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A particle moves along the parabolic pat...

A particle moves along the parabolic path `x = y^2 + 2y + 2` in such a way that Y-component of velocity vector remains `5ms^(-1)` during the motion. The magnitude of the acceleration of the particle is

A

`50 ms^(-2)`

B

`100 ms^(-2)`

C

`10(sqrt2) ms^(-2)`

D

`0.1 ms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`a_(y) = 0 = (d^2y)/(dt^2)`
`(dx)/(dt) = (2y+2)* (dy)/(dt)`
`(d^2y)/(dt^2) = (2y +2)*(d^2y)/(dt^2) + 2((dy)/(dt))^2`
`:. a_x = a = (2y +2)(0) + 2(5)^2`
`=50 m//s^2`
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