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  1. 4n1m0s1ty
    • one year ago
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    Ok this problem is pretty much the same as the last one.

  2. 4n1m0s1ty
    • one year ago
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    The only difference is that we are dealing with velocity and acceleration, rather than position. I think you probably already know that the acceleration is the change in velocity over time. So, in this problem we kind of treat velocity like position in the last problem, and acceleration like velocity previously.

  3. anonymous
    • one year ago
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    So would it start off at 2m/s?

  4. 4n1m0s1ty
    • one year ago
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    Yes

  5. 4n1m0s1ty
    • one year ago
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    Now what is the velocity at t = 1?

  6. 4n1m0s1ty
    • one year ago
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    The acceleration was 2 m/s2 at t = 0, so after 1 second, we give a boost to the initial 2m/s velocity.

  7. anonymous
    • one year ago
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    So would it be 4 since we are adding 2 to the 2m/s of t=0?

  8. 4n1m0s1ty
    • one year ago
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    yep

  9. anonymous
    • one year ago
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    t=2 would be 0

  10. 4n1m0s1ty
    • one year ago
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    Nope. What's the acceleration at t = 1?

  11. 4n1m0s1ty
    • one year ago
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    It goes from 2 m/s2 to ?

  12. anonymous
    • one year ago
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    1

  13. 4n1m0s1ty
    • one year ago
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    yes, so our velocity is at 4 m/s, and we just add 1 m/s after 1 s, so the velocity at t =2 is?

  14. anonymous
    • one year ago
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    5

  15. 4n1m0s1ty
    • one year ago
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    yep

  16. 4n1m0s1ty
    • one year ago
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    so now for t = 3. Look at the acceleration at t =2 and add is to your velocity.

  17. anonymous
    • one year ago
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    Then would t=3 be 0 since it's a line?

  18. 4n1m0s1ty
    • one year ago
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    yes the acceleration is 0.

  19. 4n1m0s1ty
    • one year ago
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    so that means the velocity doesn't change.

  20. anonymous
    • one year ago
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    Right

  21. anonymous
    • one year ago
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    Would the be negative numbers since the acceleration is in the negative now?

  22. 4n1m0s1ty
    • one year ago
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    so when the acceleration is negative, we are reducing the positive velocity. So at t =3, your velocity is 5 m/s, and we subtract 3 m/s now since its negative

  23. 4n1m0s1ty
    • one year ago
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    So at t = 4, v = ?

  24. anonymous
    • one year ago
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    why not 2?

  25. 4n1m0s1ty
    • one year ago
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    Are you saying why not v = 2 at t = 3?

  26. anonymous
    • one year ago
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    why not subtract 2?

  27. 4n1m0s1ty
    • one year ago
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    Because the acceleration is -3 right not -2?

  28. anonymous
    • one year ago
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    Sorry. Was getting confused with the +2

  29. 4n1m0s1ty
    • one year ago
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    ok so what is the velocity at t = 4

  30. anonymous
    • one year ago
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    v=2

  31. 4n1m0s1ty
    • one year ago
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    yep

  32. 4n1m0s1ty
    • one year ago
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    t = 5?

  33. 4n1m0s1ty
    • one year ago
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    the acceleration is back to 0 again right?

  34. anonymous
    • one year ago
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    v=2

  35. 4n1m0s1ty
    • one year ago
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    yep

  36. 4n1m0s1ty
    • one year ago
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    So i think thats it. Any questions?

  37. anonymous
    • one year ago
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    Okay so 0s=2 1s=4 2s=5 3s=0 4s=2 5s=2

  38. anonymous
    • one year ago
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    Sorry it took so long to type that. My browser kept freezing up on me.

  39. 4n1m0s1ty
    • one year ago
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    3s = 5, the acceleration was 0, not the velocity

  40. anonymous
    • one year ago
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    Okay so if we took the velocity at 3.41s how do we go about doing that?

  41. 4n1m0s1ty
    • one year ago
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    look between t = 3 and t = 4 on the velocity graph you just drew.

  42. anonymous
    • one year ago
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    Looks to be about 3.50 to me...does that seem reasonable?

  43. 4n1m0s1ty
    • one year ago
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    Maybe, I'm not sure how your assignment system works. I'm not sure if it wants you to calculate it exactly or just by estimation. Did your instructor give you any equations to work with.

  44. 4n1m0s1ty
    • one year ago
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    Just for my information is this a high school or college physics course?

  45. anonymous
    • one year ago
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    He didn't. It's college but this is the first time I've ever had physics.

  46. 4n1m0s1ty
    • one year ago
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    Ok well I guess its just an estimate then.

  47. anonymous
    • one year ago
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    It didn't take 3.50 so I think it's wanting more exact.

  48. 4n1m0s1ty
    • one year ago
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    Ok well, then that would require calculating the slope of the graph between t = 3 and t = 4. Do you know how to do that?

  49. 4n1m0s1ty
    • one year ago
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    Err.. nvm

  50. anonymous
    • one year ago
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    I do not.

  51. 4n1m0s1ty
    • one year ago
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    The information is already given. The slope between t = 3 and t =4 is the acceleration. So whats the acceleration in that time frame equal to (look at the accel graph)

  52. anonymous
    • one year ago
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    -3

  53. 4n1m0s1ty
    • one year ago
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    ok so the equation for constant acceleration is this: \[a = \frac{ v_{f} - v_{i} }{ t_{f} - t_{i} }\] vf is the final velocity vi is the initial velocity tf is the final time ti is the initial time

  54. 4n1m0s1ty
    • one year ago
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    and a is the acceleration

  55. 4n1m0s1ty
    • one year ago
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    so we know a is -3, vi is 5m/s, ti is 3s, and tf is 3.41s now we just need to solve for vf to find our velocity at 3.41 s

  56. anonymous
    • one year ago
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    (2m/s-5m/s)/(4s-3s)

  57. 4n1m0s1ty
    • one year ago
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    that would give you the acceleration, but we already found it from the graph.

  58. anonymous
    • one year ago
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    Right. It's -3

  59. 4n1m0s1ty
    • one year ago
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    so now we need to find the velocity at t = 3.41, \[-3 m/s^{2} = \frac{ v_{f} - 5 m/s}{ 3.41 s - 3 s }\] so solve for vf

  60. anonymous
    • one year ago
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    4.87

  61. anonymous
    • one year ago
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    *4.88

  62. 4n1m0s1ty
    • one year ago
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    I'm getting a different answer, I would check your math. Are you using -3 for the acceleration?

  63. anonymous
    • one year ago
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    Yes

  64. 4n1m0s1ty
    • one year ago
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    Are you doing this: \[(-3m/s^{2})*(3.41 s - 3 s)+5 m/s = v_{f}\] \[(-3)(0.41)+5=v_{f}\] \[-1.23 + 5 = v_{f}\]

  65. anonymous
    • one year ago
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    I see what I did now.

  66. 4n1m0s1ty
    • one year ago
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    Great

  67. anonymous
    • one year ago
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    I got 3.77 the second time I did it.

  68. 4n1m0s1ty
    • one year ago
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    yep thats what i got

  69. anonymous
    • one year ago
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    Thanks for helping out.

  70. 4n1m0s1ty
    • one year ago
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    NP

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