Question

Solve this :

Q. A particle crossing the origin of co-ordinates at t = 0, moves in xy-plane with constant acceleration a in y-direction. If its equation of motion is $y=b{x}^2(biscons\tan t)$ its velocity component in the x-direction is

$$\begin{gathered} ○\sqrt{\frac{2b}{a}} \\ ○\sqrt{\frac{2a}{b}} \\ ○\sqrt{\frac{a}{2b}} \\ ○\sqrt{\frac{a}{b}} \end{gathered}$$

Answer 1

Dear Student,


$$\begin{gathered} y=b{x}^2 \\ \frac{dy}{dt}=2bx\frac{dx}{dt} \\ {v}_y=2bx{v}_x \\ differentiateonceagain \\ a=\frac{d{v}_y}{dt}=2b{v}_x.\frac{dx}{dt} \\ a=2b{v_x}^2 \\ {v}_x=\sqrt{\frac{a}{2b}} \\ Regards \end{gathered}$$