Question

A particle moves along the curve $y=a{x}^2$, with constant speed $V$. The acceleration at the origin of co-ordinates being ( where $a$ is a constant and $v_x$ is velocity in x direction)

• A. $\frac{v_x^2}{2a}$
• B. $2a{v}_x^2$
• C. $\frac{v_x^2}{a}$
• D. $a{v}_x^2$

Answer 1

The correct option is B $2a{v}_x^2$

$y=a{x}^2\Rightarrow \frac{dy}{dt}=2ax\left(\frac{dx}{dt}\right)\Rightarrow \frac{dy}{dt}=2ax\left(v_x\right)$
We see that at origin $\frac{dy}{dt}=0$, as y component of the velocity is only changing its direction at the origin.
And as the velocity is given as constant there is acceleration only in the $y$ direction and $\frac{d^{2}x}{d{t}^2}$ would be zero.
$\Rightarrow \frac{d^{2}y}{d{t}^2}=2a{v}_x\left(\frac{dx}{dt}\right)\Rightarrow \frac{d^{2}y}{d{t}^2}=acceleration=2a{v_x}^2$