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 cab man take 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?

Answer 1

Hii dear,

# Answer- Average speed (49.28 km/h) is greater than magnitude of average velocity (21.43 km/h).


## Explaination-
# Given-
Path length d = 23 km
Displacement s = 10 km
Time taken t = 28 min = 28/60 h

# Solution-
Average velocity = Displacement/time
v = s/t
v = 10×60/28
v = 21.43 km/h

Average speed = Path length/time
v' = d/t
v' = 23×60/28
v' = 49.28 km/h

Here, v' > v
Hence, average speed (49.28 km/h) is greater than magnitude of average velocity (21.43 km/h).

Answer 2

Given :

magnitude of displacement =10 km

Total path length=23km

Time taken=28 min= 28/ 60 h= 7/15 h

Average speed = Total distance/ total time

=23/7/15 = 49.3 km/h

Magnitude of average velocity :

= Magnitude of displacement / total  time

= 10km/7/15 h

=21.43 km/

Magnitude  of displacement =10 km

Total path length=23km

Time taken=28 min= 28/ 60 h= 7/15 h

Average speed = Total distance/ total time

=23/7/15 = 49.3 km/h

Magnitude of average velocity:

= Magnitude of displacment / tota time

= 10km/7/15 h=21.43 km/h

∴ By comparing the values of average  speed and magnitude of average displacement we can say that they are not equal.