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If `v = x^2 - 5x + 4`, find the acceleration of the particle when velocity of the particle is zero.

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Correct Answer - A
`v=0rArrx^(2)-5x+4=0`
`x=1m& 4m`
`(dv)/(dt)=(2x-5)v=(2x-5)(x^(2)-5x+4)` at `x=1m` and `4m,(dv)/(dt)=0`

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