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NEET2012PhysicsMotion in One DimensionActual

The motion of a particle along a straight line is described by equation x=8+12 t-t^3 where, x is in metre and t in second. The retardation of the particle when its velocity becomes zero, is

Options

  1. A24 ~ms ⁻²
  2. Bzero
  3. C6 ~ms ⁻²
  4. D12 ~ms ⁻²

Correct answer

D. 12 ~ms ⁻²

Step-by-step solution

Given, x=8+12 t-t^3 We know v= d x d t and a= d v d t So, v=12-3 t^2 and a=-6 t At t=2 ~s y=0 and a=-6 t So, a=-12 ~m / s ^2 So, retardation of the particle =12 ~m / s ^2 .

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