Question

4.11 A passenger arriving in a new town wishes to go from the station to a hotel
10 km away on a straight road from the station. A dishonest cabman takes him al
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 ?
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Answer 1

Answer:

(a) 49.3 km/hr,  (b) 21.4 km/hr, No

Explanation:

(28 min = 28/60 = 7/15 hours)

(a) Average Speed = Distance / Time

⇒ 23 km / (7/15) hours = 49.3 km/hr              

(b) Average Velocity = Displacement / Time

= 10 km/ (7/15) = 21.4 km/hr                                                                                     (As displacement is always taken as the distance of the shortest path)

Hence, the average speed of the taxi is (approx.) 49.3 km/hr and the average velocity of the taxi is (approx.) is 21.4 km/hr.

No, the two are not equal.