Question

Abdul, while driving to school, computes the average speed for his trip to be \(20\) \(km\, h^{–1}\). On his return trip along the same route, there is less traffic and the average speed is \(30\) \(km \,h^{–1}\). What is the average speed for Abdul’s trip?

Answer 1

The distance Abdul commutes while driving from Home to School = \(S\)
Let us assume time taken by Abdul to commutes this distance = \(t_1\)
Distance Abdul commutes while driving from School to Home = \(S\)
Let us assume time taken by Abdul to commutes this distance = \(t_2\)
Average speed from home to school \(v_{1av}\) = \(20\) \(km\, h^{-1}\)
Average speed from school to home \(v_{2av}\) = \(30\) \(km\, h^{-1}\)
Also we know Time taken form Home to School \(t_1\) = \(\frac{S}{ v_{1av}}\)
Similarly Time taken form School to Home \(t_2\) = \(\frac{S}{v_{2av}}\)
Total distance from home to school and backward = \(2\) \(S\)
Total time taken from home to school and backward \((T)\) = \(\frac{S}{20}+ \frac{S}{30}\)
Therefore, Average speed \((V_{av})\) for covering total distance \((2S)\) 
= \(\frac{Total \,Dostance}{Total \,Time}\)

= \(\frac{2S }{ \bigg(\frac{S}{20} +\frac{S}{30}\bigg)}\)

= \(\frac{2S }{ \bigg[\frac{(30S+20S)}{600}\bigg]}\)

= \(\frac{1200\,S }{ 50\,S}\)

= \(24\; km\,h^{-1}\)