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The displacement of a particle moving in straight line is given as function of time as `s = ((t^(3))/(3) - (3t^(2))/(2) + 2t), s` is in m and t is in sec. The particle comes to momentary rest n times Find the value of `n`

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`s = (t^(3))/(3) - (3t^(2))/(2) + 2t`
`:. V = (ds)/(dt) = t^(2) - 3 t + 2`
Since the particle comes to rest, therefore,
`t^(2) - 3 t + 2 = 0`
`(t - 1) (t - 2) = 0`
` t = 1 s " and " 2s`
Hence, the particle comes to rest twice.
`:. n = 2`

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