Question

# A cyclist travels a distance of$1km$ in the first hour,$0.5km$ in the second hour and$0.3km$ in the third hour. What is the average speed in $\left(1\right)km/hr\phantom{\rule{0ex}{0ex}}\left(2\right)m/sec?$ ?

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Solution

## Step-1 Finding the average speed in $km/hr$Distance in one hour $=1km$Distance in second hour$=0.5km$Distance in third hour $=0.3km$We know that,$Averagespeed=\frac{Totaldis\mathrm{tan}ce}{Totaltime}$ Average speed in$km/hr$$\begin{array}{rcl}\nu & =& \frac{D}{T}\\ & =& \frac{\left(1+0.5+0.3\right)}{3}km/hr\\ & =& 0.6km/hr\end{array}$Step-2 Finding the average speed in $m/sec$Average speed in $m/sec$$\begin{array}{rcl}\nu & =& \frac{D}{T}\\ & =& \frac{\left(1+0.5+0.3\right)}{3}km/hr\\ & =& 0.6km/hr\\ & =& 0.6×\frac{5}{18}m/sec\left(1km/hr=\frac{5}{18}m/sec\right)\\ & =& 0.167m/sec\end{array}$Therefore, Average speed is$0.6km/hr$ and $0.167m/sec$.

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