#### Question

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 circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

#### Solution 1

(a) Total distance travelled = 23 km

Total time taken = 28 min = `28/60 h`

∴Average speed of the taxi = `"Total distance travelled"/"Total time taken"`

= `23/(28/60) = 49.29 "km/h"`

(b) Distance between the hotel and the station = 10 km = Displacement of the car

∴Average velocity = `10/(28/60) = 21.43 "km/h"`

Therefore, the two physical quantities (averge speed and average velocity) are not equal.

#### Solution 2

Here, actual path length travelled, s = 23 km; Displacement = 10 km;

Time taken, t = 28 min = `28/60h`

(a) Average speed of taxi = `"actual path length"/"time taken" = 23/(28/60) km/h = 49.3 "km/h"`

(b)Magnitude of average velocity = `"displacement"/"time taken" = 10/(28/60) "km/h" = 21.4 "km/h"`

The average speed is not equal to the magnitude of average velocity. The two are equal for the motion of taxi along a straight path in one direction.