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King

  • 3 years ago

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.

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  1. King
    • 3 years ago
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    s is seconds just a unit

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

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

  4. Ishaan94
    • 3 years ago
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    I can't be wrong

  5. King
    • 3 years ago
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    and please explain...

  6. King
    • 3 years ago
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    no u hav to expand then compare

  7. Ishaan94
    • 3 years ago
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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)

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

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

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

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

  12. Ishaan94
    • 3 years ago
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    Initial velocity must be U!

  13. King
    • 3 years ago
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    its not..my sir explained but i frgt

  14. King
    • 3 years ago
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    pls expand once aand see what u get

  15. Ishaan94
    • 3 years ago
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    Okay I get what you meant

  16. King
    • 3 years ago
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    ahh now xplain

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

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

  19. Ishaan94
    • 3 years ago
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    Okay!

  20. King
    • 3 years ago
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    hey explain!

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