## jaersyn 2 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

1. imron07

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

2. jaersyn

explain like i'm five? ^^"

3. jaersyn

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

4. imron07

The slope of distance vs time^2: $\frac{dy}{dt^2}= \frac{dy}{2tdt}=\frac{v}{2t}=\frac{a}{2}$ Is this clear enough?

5. inkyvoyd

$$\Large\frac{d^2s}{dt^2}=a$$ 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 $$\Large\frac{ds}{dt}=at+v_0=v$$ $$\Large s=\frac{1}{2}at^2+v_0t+s_0$$

6. inkyvoyd

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.

7. jaersyn

so I have slope y = 4.8711x

8. jaersyn

how do you find a

9. inkyvoyd

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

10. imron07

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

11. jaersyn

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

12. inkyvoyd

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

13. jaersyn

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

14. jaersyn

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

15. inkyvoyd

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

16. jaersyn

i did

17. imron07

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

18. inkyvoyd

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

19. jaersyn

yezzir

20. imron07

From the derivation before, you can see that the slope is 1/2*a. a=2*4.8711 (I was mistaken back then :) )

21. inkyvoyd

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.

22. jaersyn

assume we don't know a

23. jaersyn

or g rather

24. inkyvoyd

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

25. jaersyn

4.8711x = (1/2)at^2

26. jaersyn

do the x and t^2 cancel out somehow?

27. jaersyn

since x represents t^2 in the graph?

28. inkyvoyd

They are the same thing.

29. inkyvoyd

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

30. jaersyn

hmm

31. inkyvoyd

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 $$\Huge s(x)=\frac{1}{2}at^2$$

32. jaersyn

wtf who are you and tracks?

33. jaersyn

sorry. proceed

34. inkyvoyd

35. jaersyn

|dw:1347849594808:dw|

36. jaersyn

it feels not right canceling x and t^2

37. inkyvoyd

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

38. inkyvoyd

The x axis IS t^2.

39. jaersyn

yeahhhh but

40. jaersyn

flutterit i'll see what happens

41. jaersyn

thanks for all your help, the derivations were cool

42. inkyvoyd

Yeah, I likes dem too :D

43. jaersyn

awwwwwww yeahhhh math.