If S²=a t²+2 b t+c , then acceleration is

Step-by-step solution

We have, S²=a t²+2 b t+c ...(i) Differentiating Eq. (i) w.r.t. t , we get aligned &2 S d S d t =2 a t+2 b & d S d t = ( a t+b S ) ...(ii) aligned Differentiating Eq. (ii) w.r.t. t , we get aligned d² S d t² &= S- d d t (a t+b)-(a t+b) d S d t S² & d² S d t² = a S-(a t+b) (a t+b) S S² [using Eq. (ii)] & d² S d t² = a S²-(a t+b)² S³ aligned d² S d t² = a (a t²+2 b t+c )- [(a t)²+b²+2 a t b ] S³ d² S d t² = a² t²+2 a b t+a c-a² t²-b²-2 a t b S³ d² S d t² = a c-b² S³ Acceleration = ( d² S d t² ) 1 S³