Question

# Length of a minute hand of a clock is $$4.5\ cm$$. Find average velocity of time of the minute hand between $$6\ AM$$. to $$6.30\ A.M$$ and $$6\ A.M.$$ to $$6.30\ P.M$$.

Solution

## Given that, Time $$t = 30\ min = 1800\ sec$$ Now, the length of the minute hand is almost as long as the radius of the clock The displacement is   $$D=2\times 4.5$$  $$D=9\,cm$$  $$D=0.09\,m$$ Now, the velocity is   $$v=\dfrac{D}{t}$$  $$v=\dfrac{0.09}{1800}$$  $$v=5\times {{10}^{-5}}\,m/s$$ Hence, the average velocity is $$5\times {{10}^{-5}}\,m/s$$ Displacement during $$6:00 \ AM$$ to $$6:30\ PM$$ will be:$$D=4.5\times 2=9cm$$$$D=0.09 \ m$$Time taken:$$t=12.5$$ Hours.$$t=12.5\times 3600$$ secTherefore,Velocity will be given by:$$v=\dfrac{0.09}{12.5\times 3600}=2\times 10^{-6}\ m/s$$   Physics

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