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The velocity v of a particle is given by the equation `v=6t^(2)-6t^(3)`, where v is in `ms^(-1)`, t is the instant of time in seconds while 6 and 6 are suitable dimensional constants. At what values of t will the velocity be maximum and minimum ? Determine these maximum and minimum values of the velocity. |
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Answer» Given `v=6t^(2)-6t^(3)`. Differentiating v w.r.t. t, we have `(dv)/(dt)=12 t-18 t^(2)`. Putting `(dv)/(dt)=0` we will get the values of t at which v is maximum or minimum. Therefore, `12t-18t=0 rArr t=0,2//3 s` To the distinguish between points of maxima and minima, we need the second derivative of v. `(d^(2)v)/(dt^(2))=12-36t` Now, `(d^(2)v)/(dt^(2))|_(t=0)=12 gt0` So t=0 is a point of minima. `(d^(2)v)/(dt^(2))|_(t=2//3s)=12-36 xx(2)/(3)=-12 lt0` So, t=2/3 s is a point of maxima. Hence, the minimum value of v is `0 ms^(-1)` (by putting t=0s in v). `v_("max")=6xx(4)/(9)-6xx(8)/(27) =(8)/(3)-(16)/(9)` (by putting `t=(2)/(3)s in v`) `v_("max")=(8)/(9)ms^(-1)` |
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