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King
 4 years ago
If motion of a particle given below is
x=u(t2s)+a(t2s)^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.
King
 4 years ago
If motion of a particle given below is x=u(t2s)+a(t2s)^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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0s is seconds just a unit

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A,C and D are true :) Compare it with \(x = u\times t + \frac{1}{2}\times a \times t^2\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually ure wrong..i wont tell u the rite answer try it!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and please explain...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no u hav to expand then compare

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Na I don't have to expand since (t  2) is time only (2 is in seconds and so is t)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but its not a(t2s)^2 its supposed a/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it at^2 ok but shouldnt it be (at^2)/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}\times 2a\times (t  2)^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0initial velocity is not u C),D) is rite

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Initial velocity must be U!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its not..my sir explained but i frgt

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0pls expand once aand see what u get

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay I get what you meant

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm because it's (t2)^2 not t^2 so an extra coefficient of t is there

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yeah and btw my sir took x as xfinal  xinitial
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