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Q3.14(b)(iii):
A man walks on a straight road from his home to a market $2.5 km$ away with a speed of $5 km h^{-1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5 km h^{-1}$. What is the (b) average speed of the man over the interval of time (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

Solution :

Step 1: Establishing the Kinematic Parameters

To determine the average speed over the interval $t = 0$ to $t = 40$ minutes, we must first analyze the kinematics of the two distinct legs of the journey.

  • Distance to market ($d$): $2.5 \text{ km}$
  • Speed to market ($v_1$): $5 \text{ km h}^{-1}$
  • Speed back home ($v_2$): $7.5 \text{ km h}^{-1}$

Step 2: Calculating Temporal Constants

First, calculate the time taken to reach the market ($t_1$):

$t_1 = \frac{d}{v_1} = \frac{2.5 \text{ km}}{5 \text{ km h}^{-1}} = 0.5 \text{ hours}$

[Per the definition of average speed $v = \frac{d}{t}$]

Convert $t_1$ into minutes: $0.5 \times 60 = 30 \text{ minutes}$.

Step 3: Analyzing the Return Leg

The total duration requested is $T = 40 \text{ minutes}$. Since $t_1 = 30 \text{ minutes}$, the man spends the remaining time, $\Delta t = 40 - 30 = 10 \text{ minutes}$, walking back home.

Convert $\Delta t$ to hours: $\frac{10 \text{ min}}{60 \text{ min/h}} = \frac{1}{6} \text{ hours}$.

Calculate the distance covered during the return leg ($d_2$) within the remaining time:

$d_2 = v_2 \times \Delta t = 7.5 \text{ km h}^{-1} \times \frac{1}{6} \text{ h} = 1.25 \text{ km}$

Step 4: Calculating Total Path Length and Average Speed

The total path length ($D_{total}$) covered in 40 minutes is the sum of the distance to the market and the distance covered on the return journey:

$D_{total} = d_1 + d_2 = 2.5 \text{ km} + 1.25 \text{ km} = 3.75 \text{ km}$

[Per the principle of scalar addition of path lengths]

The total time interval is $T = \frac{40}{60} \text{ h} = \frac{2}{3} \text{ h}$.

Calculate Average Speed ($v_{avg}$):

$v_{avg} = \frac{D_{total}}{T} = \frac{3.75 \text{ km}}{2/3 \text{ h}} = 3.75 \times 1.5 = 5.625 \text{ km h}^{-1}$

Visual Representation of Motion

Home (0 km) Market (2.5 km) Leg 1: 30 min (v=5 km/h) Leg 2: 10 min (v=7.5 km/h)

Final Solution: The average speed of the man over the interval of 0 to 40 minutes is 5.625 km h⁻¹.


More Questions from Class 11 Physics Motion in a Straight Line EXERCISES


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