# A point moves in $X-Y$ plane according to $X=4 \sin \;6t$ and $Y=4(1-\cos \;6t)$.The distance traversed by the particle in 4 seconds is (x and y are in meters)

$(a)\;96 m \quad (b)\;48 m \quad (c)\;24 m \quad (d)\;108 m$

$v_x =\large\frac{dx}{dy}$$=24 \cos 6t v_y =\large\frac{dy}{dt}$$=24 \sin 6t$
$v=\sqrt {{v_x}^2+{v_y}^2}$
$\quad=24 m/s^2$
Since $v$- speed is constant.
distance traversed in 4 second is
$s=v \times t$
$s=24 \times 4$
$s=96 m$
Hence a is the correct answer.

edited Jan 25, 2014 by meena.p

+1 vote