# Calculating the average speed using only velocities

For the following question:

Pedro travels by bus to school at an average speed of $40$ km/hr. He is driven home by the same route by a friend's car at an average speed of $50$ km/hr. Which of the following is greatest:

(a) Average speed of both legs of the journey.

(b) $45$.

According to the book, the answer is (b). How did they calculate average speed here? I know that the formula for average speed is $A_v=\frac{Total~ Distance~ Covered}{Total~ Time ~Taken }$. Here we only know the velocity.

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Let $d$ be the distance from Pedro's house to school. The time it took him to travel to school was $d/50$, and the time it took him to return home was $d/40$. Thus the average speed is $$\frac{2d}{d/50+d/40}=\frac{400d}{9d}=44.\overline{4}<45$$

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You can solve this one without doing any real calculations. It takes longer to do the part of the journey going at $40$ km/h than it does to do the part of the journey going at $50$ km/h, so the majority of the time is spent going at $40$ km/h. Therefore, the average speed is going to be less than $45$ km/h (it would be $45$ km/h if the same time was spent going at both speeds).

Alternatively, you could let the distance for one leg of the journey = $s$. Then, the time taken to do the part of the journey at $40$ km/h $= \frac{s}{40}$, and the time taken to do the part of the journey at $50$ km/h $= \frac{s}{50}$. So the total time taken $= \frac{s}{40} + \frac{s}{50} = \frac{9s}{200}$. As you said, average speed $= \frac{total ~ distance}{total ~ time} = \frac{2s}{9s/200} = \frac{400}{9} = 44.4\cdots$, which is clearly less than $45$.

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Assuming they travel the same path there and back, we know that the average speed of the first journey is $40 kph = A_{v_1} = \frac{x}{t_1}$, and that the average speed of the second journey is $50 kph = A_{v_2} = \frac{x}{t_2}$.

Notice that the distance traveled, $x$, is the same in both equations. We can compute it as $x = 50 t_2 = 40 t_1$. Alternatively, $t_1 = x/40$, etc.

Therefore, the total average speed over both trips is $A_v = \frac{2x}{t_1+t_2} = \frac{2x}{x/40 + x/50}$.

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