A wheel of radius 1 m rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is? Ans is root pie sq +4 but how? When I am solving it, I get pie r but I guess here I've to use the formula on rotational motion.



Given, the radius of the wheel is 1m. so,its circumference is 2π m
when it rotates half, the initial point of contact is vertically above the final point of contact.thus,the three points form a right triangle.
by pythagoras theorem, we solve it to get the magnitude of displacement,which is the hypotenuse.
thus, d= \sqrt{4+  \pi ^{2} }
here's the answer!
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2 5 2