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3-2.Motion in Plane
medium
An athlete completes one round of a circular track of radius $10\, m$ in $40\, sec$. The distance covered by him in $2 \,min$ $20 \,sec$ is ........ $m$
A$70$
B$140$
C$110$
D$220$
Solution
(d) Time period $= 40 \,sec$
No. of revolution $ = \frac{{{\rm{Total time}}}}{{{\rm{Time period}}}}$ $ = \frac{{140\,\sec }}{{40\,\sec }} = 3.5\,Rev$.
So, distance $ = 3.5 \times 2\pi R = 3.5 \times 2\pi \times 10 = 220\,m$.
No. of revolution $ = \frac{{{\rm{Total time}}}}{{{\rm{Time period}}}}$ $ = \frac{{140\,\sec }}{{40\,\sec }} = 3.5\,Rev$.
So, distance $ = 3.5 \times 2\pi R = 3.5 \times 2\pi \times 10 = 220\,m$.
Standard 11
Physics
