A particle is moving along a straight-line path and its displacement (x) at time t is…

Physics · JEE Main · NTA ExamsKinematics

A particle is moving along a straight-line path and its displacement (x) at time t is given by             
 x2 = at2 + 2bt + c (where a, b and c are constants)
 the acceleration of the particle is:
  1. \(\frac{\mathbf{a}}{\mathbf{x}}\)
  2. \(\frac{(a t+b)^{2}}{x^{3}}\)
  3. \(\frac{a}{x}-\frac{(a t+b)^{2}}{x^{3}}\)
  4. \(\frac{a}{x}+\frac{(a t+b)^{2}}{x^{3}}\)

Answer

(C) a x - (a t+b)^ 2 x^ 3

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