InterviewSolution
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InterviewSolution
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A particle starts from point A moves along a straight line path with an acceleration given by a = p - qx where p, q are connected and x is distance from point A . The particle stops at point B. The maximum velocity of the particle isA. `p/q`B. `p/sqrt(q)`C. `q/p`D. `sqrt(q/p)` |
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Answer» Correct Answer - B Given : `a = p - qx` At maximum velocity, a = 0 `implies 0 = p - qx` or `x = p/q` `therefore` Velocity is maximum at `x = p/q` `since` Acceleration,` a = (dv)/(dt) = (dv)/(dx) (dx)/(dt) = v (dv)/(dx)` since` v = (dx)/(dt)` `therefore v (dv)/(dx) = a = p - qx` `vdv = (p - qx) dx` Integrating both sides of the above equation, we get `v^(2)/2 = px = (qx)^(2)/2` `v^(2) = 2px - qx^(2)` or v = `sqrt (2px - qx^(2)` At `x = p/q, v = v_max = sqrt (2p(p/q) - q(p/q)^(2))= p/sqrt q` |
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