
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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01) Maybe this gives you a clue: \[\frac{dv}{dt^2}=\frac{dv}{2tdt}=\frac{a}{2t}\] 2) Find the slope,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0explain like i'm five? ^^"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y = 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The slope of distance vs time^2: \[\frac{dy}{dt^2}= \frac{dy}{2tdt}=\frac{v}{2t}=\frac{a}{2} \] Is this clear enough?

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1\(\Large\frac{d^2s}{dt^2}=a\) 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 \(\Large\frac{ds}{dt}=at+v_0=v\) \(\Large s=\frac{1}{2}at^2+v_0t+s_0\)

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1So, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so I have slope y = 4.8711x

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Can you like give me what your problem that you have to solve is? :S

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x here t^2? If yes, then a=1/2*4.8711 :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it's to compute the experimental value of g (gravity) from the slope value which is y = 4.8711x

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1You're giving me an equation, but I have no idea what this equation means. Can you clarify?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0imron, how'd you know to substitute that way though?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i have this graph where the slope of the distance vs time^2 is y=4.8711x

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1you could just draw the graph, but nvm I got it now.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347849013566:dw If this graph is what you get?

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Kk. you have \(\huge s(x)=\frac{1}{2}at^2\) s(x) is the distance, a is acceleration, and t is time.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0From the derivation before, you can see that the slope is 1/2*a. a=2*4.8711 (I was mistaken back then :) )

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Well 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0assume we don't know a

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1We don't know a, but it's a constant.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do the x and t^2 cancel out somehow?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since x represents t^2 in the graph?

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1They are the same thing.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1So there you go. Cancel them both out, multiply both sides by 2, and here you have a.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1So, 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 \(\Huge s(x)=\frac{1}{2}at^2\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wtf who are you and tracks?

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1I finished already. ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347849594808:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it feels not right canceling x and t^2

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Well, you defined x as t^2. So too bad.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0flutterit i'll see what happens

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks for all your help, the derivations were cool

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah, I likes dem too :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0awwwwwww yeahhhh math.
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