Question

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

• A. $\sqrt{\frac{2b}{a}}$
• B. $\sqrt{\frac{A}{2b}}$
• C. $\sqrt{\frac{a}{b}}$
• D. $\sqrt{\frac{b}{a}}$
• E. $\sqrt{ba}$

Answer 1

Answer: (B)

$y=b{x}^2$ $\frac{dy}{dt}=2bx\frac{dx}{dt}$ ...(i)
$\frac{dy}{dt}=at$
$(\because v_y=u_y+a_{y}t)$
$at=2bx\frac{dx}{dt}$
$atdt=2bxdx$ Take integration of both sides
$\int atdt=\int 2bxdx$
$\frac{a{t}^2}{2}=b{x}^2+c$ ...(ii)
At $t=0,x=0$
$c\Rightarrow 0$
Then, $\frac{a{t}^2}{2}=b{x}^2$
$x=\sqrt{\frac{a{t}^2}{2b}}=t\sqrt{\frac{a}{2b}}$
$\therefore$ $v_x=\frac{dx}{dt}=\sqrt{\frac{a}{2b}}$