A particle starts from the origin of co-ordinates at time `t=0` and moves in the`xy` plane with a constant acceleration `alpha` in the `y`-direction. Its equation of motion is `y=betax^(2)`. Its velocity component in the `x`-direction

Options

A. `sqrt((a)/(2b))`
B. `sqrt((2a)/(b))`
C. `sqrt((a)/(b))`
D. `sqrt((2a)/(2b))`

Answer

A

Explanation

(a): Given `y=bx^(2)`
Differentiating the above relation , we get
`(dy)/(dt)=2bx(dx)/(dt)`
Acceleration is along Y-axis , hence X-component of velocity remains contant . We can differentiate the above relation once more and then can equate it with given acceleration along Y-axis.
`(d^(2)y)/(dt^(2))=2b(dx)/(dt)(dx)/(dt)`
`rArr 2b((dx)/(dt))^(2)=arArr(dx)/(dt)=sqrt((a)/(2b))`
Hence , option (a) ios correct .