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`p/q``p/sqrt(q)``q/p``sqrt(q/p)`

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BSolution :

Given : `a = p - qx` <br> At maximum velocity, a = 0 <br> `implies 0 = p - qx` or `x = p/q` <br> `therefore` Velocity is maximum at `x = p/q` <br> `since` Acceleration,` a = (dv)/(dt) = (dv)/(dx) (dx)/(dt) = v (dv)/(dx)` since` v = (dx)/(dt)` <br> `therefore v (dv)/(dx) = a = p - qx` <br> `vdv = (p - qx) dx` <br> Integrating both sides of the above equation, we get <br> `v^(2)/2 = px = (qx)^(2)/2` <br> `v^(2) = 2px - qx^(2)` or v = `sqrt (2px - qx^(2)` <br> At `x = p/q, v = v_max = sqrt (2p(p/q) - q(p/q)^(2))= p/sqrt q`