Answer:
We know,
\[=\frac{total\,\,dis\tan ce\,\,travelled\,by\,\,the\,\,car}{total\,time\,\,taken}\]
\[=\frac{S+S+S}{\frac{S}{60}+\frac{S}{20}+\frac{S}{10}}=\frac{S[3]}{S\left[ \frac{1}{60}+\frac{1}{20}+\frac{1}{10} \right]}\]
\[=\frac{3}{\frac{1+3+6}{60}}=\frac{3\times 60}{10}=18km/hr\]
\[=18\times \frac{5}{18}m/s\,s=5m{{s}^{-1}}\]
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