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The acceleration. If v(0) is themagnitu...

The acceleration. If ` v_(0)` is themagnitude the engine is cut off, is given by ` (dv)/(dt) =- k v^(2)`, wher (k) is a constant. If `v_(0)` is the magnitude of the velocity at cut off, find the magnitude of the velocity at time (t0 after the cut off.

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Given, ` (d v)/(dt) =- k v^(3) dv =- k dt`
Integration it wighin the conditions of motion i.e. as time changes from ` 0 to t`, velocity changes from ` v_(0) to v`, we have
` int _(v0)^(v) v^(-3) d v=- k int _(0)^(t) dt or [v^(-2)/(-2)](v_(0)^(v) =- k (t) _(0)^(t)`
or -1/2 [1/v^(2) -1/v_(0)^(2)] =- kt`
or 1/v^(2) =2 kt + 1/v_(0)^(2) =(2 kt v_(0)^(2) +1)/v_(0)^(2)`
or ` v=v_(0)/(sqrt 1 + 2 kt v_(0)^(2)`.
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