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ParthKohli

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.

  • 11 months ago
  • 11 months ago

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  1. ParthKohli
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    @.Sam.

    • 11 months ago
  2. myiah5766
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    The highest velocity.

    • 11 months ago
  3. ParthKohli
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    Note that the known are \(\alpha, \beta, T\).

    • 11 months ago
  4. .Sam.
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    |dw:1368898722315:dw|

    • 11 months ago
  5. ParthKohli
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    It's a particle, not a cube :-P

    • 11 months ago
  6. .Sam.
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    Is the whole journey takes time T?

    • 11 months ago
  7. ParthKohli
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    yes.

    • 11 months ago
  8. .Sam.
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    |dw:1368898894389:dw|

    • 11 months ago
  9. .Sam.
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    \[\alpha=\frac{v_1-0}{t_1} \] Highest velocity is \(v_1\) \[v_1=\alpha t_1\]

    • 11 months ago
  10. .Sam.
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    \[-\beta=\frac{v_2-v_1}{t_2}\] and also \[T=t_1+t_2\]

    • 11 months ago
  11. ParthKohli
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    Good going.

    • 11 months ago
  12. .Sam.
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    Average velocity will be just \[\frac{v_1+v_2}{2}?\]

    • 11 months ago
  13. ParthKohli
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    Careful: you don't know the average acceleration as yet.

    • 11 months ago
  14. .Sam.
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    \[a_{ave}=\frac{v_2-v_1}{t_2-t_1}\]

    • 11 months ago
  15. ParthKohli
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    Actually, the average acceleration for the whole thing is \(0\).

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

    • 11 months ago
  17. ParthKohli
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    Oh, my bad. It does stop at the end.

    • 11 months ago
  18. .Sam.
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    Hmm \[v_2=0\]

    • 11 months ago
  19. ParthKohli
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    What is \(v_1\)? The maximum velocity or the start velocity?

    • 11 months ago
  20. .Sam.
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    The diagram said it's maximum velocity

    • 11 months ago
  21. ParthKohli
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    OK. I got confused when you said v1 = v2

    • 11 months ago
  22. .Sam.
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    Nah

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

    • 11 months ago
  24. ParthKohli
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    You just eliminated \(v_1\) when you actually need it.

    • 11 months ago
  25. .Sam.
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    Right, \[v_1=\frac{\alpha(T-t_2)}{\beta}\]

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

    • 11 months ago
  27. .Sam.
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    What about the distance?

    • 11 months ago
  28. .Sam.
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    /displacement

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

    • 11 months ago
  30. ParthKohli
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    Do you want to try more?

    • 11 months ago
  31. .Sam.
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    I'm getting more variables, lol

    • 11 months ago
  32. .Sam.
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    But it would've been easier if we used calculus method

    • 11 months ago
  33. ParthKohli
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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 \(v_1 = \dfrac{\alpha\beta T}{\alpha + \beta }\)

    • 11 months ago
  34. .Sam.
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    Aw man it was getting close :P

    • 11 months ago
  35. ParthKohli
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    :-D

    • 11 months ago
  36. ParthKohli
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    Try the 2nd one maybe?

    • 11 months ago
  37. .Sam.
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    Average velocity for the whole journey is it?

    • 11 months ago
  38. ParthKohli
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    Yeah.

    • 11 months ago
  39. .Sam.
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    is it zero

    • 11 months ago
  40. ParthKohli
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    No, because the displacement isn't zero.

    • 11 months ago
  41. .Sam.
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    But the initial and final velocities are zero

    • 11 months ago
  42. ParthKohli
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    I have not asked for the average acceleration.

    • 11 months ago
  43. ParthKohli
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    :-P

    • 11 months ago
  44. ParthKohli
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    \[\overline{v} = \dfrac{\Delta s}{\Delta t}\]

    • 11 months ago
  45. .Sam.
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    I've got \[\frac{x_2}{x_3-x_2}=\frac{\beta}{\alpha}=\frac{t_1}{t_2}\]

    • 11 months ago
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