A particle starts from rest and moves along x axis. The acceleration a varies with its displacement x as a=(2-x)m/s^ 2 . Find magnitude of acceleration of particle when its displacement along positive x-axis is maximum.
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Given info : A particle starts from rest and moves along x axis. The acceleration a varies with its displacement x as a = (2-x) m/s².
To find : the magnitude of acceleration of particle when its displacement along positive x axis is maximum.
solution : a = (2 - x)
⇒ v dv/dx = ( 2 - x)
⇒∫v dv = ∫ (2 - x) dx
⇒ v² - u² = 2x - x²/2
∵ particle starts from rest, ∴ u = 0
⇒ v² = (4x - x²)/2
particle will displace until velocity of it becomes zero. after that particle starts to move backwards so the maximum distance of the particle will be when we take velocity of it equals zero.
so, v² = (4x - x²)/2 = 0
⇒ x = 4, this is the maximum distance of the particle.
now the acceleration of the particle, a = ( 2 - x) = 2 - 4 = - 2 m/s
therefore the acceleration of the particle is - 2 m/s.