Question

Question 4.11 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?

Chapter Motion In A Plane Page 86

Answer 1

Station to hotel located 10km away on the straight road .e.g displacement = 10km

Average velocity = total displacement/total time taken
= 10km/28min
= 10/{28/60} km/h
= 21.43 km/h

A dishonest cabman takes his along a circuitous path 23km along.
e.g distance covered by taxi = 23km

Average speed = total distance/ total time taken
= 23km/{28/60}h
= 49.3 km/h

Answer 2

Given that,

Difference between hotel and station = 10 km

Time = 28 min

Distance = 23 km

(a). We need to calculate the average speed

Using formula of average sped

$v_{avg}=\dfrac{D}{T}$

Where, D = distance

T = time

Put the value into the formula

$v_{avg}=\dfrac{23\times60}{28}$

$v_{avg}=49.28\ km/hr$

The average speed is 21.42 km/hr.

(b). We need to calculate the average velocity

Using formula of average sped

$v_{avg}=\dfrac{D}{T}$

Where, D = distance

T = time

Put the value into the formula

$v_{avg}=\dfrac{10\times60}{28}$

$v_{avg}=21.42\ km/hr$

The average velocity is 21.42 km/hr.

Hence, The two physical quantities (average speed and average velocity) are not equal.