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 and Explanation
a)Total distance travelled = 23 km
Total time taken = 28 min = \(\frac{28}{60}\) h
Average speed of the taxi = \(\frac{\text{Total distance travelled} }{\text{Total time taken}} \)
= \(\frac{23}{ (\frac{29}{60})}\) = 49.29 Km/h
b) Distance between the hotel and the station = 10 km = Displacement of the car
∴ Average velocity = \(\frac{10}{\frac{28}{60}} \)= 21.43 km/h
Therefore, the two physical quantities (average speed and average velocity) are not equal.
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Concepts Used:
Speed and Velocity
The rate at which an object covers a certain distance is commonly known as speed.
The rate at which an object changes position in a certain direction is called velocity.
Difference Between Speed and Velocity:
Read More: Difference Between Speed and Velocity