question_answer

A boy walks to his school at a distance of 6 km with constant speed of \[\text{2}\text{.5 km }{{\text{h}}^{\text{-1}}}\] and walks back with a constant speed of \[\text{4 km}{{\text{h}}^{\text{-1}}}\]What is average speed for the round trip in\[\text{km }{{\text{h}}^{\text{-1}}}\]?

Answer:

                Total time in going and coming back \[T=\frac{6}{2.5}+\frac{6}{4}\]; total distance travelled,  \[s=\text{6}+\text{6}=\text{12km}\]; Average speed \[=\frac{S}{T}=\frac{12}{\left( 6/2.5 \right)+\left( 6/4 \right)}=\frac{40}{13}km\,{{h}^{-1}}\]