KINEMATICS O2

Solution 2

Problem 2: Sam’s Walk to School

Logic: We can solve this by comparing the time difference relative to the school bell. The “extra” time spent is solely due to the detour (walking from the turning point back to home and then returning to the turning point).

Visualizing the Path

Home School Point P (Turn back) x Return Trip

Mathematical Approach

Let the total normal time to walk from Home to School be $T = 20$ min.

Scenario 1: Normal Walk
If Sam walks continuously, he arrives 8 minutes early relative to the bell.
$$ \text{Arrival Time}_1 = \text{Bell Time} – 8 \text{ min} $$ Since the walk takes 20 minutes, let’s say he starts at $t=0$. Then $\text{Arrival Time}_1 = 20$.
This implies $\text{Bell Time} = 28$ min (relative to start).

Scenario 2: Forgotten Notebook
Sam walks to point P, walks back to Home, and then walks the full distance to School. He arrives 10 minutes late.
$$ \text{Arrival Time}_2 = \text{Bell Time} + 10 \text{ min} $$ $$ \text{Arrival Time}_2 = 28 + 10 = 38 \text{ min} $$

Analyzing the Time Difference
The total time spent in Scenario 2 is 38 minutes.
The “useful” part of the walk (Home to School eventually) takes 20 minutes (the normal time $T$).
The “wasted” time is the time spent walking from Home to P and back to Home.
$$ \text{Wasted Time} = \text{Total Time} – \text{Normal Time} $$ $$ \text{Wasted Time} = 38 \text{ min} – 20 \text{ min} = 18 \text{ min} $$

This 18 minutes represents the round trip $H \to P \to H$. Therefore, the one-way trip from Home to P takes half of this time:

$$ t_{H \to P} = \frac{18}{2} = 9 \text{ min} $$

Calculating the Fraction

Since the speed is constant, the fraction of the distance covered is equal to the fraction of the total time covered.

$$ \text{Fraction} = \frac{\text{Time to P}}{\text{Total Time to School}} $$ $$ \text{Fraction} = \frac{9 \text{ min}}{20 \text{ min}} = \frac{9}{20} $$

Correct Option: (b)