Given
- Straight-line distance from station to hotel (displacement): \( d = 10 \,\text{km} \)
- Actual path length (distance travelled): \( s = 23 \,\text{km} \)
- Time taken: \( t = 28 \,\text{min} = \dfrac{28}{60} \,\text{h} = \dfrac{7}{15} \,\text{h} \)
(a) Average speed of the taxi
Average speed is total distance travelled divided by total time:
\( v_{\text{avg, speed}} = \dfrac{s}{t} = \dfrac{23}{7/15} = 23 \times \dfrac{15}{7} = \dfrac{345}{7} \approx 49.3 \,\text{km h}^{-1} \)
(a) Average speed of the taxi \( \approx 49.3 \,\text{km h}^{-1} \).
(b) Magnitude of average velocity
Average velocity (magnitude) is displacement divided by total time:
\( v_{\text{avg, vel}} = \dfrac{d}{t} = \dfrac{10}{7/15} = 10 \times \dfrac{15}{7} = \dfrac{150}{7} \approx 21.4 \,\text{km h}^{-1} \)
(b) Magnitude of average velocity \( \approx 21.4 \,\text{km h}^{-1} \) towards the hotel.
Comparison
- The average speed (≈ 49.3 km/h) is greater than the magnitude of the average velocity (≈ 21.4 km/h).
- They are not equal because average speed depends on the total path length (23 km), whereas average velocity depends only on the net displacement (10 km).