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King Group Title

If motion of a particle given below is x=u(t-2s)+a(t-2s)^2 A)initial velocity is u B)acceleration is a C)acceleration is 2a D)at t=2s particle is at origin More than 1 option can be right Please explain ure answer.

  • 2 years ago
  • 2 years ago

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  1. King Group Title
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    s is seconds just a unit

    • 2 years ago
  2. Ishaan94 Group Title
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    A,C and D are true :-) Compare it with \(x = u\times t + \frac{1}{2}\times a \times t^2\)

    • 2 years ago
  3. King Group Title
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    actually ure wrong..i wont tell u the rite answer try it!!

    • 2 years ago
  4. Ishaan94 Group Title
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    I can't be wrong

    • 2 years ago
  5. King Group Title
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    and please explain...

    • 2 years ago
  6. King Group Title
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    no u hav to expand then compare

    • 2 years ago
  7. Ishaan94 Group Title
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    Na I don't have to expand since (t - 2) is time only (2 is in seconds and so is t)

    • 2 years ago
  8. King Group Title
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    but its not a(t-2s)^2 its supposed a/2

    • 2 years ago
  9. King Group Title
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    it at^2 ok but shouldnt it be (at^2)/2

    • 2 years ago
  10. Ishaan94 Group Title
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    \[\frac{1}{2}\times 2a\times (t - 2)^2\]

    • 2 years ago
  11. King Group Title
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    initial velocity is not u C),D) is rite

    • 2 years ago
  12. Ishaan94 Group Title
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    Initial velocity must be U!

    • 2 years ago
  13. King Group Title
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    its not..my sir explained but i frgt

    • 2 years ago
  14. King Group Title
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    pls expand once aand see what u get

    • 2 years ago
  15. Ishaan94 Group Title
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    Okay I get what you meant

    • 2 years ago
  16. King Group Title
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    ahh now xplain

    • 2 years ago
  17. Ishaan94 Group Title
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    Hmm because it's (t-2)^2 not t^2 so an extra co-efficient of t is there

    • 2 years ago
  18. King Group Title
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    oh yeah and btw my sir took x as xfinal - xinitial

    • 2 years ago
  19. Ishaan94 Group Title
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    Okay!

    • 2 years ago
  20. King Group Title
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    hey explain!

    • 2 years ago
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