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= betax^2`. Its velocity component in the x-direction is

Options

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

Answer

B

Explanation

Here, ` y = bx^2`. Differentiating it w.r.t. t`, we get ,br. ` (dy)/(dt) = b 2 x (dx)/(dt)` or v_y = 2 bx v_x`
Again differentiating it w.r.t.(t).we get
` (dv_y)/(dt) = 2 b v_x 9dx)/(dt) + 2 b x 9dv_x)/(dt) = 2 b v_x^2 + 0`
` [ (dv _x)/(dt) =0`. becaude the particle has constant acceleration along y-direction only]` ltbRgt As per question, ` (d v_y)/(dt) =a = 2 b v_x^2`
or ` v_x^2 = a/(2 b) ` or v_x = sqrt a/(2 b)`.