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# Question :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, and (b) the magnitude of average velocity? Are the two equal?

Last updated date: 16th Jun 2024
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Hint: To find the average speed and average velocity, you'll need to calculate the total distance and total displacement, as well as the total time taken. Remember that average speed is the total distance travelled divided by the total time taken, while average velocity is the total displacement divided by the total time taken.

Step-by-Step Solutions:

(a) Average Speed:

To find the average speed, you'll need to calculate the total distance travelled and then divide it by the total time taken.

Total distance travelled = 23 km

Total time taken = 28 minutes = $\dfrac{28}{60}$ hours (since 1 hour = 60 minutes)

Average Speed $V_{avg} = \dfrac{\text{Total distance}}{\text{Total time}}$

$V_{avg} = \dfrac{23}{\dfrac{28}{60}}$ km/hours

$V_{avg} = \dfrac{23}{\dfrac{7}{15}}$ km/h

$V_{avg} = 23 \times \dfrac{15}{7}$ km/h

$V_{avg}$ ≈ 49.2857 km/h (rounded to 4 decimal places)

(b) Average Velocity:

To find the average velocity, you need to calculate the total displacement and then divide it by the total time taken. Displacement is the change in position from the starting point to the ending point.

Total displacement = Initial position - Final position

Total displacement = 10 km

Average Velocity $v_{avg} = \dfrac{\text{Total displacement}}{\text{Total time}}$

$v_{avg} = \dfrac{10}{\dfrac{28}{60}}$ km/hours

$v_{avg} = \dfrac{10}{\dfrac{7}{15}}$ km/hours

$v_{avg} = 10 \times \dfrac{15}{7}$ km/hr

$v_{avg}$ ≈ 21.4286 km/h (rounded to 4 decimal places)

Note:

(a) The average speed of the taxi is approximately 49.2857 km/h, while

(b) The magnitude of the average velocity is approximately 21.4286 km/h, and it is negative because the displacement is in the opposite direction of the initial position to the final position.

It's important to note that average speed and average velocity are not equal in this case. Average speed is the total distance divided by the total time, whereas average velocity is the total displacement divided by the total time. In this scenario, the taxi took a longer path, resulting in a higher speed than the magnitude of its velocity.