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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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s is seconds just a unit
A,C and D are true :-) Compare it with \(x = u\times t + \frac{1}{2}\times a \times t^2\)
actually ure wrong..i wont tell u the rite answer try it!!

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Other answers:

I can't be wrong
and please explain...
no u hav to expand then compare
Na I don't have to expand since (t - 2) is time only (2 is in seconds and so is t)
but its not a(t-2s)^2 its supposed a/2
it at^2 ok but shouldnt it be (at^2)/2
\[\frac{1}{2}\times 2a\times (t - 2)^2\]
initial velocity is not u C),D) is rite
Initial velocity must be U!
its sir explained but i frgt
pls expand once aand see what u get
Okay I get what you meant
ahh now xplain
Hmm because it's (t-2)^2 not t^2 so an extra co-efficient of t is there
oh yeah and btw my sir took x as xfinal - xinitial
hey explain!

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