A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitoius path 23 km long and reaches the hotel in 28 min. What are the average speed of the taxi and the magnitude of average velocity respectively (in km/h)?

Given, shortest distance between the station and the hotel = 10 km

$\therefore Displacement of the taxi=10 km$

Distance travelled by the taxi = 23 km

Time taken by the taxi = 28 m

$=\frac{28}{60}=\frac{7}{15}h$

The average speed of the taxi $=\frac{Total dis\tan ce travelled}{Total time taken}$

$=\frac{23}{\left(7/15\right)}$

$=\frac{345}{7}km/h$

= 49.3 km/h

The magnitude of average velocity 

$=\frac{Magnitude of the total displacemen}{Total time taken}$

$=\frac{10}{\left(7/15\right)}$

$=\frac{150}{7}km/h$

= 21.43 km/h

No, the average speed of the taxi is not equal to the magnitude of the average velocity of the taxi.