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Three particles have their displacement ...

Three particles have their displacement `X _(A) = 2 t , X _(B) = 3t ^(2) + 2 t + 6 and X_(C) = 5t ^(3) + 4t.` Which one of them is moving with uniform acceleration ?

Text Solution

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We are given :
`X _(A) = 2 t + 7 , X _(B) = 3t ^(2) + 2 t + 6 , X _(C) = 5 t ^(3) + 4t`
We are to calculate, Among three particles which is moving with uniform acceleration.
Using relation, `v = (dx)/(dt)`
and ` a = (dv)/(dt) implies v _(A) = (d ( X _(A)))/( dt) = (d)/(dt) ( 2t + 7) = 2`
`v _(B) = ( d ( X _(B)))/( dt ) = (d)/(dt) ( 3t + 2 t + 6) = 6 t + 2 implies v _(C) = (d (X _(C)))/( dt) = (d)/(dt) ( 5t ^(3) + 4t) = 15 t ^(2) + 4`
Now, `a _(A) = (d (V _(A)))/( dt ) = (d)/(dt) (2) = 0 impliesa_(B) = (d ( V _(B)))/(dt) = (d)/(dt) ( 6 t + 2) = 6`
`and a _(c) = (dV _(C))/( dt) = (d)/(dt) (15 t ^(2) + 4) = 30 t therefore ` body B is moving with uniform acceleration.
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