Question

The velocity of a particle moving in the $x-y$ plane is given by $\frac{d x}{d t}=8 \pi \sin 2 \pi t$ and $\frac{d y}{d t}=5 \pi \cos 2 \pi t$ where, $t=0, x=8$ and $y=0$, the path of the particle is

• A. a straight line
• B. an ellipse
• C. a circle
• D. a parabola

Answer 1

Answer: (B)

$y-x$ graph gives the shape of path of particle
$\frac{d x}{d t} =8 \pi \sin 2 \pi t$
$\int\limits_{8}^{x} d x -\int\limits_{0}^{t} 8 \pi \sin 2 \pi t d t$
$x-8 =-\frac{8 \pi}{2 \pi}[\cos 2 \pi]_{0}^{t}$
$x-8 =4(1-\cos 2 \pi t)$
$\cos 2 \pi t =\frac{x-12}{4}$
$\frac{d y}{d t} =-5 \pi \cos 2 \pi t$
$\int\limits_{0}^{y} d y =5 \pi t \int\limits_{0}^{t} \cos 2 \pi t$
$y =\frac{5}{2} \sin 2 \pi t$
$\left(\frac{x-12}{4}\right)^{2}+\frac{y^{2}}{\left(\frac{5}{2}\right)^{2}} =1$
$\frac{(x-12)^{2}}{(4)^{2}}+\frac{y^{2}}{\frac{5}{(2)^{2}}} =1$