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A car drove from town $A$ to town $B$ at an average of $60 \text{ mph}$. On the return trip from town $B$ to town $A$ the car took exactly the same route. It was foggy and the car only averaged $40 \text{ mph}$ on the return trip. What is the average speed over the entire trip?

Ross Millikan
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melrose
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1 Answers1

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Intuitively, it may seem like the answer is $50 \text{ mph}$, but it is not! This is because more time is spent on the return trip.

I'll generalize it for you. Let the average speed of the forward trip and reverse trip be $v_1$ and $v_2 \text{ mph}$ respectively. Let the distance between the two towns be $D \text{ miles}$.

Hence, we have: $$\text{Total average speed}=\frac{\text{Total Distance}}{\text{Total Time}}=\frac{D+D}{t_1+t_2}=\frac{2D}{\frac{D}{v_1}+\frac{D}{v_2}}=\frac{2v_1 v_2}{v_1+v_2}$$ You can now apply this to obtain the average speed for your specific case which is when $v_1=60$ and $v_2=40$.

projectilemotion
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    In case you were wondering about the last step: $$\frac{D}{v_1}+\frac{D}{v_2}=\frac{Dv_2}{v_1 v_2}+\frac{Dv_1}{v_1 v_2}=\frac{D(v_1+v_2)}{v_1v_2} \implies \frac{2D}{\frac{D}{v_1}+\frac{D}{v_2}}=\frac{2D}{\frac{D(v_1+v_2)}{v_1v_2}}=\frac{2D\cdot v_1 v_2}{D(v_1+v_2)}=\frac{2v_1 v_2}{v_1+v_2}$$ – projectilemotion Apr 25 '17 at 19:20