Question

The velocity of a particle moving in the $x-y$ plane is given by $\frac{dx}{dt}=8\pi \sin 2\pi t$ and $\frac{dy}{dt}=5\pi \cos 2\pi t$ 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
$\frac{dx}{dt}=8\pi \sin 2\pi t$
$\underset{8}{\overset{x}{\int }}dx-\underset{0}{\overset{t}{\int }}8\pi \sin 2\pi tdt$
$x-8=-\frac{8\pi }{2\pi }[\cos 2\pi {]}_0^t$
$x-8=4(1-\cos 2\pi t)$
$\cos 2\pi t=\frac{x-12}{4}$
$\frac{dy}{dt}=-5\pi \cos 2\pi t$
$\underset{0}{\overset{y}{\int }}dy=5\pi t\underset{0}{\overset{t}{\int }}\cos 2\pi t$
$y=\frac{5}{2}\sin 2\pi t$
${\left(\frac{x-12}{4}\right)}^2+\frac{y^2}{{\left(\frac{5}{2}\right)}^2}=1$
$\frac{(x-12{)}^2}{(4{)}^2}+\frac{y^2}{\frac{5}{(2{)}^2}}=1$