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ParthKohli

  • one year ago

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.

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  1. ParthKohli
    • one year ago
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    @.Sam.

  2. myiah5766
    • one year ago
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    The highest velocity.

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

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

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

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

  7. ParthKohli
    • one year ago
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    yes.

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

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

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

  11. ParthKohli
    • one year ago
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    Good going.

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

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

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

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

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

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

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

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

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

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

  22. .Sam.
    • one year ago
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    Nah

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

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

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

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

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

  28. .Sam.
    • one year ago
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    /displacement

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

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

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

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

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

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

  35. ParthKohli
    • one year ago
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    :-D

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

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

  38. ParthKohli
    • one year ago
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    Yeah.

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

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

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

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

  43. ParthKohli
    • one year ago
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    :-P

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

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

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