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If s=2t^(3)+3t^(2)+2t+8 then find time a...

If `s=2t^(3)+3t^(2)+2t+8` then find time at which acceleration is zero.

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(a) `v = (ds)/(dt) = 6t^(2) + 6t+ 2 rArr a = (dv)/(dt) = 12t = 6 =0 rArr t =-(1)/(2)` which is impossible. There acceleration can never be zero.
(b) `because (dx)/(dt) = v therefore x = int v dt = overset4 underset2 int 4tdt = [2t^(2) ] _(2)^(4) = 2(4)^(2)- 2(2)^(2) = 32 - 8 = 24` m
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