A particle moves in space along the path z = a x 3 + b y 2 in such a way that d ⁡ x d ⁡ t = c = d ⁡ y d ⁡ t where a , b and c are constants. The acceleration of the particle is

Step-by-step solution

Given that d ⁡ x d ⁡ t = d ⁡ y d ⁡ t = c ∴ d ⁡ 2 x d ⁡ t 2 = d ⁡ 2 y d ⁡ t 2 = 0 Further z = ax 3 + by 2 ∴ d ⁡ z d ⁡ t = 3 a x 2 d ⁡ x d ⁡ t + 2 b y d ⁡ y d ⁡ t = 3 a c x 2 + 2 b c y d ⁡ x d ⁡ t = c = d ⁡ y d ⁡ t ∴ d ⁡ 2 z d ⁡ t 2 = 6 a c x d ⁡ x d ⁡ t + 2 b c d ⁡ y d ⁡ t = 6ac 2 x + 2bc 2 Now acceleration of particle is a → = d ⁡ 2 x d ⁡ t 2 i ^ + d ⁡ 2 y d ⁡ t 2 j ^ + d ⁡ 2 z d ⁡ t 2 k ^ = 6 a c 2 x + 2 b c 2 k ^