Solution to Question 19
This alternative method simplifies the problem by considering the physics of the round trip relative to the ground. Regardless of where the boys meet, they always start at a fixed point (pulley), travel some ground distance $x$ to a meeting point, and return the same ground distance $x$ to the start.
1. Ground Velocities:
- Going along with the belt: $v_1 = 3 + 1 = 4$ m/s
- Going opposite to the belt: $v_2 = 3 – 1 = 2$ m/s
2. Average Speed Calculation:
Since the distance traveled “out” ($x$) equals the distance traveled “back” ($x$) for every single loop, we can calculate the average speed $\langle V \rangle$ for the entire motion:
$$ \langle V \rangle = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{x + x}{t_1 + t_2} = \frac{2x}{\frac{x}{v_1} + \frac{x}{v_2}} $$
$$ \langle V \rangle = \frac{2x}{\frac{x}{4} + \frac{x}{2}} = \frac{2x}{\frac{3x}{4}} = \frac{8}{3} \text{ m/s} $$
3. Total Distance:
$$ S = \langle V \rangle \times T = \frac{8}{3} \times 300 = \mathbf{800 \text{ m}} $$
Detailed Analysis
Two boys walk back and forth on a 100m conveyor belt. They turn around instantaneously whenever they meet.
- Belt Speed: $v_b = 1$ m/s (moves Left $\to$ Right)
- Walking Speed: $v_m = 3$ m/s (relative to belt)
- Total Time: $T = 300$ s
We calculate velocities relative to the ground ($v_g = v_{belt} + v_{walk}$):
| Boy | Direction | Calculation | Ground Velocity |
|---|---|---|---|
| Boy A (Starts Left) | Moving Right ($\to$) | $3 + 1$ | $+4$ m/s |
| Moving Left ($\leftarrow$) | $-3 + 1$ | $-2$ m/s | |
| Boy B (Starts Right) | Moving Left ($\leftarrow$) | $-3 + 1$ | $-2$ m/s |
| Moving Right ($\to$) | $3 + 1$ | $+4$ m/s |
Because their ground speeds are asymmetric ($4$ vs $2$), they return to their starting points at different times. This causes the meeting point to shift. However, the system resets completely every 75 seconds.
Phase 1: First Meeting & Asymmetric Return ($0 – 50$s)
-
The Meeting ($t=0 \to 16.67$s):
They approach at relative speed $6$ m/s. Meeting time $t_1 = 100/6 = 50/3$ s.
Location: A travels $4 \times (50/3) = 200/3$ m (approx 66.7m). -
Boy B Returns Fast ($t=16.67 \to 25$s):
B returns to the Right end ($100$m). Distance $100/3$ m at speed $4$ m/s.
Time taken: $25/3$ s. Total time: $\mathbf{25 \text{ s}}$.
Boy B turns around immediately and starts walking Left again. -
Boy A Returns Slow ($t=16.67 \to 50$s):
A returns to the Left end ($0$m). Distance $200/3$ m at speed $2$ m/s.
Time taken: $100/3$ s. Total time: $\mathbf{50 \text{ s}}$.
Phase 2: Second Meeting & Resynchronization ($50 – 75$s)
At $t=50$s, Boy A is at $0$. Boy B has been walking Left for 25 seconds (since $t=25$), so he is at position $100 – (2 \times 25) = 50$m.
-
The Second Meeting:
Gap is $50$m. Relative speed $6$ m/s. Time to meet: $50/6 = 25/3$ s.
Total time: $50 + 8.33 = \mathbf{58.33 \text{ s}}$.
Location: A travels $4 \times (25/3) = 100/3$ m (approx 33.3m). -
Final Return:
Boy A returns to 0: Distance $100/3$, Speed $2$. Time $50/3$ s.
Boy B returns to 100: Distance $200/3$, Speed $4$. Time $50/3$ s.
Both arrive exactly at $58.33 + 16.67 = \mathbf{75 \text{ s}}$.
The graph below shows one complete 75s cycle. Notice the “Butterfly” shape:
Boy A does a Large Loop then a Small Loop.
Boy B does a Small Loop then a Large Loop.
Total Distance Calculation
In one 75s cycle, both boys walk exactly the same total ground distance:
- Boy A: Leg 1 ($400/3$ m) + Leg 2 ($200/3$ m) = 200 m
- Boy B: Leg 1 ($200/3$ m) + Leg 2 ($400/3$ m) = 200 m
Final Result:
Total Time $T = 300$ s.
Number of Cycles $N = 300 / 75 = 4$ cycles.
Total Distance $= 4 \times 200 \text{ m} = \mathbf{800 \text{ m}}$.