Question

A particle moves in the plane $xy$ with constant acceleration $\omega$ directed along the negative $y$ axis. The equation of motion of the particle has the form $y=ax-b{x}^2$, where $a$ and $b$ are positive constants. Find the magnitude of velocity of the particle $v_0$ at the origin of coordinates.

• A. $\sqrt{(1+a^2)\frac{\omega }{2b}}$
• B. $\sqrt{(1+a^3)\frac{\omega }{2b}}$
• C. $\sqrt{(1+a^2)\frac{\omega }{4b}}$
• D. $\sqrt{(1+a^2)\frac{\omega }{3b}}$

Answer 1

The correct option is A $\sqrt{(1+a^2)\frac{\omega }{2b}}$

According to the problem
$\overrightarrow{w}=w(-\overrightarrow{j})$
So, $w_x=\frac{d{v}_x}{dt}=0$ and $w_y=\frac{d{v}_y}{dt}=-w$ ...........$\left(1\right)$
Differentiating equation of trajectory, $y=ax-b{x}^2$, with respect to time
$\frac{dy}{dt}=\frac{adx}{dt}-2bx\frac{dx}{dt}$ ...................$\left(2\right)$
So, $\frac{dy}{dt}{|}_{(}x=0)=a\frac{dx}{dt}{|}_{(}x=0)$
Again differentiating with respect to time
$\frac{d^{2}y}{d{t}^2}=\frac{a{d}^{2}x}{d{t}^2}-2b{\left(\frac{dx}{dt}\right)}^2-2bx\frac{d^{2}x}{d{t}^2}$
or, $-w=a\left(0\right)-2b{\left(\frac{dx}{dt}\right)}^2-2bx\left(0\right)$ (using $1$)
or, $\frac{dx}{dt}=\sqrt{\frac{w}{2b}}$ (using 1) ..............$\left(3\right)$
Using (3) in (2) $\frac{dy}{dt}{|}_{x=0}=a\sqrt{\frac{w}{2b}}$............. $\left(4\right)$
Hence, the velocity of the particle at the origin
$v=\sqrt{{\left(\frac{dx}{dt}\right)}_{x=0}^2+{\left(\frac{dy}{dt}\right)}_{x=0}^2}=\sqrt{\frac{w}{2b}+a^2\frac{w}{2b}}$ (using equations $\left(3\right)$ and $\left(4\right)$)
Hence, $v_0=\sqrt{\frac{w}{2b}(1+a^2)}$