Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. imron07
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  2. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    explain like i'm five? ^^"

  3. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  4. imron07
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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?

  5. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 \(\Large\frac{ds}{dt}=at+v_0=v\) \(\Large s=\frac{1}{2}at^2+v_0t+s_0\)

  6. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  7. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so I have slope y = 4.8711x

  8. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how do you find a

  9. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  10. imron07
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  11. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  13. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  15. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  16. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i did

  17. imron07
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  18. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yezzir

  20. imron07
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  22. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    assume we don't know a

  23. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    or g rather

  24. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  25. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    4.8711x = (1/2)at^2

  26. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    do the x and t^2 cancel out somehow?

  27. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    since x represents t^2 in the graph?

  28. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    They are the same thing.

  29. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  30. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm

  31. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 \(\Huge s(x)=\frac{1}{2}at^2\)

  32. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wtf who are you and tracks?

  33. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry. proceed

  34. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I finished already. ;)

  35. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1347849594808:dw|

  36. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it feels not right canceling x and t^2

  37. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  38. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The x axis IS t^2.

  39. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeahhhh but

  40. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    flutterit i'll see what happens

  41. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks for all your help, the derivations were cool

  42. inkyvoyd
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeah, I likes dem too :D

  43. jaersyn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    awwwwwww yeahhhh math.

  44. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.