Q.A passenger arriving in a new town wishes to go from the station to a hotel located $10\ \text{km}$ away on a straight road from the station. A dishonest cabman takes him along a circuitous path $23\ \text{km}$ long and reaches the hotel in $28\ \text{min}$. What is

Concept: Average speed depends on total distance traveled; average velocity depends on net displacement.

The taxi covers a distance of $23\ \text{km}$ along the actual (circuitous) path, but the displacement—the straight-line separation between station and hotel—is only $10\ \text{km}$. Time taken is $28\ \text{min} = \frac{28}{60}\ \text{h} = \frac{7}{15}\ \text{h}$.

1. Average speed = $\dfrac{\text{total distance}}{\text{time}} = \dfrac{23}{\frac{7}{15}} = 23 \times \frac{15}{7} = \frac{345}{7} \approx 49.3\ \text{km/h}$.
2. Magnitude of average velocity = $\dfrac{\text{displacement}}{\text{time}} = \dfrac{10}{\frac{7}{15}} = 10 \times \frac{15}{7} = \frac{150}{7} \approx 21.4\ \text{km/h}$. …

Why average speed and average velocity differ

Speed is a scalar—it cares only about how much ground you cover, regardless of direction. Velocity is a vector—it cares about where you end up relative to where you started. When a path is not straight, the distance travelled exceeds the straight-line displacement, so average speed will be larger than the magnitude of average velocity.

In this problem, the cabman takes a winding $23\,\text{km}$ route to cover a straight-line distance of only $10\,\text{km}$. The passenger pays for the longer journey, but physically he's displaced by much less.

Step-by-step solution

1. Convert the time to consistent units

The journey takes $28\,\text{min}$. Since speeds are typically expressed in $\text{km/h}$, convert:

2. Calculate the average speed

Average speed is defined as the total distance travelled divided by the total time:

3. Calculate the magnitude of average velocity

Average velocity is defined as the displacement (straight-line distance from start to finish) divided by time. The hotel is $10\,\text{km}$ from the station along a straight road, so the displacement is $10\,\text{km}$:

4. Compare the two …

Concept: Work Entirely in SI Units First, Convert to km/h Only at the End

Method: Metres-and-Seconds Route (No Fractional Hours), Using the $\times 3.6$ Shortcut Last

Both existing solutions convert the $28\ \text{min}$ trip time into a fraction of an hour ($7/15\ \text{h}$) and keep every calculation in kilometres and hours throughout. This method instead converts everything to base SI units (metres, seconds) first, computes both speeds cleanly in m/s, and only converts to km/h as the very last step — avoiding fractional-hour arithmetic entirely.

Step 1 — Convert every given quantity to SI units

Step 2 — Average speed, directly in m/s

Step 3 — Magnitude of average velocity, directly in m/s

Step 4 — Convert both to km/h in one shot, at the very end

The conversion factor from m/s to km/h is exactly $3.6$ (i.e. $\times\frac{18}{5}$):

matching $\dfrac{345}{7}\approx49.3\ \text{km/h}$ and $\dfrac{150}{7}\approx21.4\ \text{km/h}$ exactly.

Step 5 — Compare …