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ParthKohli

  • 2 years 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
    • 2 years ago
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    @.Sam.

  2. myiah5766
    • 2 years ago
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    The highest velocity.

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

  4. .Sam.
    • 2 years ago
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    |dw:1368898722315:dw|

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

  6. .Sam.
    • 2 years ago
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    Is the whole journey takes time T?

  7. ParthKohli
    • 2 years ago
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    yes.

  8. .Sam.
    • 2 years ago
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    |dw:1368898894389:dw|

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

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

  11. ParthKohli
    • 2 years ago
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    Good going.

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

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

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

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

  16. .Sam.
    • 2 years 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
    • 2 years ago
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    Oh, my bad. It does stop at the end.

  18. .Sam.
    • 2 years ago
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    Hmm \[v_2=0\]

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

  20. .Sam.
    • 2 years ago
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    The diagram said it's maximum velocity

  21. ParthKohli
    • 2 years ago
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    OK. I got confused when you said v1 = v2

  22. .Sam.
    • 2 years ago
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    Nah

  23. .Sam.
    • 2 years 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
    • 2 years ago
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    You just eliminated \(v_1\) when you actually need it.

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

  26. ParthKohli
    • 2 years 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.
    • 2 years ago
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    What about the distance?

  28. .Sam.
    • 2 years ago
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    /displacement

  29. ParthKohli
    • 2 years 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
    • 2 years ago
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    Do you want to try more?

  31. .Sam.
    • 2 years ago
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    I'm getting more variables, lol

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

  33. ParthKohli
    • 2 years 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.
    • 2 years ago
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    Aw man it was getting close :P

  35. ParthKohli
    • 2 years ago
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    :-D

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

  37. .Sam.
    • 2 years ago
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    Average velocity for the whole journey is it?

  38. ParthKohli
    • 2 years ago
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    Yeah.

  39. .Sam.
    • 2 years ago
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    is it zero

  40. ParthKohli
    • 2 years ago
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    No, because the displacement isn't zero.

  41. .Sam.
    • 2 years ago
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    But the initial and final velocities are zero

  42. ParthKohli
    • 2 years ago
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    I have not asked for the average acceleration.

  43. ParthKohli
    • 2 years ago
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    :-P

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

  45. .Sam.
    • 2 years 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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