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shayanreloaded

Kinematics -graphs

  • one year ago
  • one year ago

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

    • one year ago
  2. shayanreloaded
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    |dw:1342512503350:dw| which curves(of the 3) will be steeper?

    • one year ago
  3. shayanreloaded
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    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

    • one year ago
  4. Vaidehi09
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    ok, so u want to know how will the position-time graph look like?

    • one year ago
  5. shayanreloaded
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    yeah

    • one year ago
  6. Vaidehi09
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    I'm not sure, but maybe it looks like this: |dw:1342513483775:dw| correct me if i'm wrong.

    • one year ago
  7. shayanreloaded
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    |dw:1342513648649:dw| ace remains positive always

    • one year ago
  8. Vaidehi09
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    yea, i was going to type in that right now. the max point should not be reached. since V is not 0 anywhere.

    • one year ago
  9. Vaidehi09
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    the graph should be steeper from origin to t1 than in the interval t1 - t2.

    • one year ago
  10. Vaidehi09
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    *t1 to t2

    • one year ago
  11. Vaidehi09
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    @Ishaan94 the velocity is NOT constant from t1 to t2. accel is.

    • one year ago
  12. shayanreloaded
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    velocity is NEVER constant

    • one year ago
  13. shayanreloaded
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    |dw:1342513895634:dw| The graph should look somewhat like this I want to know the order of their steepness.

    • one year ago
  14. Vaidehi09
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    i think the graph should be of this form: |dw:1342513986456:dw|

    • one year ago
  15. Vaidehi09
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    1 will be more steep than 2.

    • one year ago
  16. shayanreloaded
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    the ace is always positive so the velocity always increases and so the slope of xt graph always increases

    • one year ago
  17. Vaidehi09
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    sry my bad! i saw the slop of accel!

    • one year ago
  18. shayanreloaded
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    ace decreases t2 onward but stays positive notice

    • one year ago
  19. Vaidehi09
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    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.

    • one year ago
  20. shayanreloaded
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    the rate of increase of velocity is max from t1 to t2 because ace is max(constant)

    • one year ago
  21. shayanreloaded
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    and from t2 the velocity increases but at a decreasing rate

    • one year ago
  22. shayanreloaded
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    @Ishaan94 you lost me

    • one year ago
  23. shayanreloaded
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    even after t2 ace is positive so slope cant be curving down right?

    • one year ago
  24. Vaidehi09
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    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.

    • one year ago
  25. shayanreloaded
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    but if ace is positive xt curve never curves down

    • one year ago
  26. Vaidehi09
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    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.

    • one year ago
  27. shayanreloaded
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    |dw:1342514919696:dw|

    • one year ago
  28. shayanreloaded
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    but that works only for constant ace part what about the rest?

    • one year ago
  29. shayanreloaded
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    and i just want the order @Ishaan94

    • one year ago
  30. Vaidehi09
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    @Ishaan94 yea, i think we can leave the precision out for this sum. just an approximation ill suffice.

    • one year ago
  31. Vaidehi09
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    okay, here's my final guess: 2 > 1 > 3.

    • one year ago
  32. Vaidehi09
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    or should it be 2 > 1 = 3 ?

    • one year ago
  33. shayanreloaded
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    guess?

    • one year ago
  34. Vaidehi09
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    *answer!

    • one year ago
  35. shayanreloaded
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    Lets keep the question open for more opinions

    • one year ago
  36. Vaidehi09
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    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?

    • one year ago
  37. Ishaan94
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    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

    • one year ago
  38. Ishaan94
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    Oops wrong curve, from t1 to t2, it should be |dw:1342517034897:dw|A parabola.

    • one year ago
  39. Vaidehi09
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    so the answer is 2 > 1 > 3 right?

    • one year ago
  40. Ishaan94
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    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?

    • one year ago
  41. Ishaan94
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    Darwin... this is so embarrassing :(

    • one year ago
  42. Vaidehi09
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    it tends to go downwards. but doesn't do so completely.

    • one year ago
  43. Ishaan94
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    @mukushla

    • one year ago
  44. Ishaan94
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    so my graph from t2 isn't right either

    • one year ago
  45. Ishaan94
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    i hope my math is right

    • one year ago
  46. mukushla
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    i go with @Ishaan94

    • one year ago
  47. Ishaan94
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    mukushla just to be sure of my answer. can you draw an approximate graph? please.

    • one year ago
  48. Vincent-Lyon.Fr
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    Velocity-time is: |dw:1342520614050:dw| with 2 half-parabolas between 0 and t1 and between 0 t2 and tfinal

    • one year ago
  49. Vincent-Lyon.Fr
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    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

    • one year ago
  50. Ishaan94
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    Seems like I made a fool of myself. But Vincent, slope isn't always increasing. It's constant from \(t_1\) to \(t_2\).

    • one year ago
  51. Ishaan94
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    And it's decreasing from \(t_2\). Correct me if I am wrong. :|

    • one year ago
  52. Ishaan94
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    Oh silly me :(

    • one year ago
  53. Ishaan94
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    Please Ignore me :(

    • one year ago
  54. Vincent-Lyon.Fr
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    @Ishaan what you say is true of velocity, but wrong about displacement

    • one year ago
  55. Ishaan94
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    I know :/ I got confused :/

    • one year ago
  56. shayanreloaded
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    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??

    • one year ago
  57. shayanreloaded
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    @Vincent-Lyon.Fr

    • one year ago
  58. Vincent-Lyon.Fr
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    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

    • one year ago
  59. shayanreloaded
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    Thanks

    • one year ago
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