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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 ?
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An0nym0us
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Qalaxia Master Bot
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I found an answer from www.khanacademy.org

Distance and displacement introduction (video) | Khan Academy


An introduction to the difference between distance traveled and displacement. Created by Sal Khan. ... Average velocity and average speed. Sort by: Top Voted  ...


For more information, see Distance and displacement introduction (video) | Khan Academy

Krishna
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The following formulas and definitions are essential to address the questions.

Displacement is defined as the shortest distance between a particle's initial and final positions.

Average speed V_a = \frac{\text{ total distance }}{\text{time }}

Average velocity v_a = \frac{\text{ displacement }}{\text{time }}


Given that

Displacement = 10 km

Total distance = 23 km

Total time taken by the taxi to reach hotel, t = 28 min

a)

Average speed V_a = \frac{\text{ total distance }}{\text{time }}

V_a = \frac{23 km}{28 min}

V_a = 0.82 km/min

V_a = 0.82 * 60 = 49.29 km/h

b)

Average velocity v_a = \frac{\text{ displacement }}{\text{time }}

v_a = \frac{10}{28}

v_a = 0.357 km/min

v_a = 0.357 *60 = 21.43 Km/h

Hence, The value of the average speed is not equal to the average velocity.