Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

shayanreloaded

  • 2 years ago

Kinematics -graphs

  • This Question is Closed
  1. shayanreloaded
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Acceleration-time graph |dw:1342512372407:dw| Velocity time graph |dw:1342512427429:dw| Now about the position time graph

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

    |dw:1342512503350:dw| which curves(of the 3) will be steeper?

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

    ace increases from origin to t1 constant from t1 to t2 and decreases from t2 onward. but ace is max from t1 to t2. I hope im understandable

  4. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok, so u want to know how will the position-time graph look like?

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

    yeah

  6. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I'm not sure, but maybe it looks like this: |dw:1342513483775:dw| correct me if i'm wrong.

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

    |dw:1342513648649:dw| ace remains positive always

  8. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yea, i was going to type in that right now. the max point should not be reached. since V is not 0 anywhere.

  9. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the graph should be steeper from origin to t1 than in the interval t1 - t2.

  10. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    *t1 to t2

  11. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @Ishaan94 the velocity is NOT constant from t1 to t2. accel is.

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

    velocity is NEVER constant

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

    |dw:1342513895634:dw| The graph should look somewhat like this I want to know the order of their steepness.

  14. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think the graph should be of this form: |dw:1342513986456:dw|

  15. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1 will be more steep than 2.

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

    the ace is always positive so the velocity always increases and so the slope of xt graph always increases

  17. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sry my bad! i saw the slop of accel!

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

    ace decreases t2 onward but stays positive notice

  19. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think the order of steepness should be 1 > 2 > 3. i'm just using logic here: accel is increasing from 0 to t1 hence vel is increasing at the fastest rate in this interval. from t1 to t2, accel is constant. so vel is increasing at a constant rate ( this rate < the rate from 0 to t1). and from t2 to the end, accel is decreasing. so the vel is anyway decreasing. so less distance will be covered as compared to the former 2 intervals.

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

    the rate of increase of velocity is max from t1 to t2 because ace is max(constant)

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

    and from t2 the velocity increases but at a decreasing rate

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

    @Ishaan94 you lost me

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

    even after t2 ace is positive so slope cant be curving down right?

  24. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think it should be curving down. because now the distance covered per unit time is decreasing. so 3 should be the least steep of all.

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

    but if ace is positive xt curve never curves down

  26. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    right now, the interval we're considering is this: |dw:1342514762867:dw| so at some point the graph will curve downwards, but we don't have that within our interval.

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

    |dw:1342514919696:dw|

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

    but that works only for constant ace part what about the rest?

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

    and i just want the order @Ishaan94

  30. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @Ishaan94 yea, i think we can leave the precision out for this sum. just an approximation ill suffice.

  31. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    okay, here's my final guess: 2 > 1 > 3.

  32. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    or should it be 2 > 1 = 3 ?

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

    guess?

  34. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    *answer!

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

    Lets keep the question open for more opinions

  36. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    tell me this: accel is increasing and decreasing at the same rate from 0 to t1 and from t2 to the end. so what happens to the distance covered per unit time during these intervals? will the be equal?

  37. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I am sorry for deleting my previous posts. Acceleration is positive from t2 and on wards but I presumed it to be negative :( My latest attempt. I know it sucks but still

  38. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oops wrong curve, from t1 to t2, it should be |dw:1342517034897:dw|A parabola.

  39. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so the answer is 2 > 1 > 3 right?

  40. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I am sorry. I can't be sure from what it seems my graph from t2 can't be right. Velocity-Time graph form t2 is downward parabola right? a = -dv/dt?

  41. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Darwin... this is so embarrassing :(

  42. Vaidehi09
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it tends to go downwards. but doesn't do so completely.

  43. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @mukushla

  44. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so my graph from t2 isn't right either

  45. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i hope my math is right

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

    i go with @Ishaan94

  47. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    mukushla just to be sure of my answer. can you draw an approximate graph? please.

  48. Vincent-Lyon.Fr
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Velocity-time is: |dw:1342520614050:dw| with 2 half-parabolas between 0 and t1 and between 0 t2 and tfinal

  49. Vincent-Lyon.Fr
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Displacement-time is: |dw:1342520737774:dw| The curve is always increasing as well as increasing in slope. Only the curvature becomes 0 at the end (would go on in a straight line if acceleration remains 0 or would reach an inflexion point if acceleration becomes negative after tfinal

  50. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Seems like I made a fool of myself. But Vincent, slope isn't always increasing. It's constant from \(t_1\) to \(t_2\).

  51. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    And it's decreasing from \(t_2\). Correct me if I am wrong. :|

  52. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh silly me :(

  53. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Please Ignore me :(

  54. Vincent-Lyon.Fr
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    @Ishaan what you say is true of velocity, but wrong about displacement

  55. Ishaan94
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I know :/ I got confused :/

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

    so whats the order of steepness?? 3>2>1 ? or 1=2<3 ? does it depend on ace magnitude or rate of change of ace??

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

    @Vincent-Lyon.Fr

  58. Vincent-Lyon.Fr
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    The steepness (of displacement curve) is ever increasing because velocity is; so 3>2>1 But, as velocity is a continuous function, the 3 parts of the curve will smoothly link together, without any angle at t1 and t2

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

    Thanks

  60. 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.