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BITSAT2024PhysicsMotion in One DimensionActual

A particle is moving in a straight line. The variation of position ' x ' as a function of time ' t ' is given as x= (t^3-6 t^2+20 t+15 ) m . The velocity of the body when its acceleration becomes zero is:

Options

  1. A6 ~m / s
  2. B10 ~m / s
  3. C8 ~m / s
  4. D4 ~m / s

Correct answer

C. 8 ~m / s

Step-by-step solution

Displacement, x=t^3-6 t^2+20 t+15 Velocity, v= dx dt =3 t ^2-12 t +20 Acceleration, a= d v d t =6 t-12 When a=0 6 t -12=0 t=2 ~s At t =2 ~s , v=3(2)^2-12(2)+20=8 ~m / s

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