Question

The travel times of three vehicles on a 2 km stretch of road are 4, 5, and 8 minutes. Assuming the speed of each vehicle to be constant in this stretch, the Space Mean Speed (in km/h) of the vehicles is (rounded off to two decimal places).

Hint:

Space Mean Speed is the harmonic mean of speeds, equal to total distance over total time.

Answer 1

Correct Answer: 21.18

Space Mean Speed is found from the harmonic mean of the individual vehicle speeds, since it reflects the average speed of vehicles actually present over a stretch of road at one time, not the average of speeds clocked at a single point. This route finds each vehicle's speed first, then applies the harmonic mean formula term by term instead of going through total distance and total time.

Convert each travel time on the 2 km stretch to a speed in km/h. A time of $t$ minutes over 2 km gives a speed of $v = 2/(t/60) = 120/t$ km/h.

\[ v_1 = \frac{120}{4} = 30 \ \text{km/h}, \quad v_2 = \frac{120}{5} = 24 \ \text{km/h}, \quad v_3 = \frac{120}{8} = 15 \ \text{km/h} \]

The Space Mean Speed formula for $n$ vehicles is the harmonic mean of their speeds:

\[ SMS = \frac{n}{\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}} \]

Find each reciprocal:

\[ \frac{1}{v_1} = \frac{1}{30} = 0.03333, \quad \frac{1}{v_2} = \frac{1}{24} = 0.04167, \quad \frac{1}{v_3} = \frac{1}{15} = 0.06667 \]

Add them up:

\[ \sum \frac{1}{v_i} = 0.03333 + 0.04167 + 0.06667 = 0.14167 \]

So the Space Mean Speed is

\[ SMS = \frac{3}{0.14167} = 21.18 \ \text{km/h} \]

The slowest vehicle (15 km/h) pulls the harmonic mean down more than the fastest vehicle (30 km/h) pulls it up, which is why the Space Mean Speed of 21.18 km/h sits closer to the low end of the three speeds rather than at their plain average of 23 km/h.

\[ \boxed{21.18 \ \text{km/h}} \]