Question

A particle moves in space along the path z=ax^3+by^2? in such a way that,
dx/dy=c=dy/dt
Where a, b and
C are constants. The acceleration of the particle is
(A) (6ac²x + 2bc2) Â
(B) (2ax2 + 6by2)
(C)(4bc²x + 6ac2) Â
(D) (bc?x + 2by) Â​

Answer 1

Answer:

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