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KCET2023MathematicsApplication of Derivatives

The distance ' s ' in meters travelled by a particle in ' t ' seconds is given by s= 2 t^3 3 -18 t+ 5 3 . The acceleration when the particle comes to rest is

Options

  1. A10 ~m ^2 / s
  2. B12 ~m ^2 / s
  3. C18 ~m ^2 / s
  4. D3 ~m ^2 / s

Correct answer

B. 12 ~m ^2 / s

Step-by-step solution

Given, aligned & s=2 t^3-18 t+ 5 3 & d s d t =6 t^2-18 d^2 s d t^2 & =12 t aligned We know that array ll & v=u+a t & 0=6 t^2-18+12 t^2 18 t^2 & =18 & t=1 ~s array After 1 ~s , particles comes to rest So, a=12 t [From Eq. (i)] aligned & a=-12 1 ~m ^2 / s & a=12 ~m ^2 / s aligned

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