Answer:
Distance covered by cyclist
= Quadrant of circle
\[=\frac{1}{4}\times Circumference\,\,of\,\,circle\]
\[=\frac{1}{4}\times 2\pi r=\frac{\pi r}{2}\]
Initial position\[=x,\] Final position = z
Magnitude of displacement\[=xz=?\]
Considering right triangle oxz, we have
\[xz=\sqrt{{{(ox)}^{2}}+{{(oz)}^{2}}}\]
\[=\sqrt{{{(r)}^{2}}+{{(r)}^{2}}}\] (from Pythagoras theorem)
\[=\sqrt{2{{r}^{2}}}=\sqrt{2}r\]
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