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Q2.10(a)(ii):
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 (a) magnitude of average velocity over the interval of time (ii) 0 to 50 min ?

Solution :

Step 1: Given Variables and Initial Setup

To determine the magnitude of the average velocity, we define the following parameters based on the motion profile provided:

  • Distance to market ($d$): $2.5 \text{ km}$
  • Speed to market ($v_1$): $5 \text{ km h}^{-1}$
  • Speed back home ($v_2$): $7.5 \text{ km h}^{-1}$
  • Time interval ($\Delta t$): $0$ to $50 \text{ min} = \frac{50}{60} \text{ h} = \frac{5}{6} \text{ h}$

Step 2: Calculating Temporal Milestones

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

$t_1 = \frac{d}{v_1} = \frac{2.5 \text{ km}}{5 \text{ km h}^{-1}} = 0.5 \text{ h} = 30 \text{ min}$

Since the total interval is $50 \text{ min}$, the man spends the remaining time moving back toward home:

$t_{\text{return}} = 50 \text{ min} - 30 \text{ min} = 20 \text{ min} = \frac{20}{60} \text{ h} = \frac{1}{3} \text{ h}$

Step 3: Determining Net Displacement

Displacement ($\Delta x$) is a vector quantity defined as the change in position. [Per the definition of displacement in kinematics: $\Delta x = x_f - x_i$]

Distance covered during return journey ($d_r$):

$d_r = v_2 \times t_{\text{return}} = 7.5 \text{ km h}^{-1} \times \frac{1}{3} \text{ h} = 2.5 \text{ km}$

The man reaches the market at $30 \text{ min}$. By $50 \text{ min}$, he has traveled $2.5 \text{ km}$ back toward home. Since the total distance home is also $2.5 \text{ km}$, he has returned exactly to the starting point.

$\Delta x = 0 \text{ km}$

Step 4: Calculation of Average Velocity

Average velocity ($v_{avg}$) is defined as the total displacement divided by the total time interval. [Per the kinematic formula: $v_{avg} = \frac{\Delta x}{\Delta t}$]

$v_{avg} = \frac{0 \text{ km}}{50/60 \text{ h}} = 0 \text{ km h}^{-1}$

Visual Representation of Displacement

Home (x=0) Market (x=2.5) Net Displacement = 0

Final Solution: The magnitude of average velocity over the interval 0 to 50 minutes is $0 \text{ km h}^{-1}$.


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