A particle starts from point A moves along a straight line path with an aceleration given by a = p - qx where p,q are constants and x is distance from point A. The particle stops at point B. The maximum velocity of the paticle is
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answer is B.
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Detailed Solution
Acceleration a = (p - qx)The velocity is maximum when dv/dx = 0Now, velocity is zero when a = p - qx = 0 i.e. x = p/qNow, v. dv = (p - qx)dx⇒∫0xmv dv =p∫0pq dv -q∫0pqxdx ⇒vm=pq