A particle starts from the origin of coordinates 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 = beta x^(2)`. Its velocity component in the x-directon is

Options

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

Answer

B

Explanation

`y = bx^(2)`
Differentiating w.r.t t on both sides, we get
`(dy)/(dt) = b2x (dx)/(dt) " or " v_y = 2bxv_x`
Again , diffenentiating w.r.t t on both sides , we get
`(dv_y)/(dt)=2bv_x(dx)/(dt)+2bx (dv_x)/(dt)=2bv_x^2+0`
[ `(dv_X)/(dt)=0` , because the contents acceleration along y-direction]
As per question `(dv_y)/(dt)=a =2bv_x^2" or " v_x =sqrt((a)/(2b))`