A point moves x-y plane according to the law x=4 sin 6t and y= 4(1-cos 6t). The distance traversed by the particle in 4s is (x and y are in metre)

From the trajectory of the particle we can say that the particle is moving in a circle around (0,4) point.

Since, ${\mathrm{x}}^2+{\left(\mathrm{y}-4\right)}^2=4^2$

${\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{dx}}{\mathrm{dt}}=24\cos 6\mathrm{t}$

${\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}=24\sin 6\mathrm{t}$

$\mathrm{v}=\sqrt{{\mathrm{v}}_{\mathrm{x}}^2+{\mathrm{v}}_{\mathrm{y}}^2}=24\mathrm{m}/\mathrm{s}$

i.e., speed of the particle is constant. 

Hence, the distance traversed by the particle in 4s is

$\mathrm{s}=\mathrm{vt}=(24\times 4)\mathrm{m}=96\mathrm{m}$