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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
 one year ago
 one year 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
 one year ago
 one year ago

This Question is Closed

imron07Best ResponseYou've already chosen the best response.1
1) Maybe this gives you a clue: \[\frac{dv}{dt^2}=\frac{dv}{2tdt}=\frac{a}{2t}\] 2) Find the slope,
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
explain like i'm five? ^^"
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
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
 one year ago

imron07Best ResponseYou've already chosen the best response.1
The slope of distance vs time^2: \[\frac{dy}{dt^2}= \frac{dy}{2tdt}=\frac{v}{2t}=\frac{a}{2} \] Is this clear enough?
 one year ago

inkyvoydBest 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 \(\Large\frac{ds}{dt}=at+v_0=v\) \(\Large s=\frac{1}{2}at^2+v_0t+s_0\)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
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.
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
so I have slope y = 4.8711x
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
Can you like give me what your problem that you have to solve is? :S
 one year ago

imron07Best ResponseYou've already chosen the best response.1
x here t^2? If yes, then a=1/2*4.8711 :D
 one year ago

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

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

jaersynBest ResponseYou've already chosen the best response.0
imron, how'd you know to substitute that way though?
 one year ago

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

inkyvoydBest ResponseYou've already chosen the best response.1
you could just draw the graph, but nvm I got it now.
 one year ago

imron07Best ResponseYou've already chosen the best response.1
dw:1347849013566:dw If this graph is what you get?
 one year ago

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

imron07Best ResponseYou've already chosen the best response.1
From the derivation before, you can see that the slope is 1/2*a. a=2*4.8711 (I was mistaken back then :) )
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
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.
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
assume we don't know a
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
We don't know a, but it's a constant.
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
do the x and t^2 cancel out somehow?
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
since x represents t^2 in the graph?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
They are the same thing.
 one year ago

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

inkyvoydBest ResponseYou've already chosen the best response.1
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 \(\Huge s(x)=\frac{1}{2}at^2\)
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
wtf who are you and tracks?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
I finished already. ;)
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
it feels not right canceling x and t^2
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
Well, you defined x as t^2. So too bad.
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
flutterit i'll see what happens
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
thanks for all your help, the derivations were cool
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
Yeah, I likes dem too :D
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
awwwwwww yeahhhh math.
 one year ago
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