$A$ 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 $= \frac{\text{Total distance travelled}}{\text{Total time taken}} = \frac{23}{(28/60)} \approx 49.29 \;km/h$.
$(b)$ Displacement $= 10 \;km$ (straight line distance between station and hotel).
$\therefore$ Average velocity $= \frac{\text{Displacement}}{\text{Total time taken}} = \frac{10}{(28/60)} \approx 21.43 \;km/h$.
Since the total distance travelled is greater than the magnitude of displacement,the average speed and average velocity are not equal.
Total time taken $= 28 \;min = \frac{28}{60} \;h$.
$\therefore$ Average speed of the taxi $= \frac{\text{Total distance travelled}}{\text{Total time taken}} = \frac{23}{(28/60)} \approx 49.29 \;km/h$.
$(b)$ Displacement $= 10 \;km$ (straight line distance between station and hotel).
$\therefore$ Average velocity $= \frac{\text{Displacement}}{\text{Total time taken}} = \frac{10}{(28/60)} \approx 21.43 \;km/h$.
Since the total distance travelled is greater than the magnitude of displacement,the average speed and average velocity are not equal.
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