Match the velocity-position ( v-x ) relationships of a particle in Column-I with the corresponding acceleration-position ( a-x ) relationships in Column-II. Assume v₀ and k are positive constants. Column-I ( v-x relation) Column-II ( a-x relation) (A) v = kx (I) a = 0 (B) v = v₀ - kx (II) a = k 2 (C) v = kx (III) a = k^2 x (D) v = v₀ (IV) a = k^2 x - k v₀ Choose the correct answer from the options given below:

Step-by-step solution

The acceleration a of a particle can be written in terms of velocity v and position x as a = v dv dx . (A) For v = kx , dv dx = k . Thus, a = (kx)(k) = k^2 x . (Matches III) (B) For v = v₀ - kx , dv dx = -k . Thus, a = (v₀ - kx)(-k) = k^2 x - k v₀ . (Matches IV) (C) For v = kx , dv dx = k 2 kx . Thus, a = ( kx ) ( k 2 kx ) = k 2 . (Matches II) (D) For v = v₀ , dv dx = 0 . Thus, a = (v₀)(0) = 0 . (Matches I) Therefore, the correct matching is A-III, B-IV, C-II, D-I. Answer: A-III, B-IV, C-II, D-I