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A particle moves in space along the path...

A particle moves in space along the path `z=ax^(3)+by^(2)` in such a way that `(dx)/(dt)=c=(dy)/(dt)` where `a,b` and `c` are constants. The acceleration of the particle is

A

`(6ac^(2)x+2bc^(2))hatk`

B

`(2ax^(2)+6by^(2))hatk`

C

`(4bc^(2)+3ac^(2))hatk`

D

`(bc^(2)x+2by)hatk`

Text Solution

Verified by Experts

The correct Answer is:
A
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