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jaersyn
Group Title
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
 2 years ago
 2 years ago
jaersyn Group Title
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
 2 years ago
 2 years ago

This Question is Closed

imron07 Group TitleBest 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,
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
explain like i'm five? ^^"
 2 years ago

jaersyn Group TitleBest 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
 2 years ago

imron07 Group TitleBest 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?
 2 years ago

inkyvoyd Group TitleBest 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 \(\Large\frac{ds}{dt}=at+v_0=v\) \(\Large s=\frac{1}{2}at^2+v_0t+s_0\)
 2 years ago

inkyvoyd Group TitleBest 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.
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
so I have slope y = 4.8711x
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
how do you find a
 2 years ago

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

imron07 Group TitleBest ResponseYou've already chosen the best response.1
x here t^2? If yes, then a=1/2*4.8711 :D
 2 years ago

jaersyn Group TitleBest 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
 2 years ago

inkyvoyd Group TitleBest 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?
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
imron, how'd you know to substitute that way though?
 2 years ago

jaersyn Group TitleBest 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
 2 years ago

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

imron07 Group TitleBest ResponseYou've already chosen the best response.1
dw:1347849013566:dw If this graph is what you get?
 2 years ago

inkyvoyd Group TitleBest 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.
 2 years ago

imron07 Group TitleBest 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 :) )
 2 years ago

inkyvoyd Group TitleBest 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.
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
assume we don't know a
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
or g rather
 2 years ago

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

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
4.8711x = (1/2)at^2
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
do the x and t^2 cancel out somehow?
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
since x represents t^2 in the graph?
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
They are the same thing.
 2 years ago

inkyvoyd Group TitleBest 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.
 2 years ago

inkyvoyd Group TitleBest 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 \(\Huge s(x)=\frac{1}{2}at^2\)
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
wtf who are you and tracks?
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
sorry. proceed
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
I finished already. ;)
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
dw:1347849594808:dw
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
it feels not right canceling x and t^2
 2 years ago

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

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
The x axis IS t^2.
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
yeahhhh but
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
flutterit i'll see what happens
 2 years ago

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
thanks for all your help, the derivations were cool
 2 years ago

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

jaersyn Group TitleBest ResponseYou've already chosen the best response.0
awwwwwww yeahhhh math.
 2 years ago
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