## 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 one year ago one year ago

1. imron07 Group Title

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

2. jaersyn Group Title

explain like i'm five? ^^"

3. jaersyn Group Title

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 Group Title

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 Group Title

$$\Large\frac{d^2s}{dt^2}=a$$ 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 Group Title

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 Group Title

so I have slope y = 4.8711x

8. jaersyn Group Title

how do you find a

9. inkyvoyd Group Title

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

10. imron07 Group Title

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

11. jaersyn Group Title

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

12. inkyvoyd Group Title

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

13. jaersyn Group Title

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

14. jaersyn Group Title

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

15. inkyvoyd Group Title

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

16. jaersyn Group Title

i did

17. imron07 Group Title

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

18. inkyvoyd Group Title

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 Group Title

yezzir

20. imron07 Group Title

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 Group Title

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 Group Title

assume we don't know a

23. jaersyn Group Title

or g rather

24. inkyvoyd Group Title

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

25. jaersyn Group Title

4.8711x = (1/2)at^2

26. jaersyn Group Title

do the x and t^2 cancel out somehow?

27. jaersyn Group Title

since x represents t^2 in the graph?

28. inkyvoyd Group Title

They are the same thing.

29. inkyvoyd Group Title

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

30. jaersyn Group Title

hmm

31. inkyvoyd Group Title

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 Group Title

wtf who are you and tracks?

33. jaersyn Group Title

sorry. proceed

34. inkyvoyd Group Title

35. jaersyn Group Title

|dw:1347849594808:dw|

36. jaersyn Group Title

it feels not right canceling x and t^2

37. inkyvoyd Group Title

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

38. inkyvoyd Group Title

The x axis IS t^2.

39. jaersyn Group Title

yeahhhh but

40. jaersyn Group Title

flutterit i'll see what happens

41. jaersyn Group Title

thanks for all your help, the derivations were cool

42. inkyvoyd Group Title

Yeah, I likes dem too :D

43. jaersyn Group Title

awwwwwww yeahhhh math.