Question

A particle moves in a plane along an elliptical path given by $\frac{x^2}{9}+\frac{y^2}{4}=1$. At point $(0,2)$, x - component of velocity is $9m/s$. The magnitude of y - component of acceleration at this point is

Answer 1

Given,
$\frac{x^2}{9}+\frac{y^2}{4}=1..........\left(1\right)$
Given $U_x=\frac{dx}{dt}=9m/s,$ at (0,2)
Differentiating (1) wrt 't'

$\frac{2x}{9}\frac{dx}{dt}+\frac{2y}{4}\frac{dy}{dt}=0..........\left(2\right)$
At $(0,2)$
$\frac{2\left(0\right)}{9}\frac{dx}{dt}+\frac{2\left(2\right)}{4}\frac{dy}{dt}=0$
$\Rightarrow \frac{dy}{dt}=0$
Differentiating (2) wrt 't'
$\frac{1}{9}[{\left(\frac{dx}{dt}\right)}^2+x\frac{d{x}^2}{d{t}^2}]+\frac{1}{4}[{\left(\frac{dy}{dt}\right)}^2+y\left(\frac{d^{2}y}{d{t}^2}\right)]=0$
$\Rightarrow \frac{1}{9}[9^2+0]+\frac{1}{4}[0^2+ay]=0$
$[\text{As}\frac{dx}{dt}=U_x=9,x=0,y=2;\frac{dy}{dt}=U_y=0,\frac{d^{2}y}{d{x}^2}=a_y]$
$\Rightarrow a_y=-2\times \frac{9^2}{9}=-18m/s^2$
$|a_y|=18m/s^2$