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Motion in a Plane

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Question : 11 of 32
Marks: +1, -0
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:  
Magnitude of the displacement = 10 km
Distance covered = 23 km
Time taken = 28 min
(a) Average speed of the taxi is given by
=Total distance coveredTotal time taken = \frac{\text{Total distance covered}}{\text{Total time taken }} =23(715)=23×157= \frac{23}{\left( \frac{7}{15} \right)} = \frac{23 \times 15}{7} =49.3km h−1= 49.3 \text{km h}^{-1}
(b) Magnitude of average velocity
=Magnitude of the displacementTotal time taken = \frac{\text{Magnitude of the displacement}}{\text{Total time taken }} =10(715)=1507= \frac{10}{\left( \frac{7}{15} \right)} = \frac{150}{7} =21.43km h−1= 21.43 \text{km h}^{-1}
Clearly, the average speed and the magnitude of average velocity are not equal. They are equal only for a straight path.
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