A toy cyclist completes one round of square track of side 4 m in 80 seconds. What will be the displa

Question

A toy cyclist completes one round of square track of side 4 m in 80 seconds. What will be the displacement at the end of 6 minutes?\

Professor PC Thomas Classes · Answer

Total time of motion on square track\ $=6 \mathrm{~min}=6 	imes 60=360 \mathrm{~s}$\ Time period of revolution $=80 \mathrm{~s}$\ $	herefore$ Displacement in time $(=4 	imes 80=320 \mathrm{~s})=0$ as, toy cyclist will be reaching at the starting point.\ Thus displacement in 360s = displacement in 40s\ $=\sqrt{(4)^{2}+(4)^{2}}=4 \sqrt{2} \mathrm{~m}$\