A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

blackstreet23

  • one year ago

Shakina and Juliette discuss the relationship between the velocity vs. time graph and the displacement of a particle. Which of the following statements by Shakina is true? 1- "The displacement of a particle is the area between the curve of the velocity vs. time graph and the horizontal axis for a given time interval, with areas above the horizontal axis counted as positive and areas below the horizontal axis counted as negative." 2- "The displacement of a particle is the slope of the velocity vs. time graph." 3- "The displacement of a particle is the area in the first quadrant between t

  • This Question is Closed
  1. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The question above is incomplete. Shakina and Juliette discuss the relationship between the velocity vs. time graph and the displacement of a particle. Which of the following statements by Shakina is true? 1- "The displacement of a particle is the area between the curve of the velocity vs. time graph and the horizontal axis for a given time interval, with areas above the horizontal axis counted as positive and areas below the horizontal axis counted as negative." 2- "The displacement of a particle is the slope of the velocity vs. time graph." 3- "The displacement of a particle is the area in the first quadrant between the curve of the velocity vs. time graph and the horizontal axis for a given time interval." 4- "The displacement of a particle is the area in the third quadrant between the curve of the velocity vs. time graph and the horizontal axis for a given time interval."

  2. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @abhisar @Shalante @freckles

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Are you aware of the relationships between displacement, velocity, and acceleration given by calculus?

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm just curious because if you are not, then there isn't much more I can do but just tell you the answer which I would prefer not to do if I don't have to. Stating answers without explanations may save you on this problem, but in the long run lack of understanding will cause you to get things wrong on tests where it matters.

  5. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes. Displacement is the integral of velocity right?

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Correct, what does the integral of a function mean?

  7. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    area under the curve

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Why does it mean that? Pretend I don't know and explain to me where that idea comes from please :D

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Think first, how do you calculate the area under a curve?

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Please just state what you know, dont go referencing books or google.... consider me an interactive google in a sense on this topic :D

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't see you typing which means you are looking up an answer, please I am curious to know where you stand currently

  12. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no, I was answering a friend's text. Sorry

  13. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So the area under the curve can be found by obtaining the integral of a function when is above the x -axis and continuos

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh no problems :D :D

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Correct... how do you integrate though? It isnt a magical process (strictly speaking).... do recall how an integral is defined?

  16. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    as the summation of ....

  17. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    taking the limit

  18. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sigma

  19. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    delta xk

  20. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i dont remember totally lol

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You have the pieces lets put it all together: \[\int\limits_{a}^{b} f(x) dx \approx \sum_{n=0}^{ N } f(x^{*}_{n}) \Delta x_{n}\] (one way it can be defined at least) Now the limit process you allude to would result in the width (delta x) tending to zero which implies the number of sub intervals (N) goes to infinity. Intuitively, this results from the fact that as each sub division decreases in size it takes more subdivisions to make up the entire interval.

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    But the formal limiting process isnt strictly needed here for the explaination because the main concept here lies in that approximate definition. Please tell me what f(x) stands for here.

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    And for simplicity the * on the x we will just take to mean the function evaluated at the midpoint of the interval delta x

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What does it stand for in terms of the problem at hand?

  25. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    f(x)

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  27. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the change of velocity over time?

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    correct

  29. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    or an exact point of velocity in a certain time?

  30. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so acceleration?

  31. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh wait no sorry I just saw velocity

  32. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but i thought that the slope of the graph meant acceleration which is f'(x). I am confused

  33. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It stands for velocity i.e. the rate of change of position over time

  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well formally speaking f(x) can stand for any function and as you noted the displacement is the integral of velocity.

  35. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So for this problem f(x) stands for velocity... and also I probably should be using f(t) to avoid confusion... I apologize... Please substitute t for all the x's in the formula above

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Also, your second velocity comment was correct.... since f(t) stands for the velocity then.... \[f(t^{*}_{n})\] stands for the velocity evaluated at a certain point of time

  37. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sooooooo. f(t) in the graph of velocity vs time. Is still velocity?

  38. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes

  39. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if f(t) gives is velocity and t is time then, the following tells me how velocity changes over time:

  40. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but that is the definition of acceleration

  41. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how velocity changes over time

  42. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1442106917949:dw|

  43. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I guess it would be instantaneous velocity right?

  44. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Sorry it is really hard to draw using this thing

  45. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I mean velocity at a certain point

  46. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but so far with your explaination. I guess the answer is 1 lol

  47. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The definition of acceleration is the rate instantaneous rate of change of velocity with respect to time.... i.e. the derivative of velocity curve at a given point

  48. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    kind of makes sense. I mean which what i know of integrals

  49. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i was refering to answer #1

  50. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok then so it is clear the meaning of f here is the velocity curve correct?

  51. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes but I want to connect the concepts for you while I have you here.... to give you something to think about next time you face something like this so please dont just be happy with the answer

  52. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you for the medal btw

  53. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    umm. I mean the displacement is the integral of velocity. And when an integral in above the x-axis is positive and when is below is negative

  54. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right?

  55. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes but why is area under the curve (aka the integral of the velocity function) equal to the net displacement?

  56. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wait 2 medals? lol who else is in here?!?!

  57. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I had to refresh because my backspace key ceased to function for some reason.

  58. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    umm. Could you repeat the question? sorry lol

  59. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i do not understand what you are asking me :/

  60. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Its ok... You are correct that the integral of velocity is equal to the area under the velocity curve, where it is + above the x-axis and - below the x-axis, and that this represents the net displacement. My question is, how do you know this is true? Why is this the case? Think physically, based on the definition I gave above.

  61. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    because the velocity is a vector, so if the velocity is negative its integral is negative right?

  62. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int\limits_{a}^{b} f(t) dt \approx \sum_{n=0}^{N} f(t^{*}_{n}) \Delta t_{n}\]

  63. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am actually lost with the formula lol

  64. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes velocity is a vector, but if we were just in one-dimension (i.e. if we only had an x-velocity) the fact that it is a vector doesn't matter in the question I asked you. Therefore, higher dimensions it similarly doesn't affect the answer.

  65. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok no this is good we have identified your hang up :D feel good this means progress can ensue

  66. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but the answer is #1 right?

  67. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i'll submit

  68. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok so the first term: \[ f(t^{*}_{n}) \] represents the velocity evaluated at the time\[ t^{*} \] on the nth subinterval

  69. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes because #2 is a derivative, #3 would ignore the potential "negative" area in the 4th quadrant and answer #4 would ignore the "positive" area in the second quadrant

  70. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohhh i see

  71. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok back to the explaination.... so that \[ t^* \] I have definied to simply mean the time at the halfway point on the sub interval.... ill draw it for N=2

  72. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yuhu! yeah it was correct!

  73. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but yeah. I think I got it

  74. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1442108030512:dw|

  75. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok sorry the drawing is a mess it is so hard to draw in this thing

  76. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So this drawing shows you i am APPROXIMATING the area under the curve by drawing N=2 rectangles that have a height of: \[ f(t^{*}_{n}) \] and a width of \[ \Delta t \] where I am assuming that all the widths are equal (aka all time intervals are the same amount)

  77. blackstreet23
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hey btw could you help me with another problem?

  78. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Therefore the area of the rectangle is given by width * height which is: \[\ f(t^{*}_{n}) \Delta t = velocity * time = distance \]

  79. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Does this make sense?

  80. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Applying the formalism, taking the limit, etc. just refines this definition to instantaneous points on the curve and the infinite amount of resulting rectangles makes the approximation become an equality

  81. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The easiest way to become fully convinced is draw the graph of a constant velocity starting out at v=0 at t=0 and going to some value v=V at time t=T

  82. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This graph will be a right triangle, which you can caluclate the area exactley by applying the formula 1/2 base times height

  83. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Checking the units wil reveal that this is indeed a unit of distance

  84. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Does this make sense to you?

  85. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Sure go ahead

  86. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Actually if you dont mind could you start a new problem thread for your question (if it is unrelated) since this one is already really long? Also, so if my answer is good I can get another medal :D

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.