A wheel of circumference C is at rest on the ground. When the wheel rolls forward through half a revolution, then the displacement of point of contact will be

During half revolution of the wheel the point A covers AC =$\mathrm{\pi }$R horizontal distance and
 

BC = 2R vertical distance
x = $\mathrm{\pi }$R , y= 2R , Here circumference(C) =2$\mathrm{\pi }$R
Displacement $\mathrm{S}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}$
$\mathrm{x}=\mathrm{\pi }\frac{\mathrm{C}}{2\mathrm{\pi }}=\frac{\mathrm{C}}{2}\mathrm{y}=\frac{\mathrm{C}}{\mathrm{\pi }}$

$\mathrm{S}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}$

$\mathrm{S}=\sqrt{\frac{{\mathrm{C}}^2}{4}+\frac{{\mathrm{C}}^2}{{\mathrm{\pi }}^2}}\Rightarrow \mathrm{S}=\mathrm{C}\sqrt{\frac{1}{4}+\frac{1}{{\mathrm{\pi }}^2}}$