A car travels the first half of a straight route at a constant 50 km/h and the second half at a constant 60 km/h. Find the average speed for the whole route.

Step-by-step solution

1. 1

   Start from the definition, not from intuition. Average speed is always

   $v_{\text{avg}}=\frac{\text{total distance}}{\text{total time}}$

   It is not the average of the two speeds. Averaging $50$ and $60$ to get $55$ silently assumes the car spends equal time at each speed, but here it covers equal distance — and it spends longer on the slower half, which drags the average down.

2. 2

   Name the distance so it can cancel. Let each half be $d$, so the total distance is $2d$. The time for each half is distance over speed:

   $t_1=\frac{d}{50},\qquad t_2=\frac{d}{60}$

3. 3

   Add the times over a common denominator.

   $t_1+t_2=\frac{d}{50}+\frac{d}{60}=d\left(\frac{6}{300}+\frac{5}{300}\right)=\frac{11d}{300}$

4. 4

   Divide total distance by total time.

   $v_{\text{avg}}=\frac{2d}{\dfrac{11d}{300}}=2d\cdot\frac{300}{11d}=\frac{600}{11}\approx54.5\ \text{km/h}$

   The unknown $d$ cancels, which is why the answer does not depend on how long the road is.

5. 5

   Recognise the general formula. Repeating the algebra with symbols gives the harmonic mean of the two speeds:

   $v_{\text{avg}}=\frac{2v_1v_2}{v_1+v_2}=\frac{2(50)(60)}{50+60}=\frac{6000}{110}=54.\overline{54}\ \text{km/h}$

   The harmonic mean is always less than the arithmetic mean unless the two speeds are equal, so a result slightly under $55$ is the built-in sanity check.

6. 6

   Test it with a concrete number. Take each half to be $300\ \text{km}$: the first half takes $6$ h and the second $5$ h, so $600\ \text{km}$ in $11$ h, giving $600/11\approx54.5\ \text{km/h}$ — exactly the formula's answer.