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Ruchi.

  • 3 years ago

displacement time equation of a particle moving along x-axis is x=20+t3-12t. a)find velocity of particle at time t=0 b)state whether the motion is uniformly acceleration or nt. c)find position of the particle when v=0.

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  1. apoorvk
    • 3 years ago
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    if that^ is the equation of the displacement, how would you get the equation for velocity?

  2. apoorvk
    • 3 years ago
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    wrt 'time' ofcourse.

  3. Aadarsh
    • 3 years ago
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    "Substitute" the value of "t" and find.

  4. yakeyglee
    • 3 years ago
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    Hint for part (a): \(v=\frac{dx}{dt}\) Hint for part (b): \(a=\frac{dv}{dt} = \frac{d^2x}{dt^2}\). If this is a function of time, then it is not constant. Hint for part (c): What \(t\) values satisfies this condition?

  5. Ruchi.
    • 3 years ago
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    v=dx/dt

  6. Aadarsh
    • 3 years ago
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    Sorry. @yakeyglee is absolutely correct. Follow his steps and ur answer is done

  7. Ruchi.
    • 3 years ago
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    as v=0,we 0=3t2-12

  8. Ruchi.
    • 3 years ago
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    am i right.

  9. Aadarsh
    • 3 years ago
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    Yeah, perfect @Ruchi.

  10. Aadarsh
    • 3 years ago
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    So, we get|dw:1338103103383:dw|

  11. Aadarsh
    • 3 years ago
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    A) answer is 12

  12. yakeyglee
    • 3 years ago
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    More specifically, \(t=\pm 2\).

  13. Ruchi.
    • 3 years ago
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    a)=-12m/s

  14. Aadarsh
    • 3 years ago
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    But, "t" cannot be negative, so t = 2. so, c) 2

  15. Aadarsh
    • 3 years ago
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    A) -12m/s correct. sorry for my error.

  16. Aadarsh
    • 3 years ago
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    On double differentiation, we get, a = 6t (I think I am correct).

  17. Aadarsh
    • 3 years ago
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    So, unifrom acceleration

  18. Aadarsh
    • 3 years ago
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    @Ishaan94 , @apoorvk @him1618 please help me and correct me if I am wqrong.

  19. him1618
    • 3 years ago
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    Acceleration is not uniform

  20. Aadarsh
    • 3 years ago
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    But how? My steps r wrong or rite? Please help me.

  21. apoorvk
    • 3 years ago
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    No as a=6t, acceleration is dependent on time. So it's not uniform.

  22. yakeyglee
    • 3 years ago
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    \(t\) most certainly can be negative. It simply corresponds to be fore \(t=0\).

  23. him1618
    • 3 years ago
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    Bt that isnt in the domain@ yakeyglee

  24. Aadarsh
    • 3 years ago
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    So @apoorvk bhai, when can we say that a is uniform? in what relation?

  25. apoorvk
    • 3 years ago
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    And put in the value of 't' when v=0 from the first part, in the equation of displacement, that is put in t=2. That should give you the displacement at the time when v=0.

  26. him1618
    • 3 years ago
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    When the accn is not a function of time

  27. apoorvk
    • 3 years ago
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    uniform means constant - it isn't constant is it?

  28. him1618
    • 3 years ago
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    In this case accn is a function of time...so its not uniform

  29. Aadarsh
    • 3 years ago
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    Thanks all, I understood.

  30. Aadarsh
    • 3 years ago
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    But, plz show me what will be the answer of (a), (b) and (c)

  31. him1618
    • 3 years ago
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    velocity is dx/dt

  32. yakeyglee
    • 3 years ago
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    @him1618 I don't see any statement of the domain...it's usually assumed \(t\in\mathbb R\), which includes negative values.

  33. Aadarsh
    • 3 years ago
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    So, (a) = -12, (c) = 2, (b) = not uniform. Is this the final answer? Please say.

  34. him1618
    • 3 years ago
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    it is so my man..bt not here..this is simple/...were not into time domain circuits here :P

  35. him1618
    • 3 years ago
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    c=4

  36. Aadarsh
    • 3 years ago
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    Is (a) and (b) what i gave the answer rite? How is (c) = 4?????????

  37. him1618
    • 3 years ago
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    v=0 means 3t^2 - 12 =0 t=2 Put t=2 in x(t)

  38. Aadarsh
    • 3 years ago
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    Still bhaiya, we get, 20 + 2*3 - 12*2 = 2. Am I rite????????

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