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ParthKohli Group Title

OK, here's a nice question that I was just checking out. Try it :-D A particle is at rest and starts to move at a constant acceleration \(\alpha\). After some time, it starts decelerating at \(\beta\) and returns to rest. Given that the particle takes time \(T\) to do so, find in the terms of the known: 1: The highest velocity. 2: The average velocity.

  • one year ago
  • one year ago

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  1. ParthKohli Group Title
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    @.Sam.

    • one year ago
  2. myiah5766 Group Title
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    The highest velocity.

    • one year ago
  3. ParthKohli Group Title
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    Note that the known are \(\alpha, \beta, T\).

    • one year ago
  4. .Sam. Group Title
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    |dw:1368898722315:dw|

    • one year ago
  5. ParthKohli Group Title
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    It's a particle, not a cube :-P

    • one year ago
  6. .Sam. Group Title
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    Is the whole journey takes time T?

    • one year ago
  7. ParthKohli Group Title
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    yes.

    • one year ago
  8. .Sam. Group Title
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    |dw:1368898894389:dw|

    • one year ago
  9. .Sam. Group Title
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    \[\alpha=\frac{v_1-0}{t_1} \] Highest velocity is \(v_1\) \[v_1=\alpha t_1\]

    • one year ago
  10. .Sam. Group Title
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    \[-\beta=\frac{v_2-v_1}{t_2}\] and also \[T=t_1+t_2\]

    • one year ago
  11. ParthKohli Group Title
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    Good going.

    • one year ago
  12. .Sam. Group Title
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    Average velocity will be just \[\frac{v_1+v_2}{2}?\]

    • one year ago
  13. ParthKohli Group Title
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    Careful: you don't know the average acceleration as yet.

    • one year ago
  14. .Sam. Group Title
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    \[a_{ave}=\frac{v_2-v_1}{t_2-t_1}\]

    • one year ago
  15. ParthKohli Group Title
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    Actually, the average acceleration for the whole thing is \(0\).

    • one year ago
  16. .Sam. Group Title
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    Hmm, lol, but it didn't say it stops at the end and usually when its average acceleration \(\alpha=\beta\)

    • one year ago
  17. ParthKohli Group Title
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    Oh, my bad. It does stop at the end.

    • one year ago
  18. .Sam. Group Title
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    Hmm \[v_2=0\]

    • one year ago
  19. ParthKohli Group Title
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    What is \(v_1\)? The maximum velocity or the start velocity?

    • one year ago
  20. .Sam. Group Title
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    The diagram said it's maximum velocity

    • one year ago
  21. ParthKohli Group Title
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    OK. I got confused when you said v1 = v2

    • one year ago
  22. .Sam. Group Title
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    Nah

    • one year ago
  23. .Sam. Group Title
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    So we'll mix them up \[v_1=\alpha t_1 ~~~~~~ -\beta t_2=-v_1 \] Getting \[\beta t_2= \alpha t_1\]

    • one year ago
  24. ParthKohli Group Title
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    You just eliminated \(v_1\) when you actually need it.

    • one year ago
  25. .Sam. Group Title
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    Right, \[v_1=\frac{\alpha(T-t_2)}{\beta}\]

    • one year ago
  26. ParthKohli Group Title
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    But \(t_2\) is not a known thing. You defined it. Express in terms of alpha, beta and T.

    • one year ago
  27. .Sam. Group Title
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    What about the distance?

    • one year ago
  28. .Sam. Group Title
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    /displacement

    • one year ago
  29. ParthKohli Group Title
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    You have to calculate that. Both 1 and 2 are to be given in terms of alpha, beta and T.

    • one year ago
  30. ParthKohli Group Title
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    Do you want to try more?

    • one year ago
  31. .Sam. Group Title
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    I'm getting more variables, lol

    • one year ago
  32. .Sam. Group Title
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    But it would've been easier if we used calculus method

    • one year ago
  33. ParthKohli Group Title
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    OK, so like \(v_1 = \alpha t_1 \Rightarrow t_1 = \dfrac{v_1}{\alpha}\). Similarly \(0 = v_1 - \beta t_2 \Rightarrow t_2 = \dfrac{v_1}{\beta}\). Now\[T = t_1 + t_2 = \frac{v_1}{\alpha} + \frac{v_1}{\beta}\]Solving for \(v_1 \), we get \(v_1 = \dfrac{\alpha\beta T}{\alpha + \beta }\)

    • one year ago
  34. .Sam. Group Title
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    Aw man it was getting close :P

    • one year ago
  35. ParthKohli Group Title
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    :-D

    • one year ago
  36. ParthKohli Group Title
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    Try the 2nd one maybe?

    • one year ago
  37. .Sam. Group Title
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    Average velocity for the whole journey is it?

    • one year ago
  38. ParthKohli Group Title
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    Yeah.

    • one year ago
  39. .Sam. Group Title
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    is it zero

    • one year ago
  40. ParthKohli Group Title
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    No, because the displacement isn't zero.

    • one year ago
  41. .Sam. Group Title
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    But the initial and final velocities are zero

    • one year ago
  42. ParthKohli Group Title
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    I have not asked for the average acceleration.

    • one year ago
  43. ParthKohli Group Title
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    :-P

    • one year ago
  44. ParthKohli Group Title
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    \[\overline{v} = \dfrac{\Delta s}{\Delta t}\]

    • one year ago
  45. .Sam. Group Title
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    I've got \[\frac{x_2}{x_3-x_2}=\frac{\beta}{\alpha}=\frac{t_1}{t_2}\]

    • one year ago
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