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 ?
$(a)$ the average speed of the taxi,
$(b)$ the magnitude of average velocity ? Are the two equal ?
$(a)$ Total distance travelled $=23 \,km$
Total time taken $=28\, min =\frac{28}{60} \,h$
$\therefore$ Average speed of the taxi $=\frac{\text { Total distance travelled }}{\text { Total time taken }}=\frac{23}{\left(\frac{28}{60}\right)}=49.29 \,km / h$
$(b)$ Distance between the hotel and the station $=10\, km =$ Displacement of the car
$\therefore$ Average velocity $=\frac{10}{\frac{28}{60}}=21.43\, km / h$
Therefore, the two physical quantities (average speed and average velocity) are not equal.
Total time taken $=28\, min =\frac{28}{60} \,h$
$\therefore$ Average speed of the taxi $=\frac{\text { Total distance travelled }}{\text { Total time taken }}=\frac{23}{\left(\frac{28}{60}\right)}=49.29 \,km / h$
$(b)$ Distance between the hotel and the station $=10\, km =$ Displacement of the car
$\therefore$ 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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