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 is

Options

A. `p/q`
B. `p/sqrt(q)`
C. `q/p`
D. `sqrt(q/p)`

Answer

B

Explanation

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`