KINEMATICS O15

Solution 15

Solution: Traffic Signal and Merging Groups

1. Understanding the Mechanism of “Merging”

The traffic light creates gaps between groups of vehicles. A group becomes “indistinguishable” from the next when the gap between them closes completely.

  • Leading Group (Group 1): The “tail” of this group is determined by the slowest vehicle ($v_{min}$) departing at the very end of the Green light.
  • Trailing Group (Group 2): The “head” of this group is determined by the fastest vehicle ($v_{max}$) departing at the very beginning of the next Green light (immediately after the Red light ends).

The groups merge when the “Head of Group 2” catches up to the “Tail of Group 1”.

Distance (x) Time (t) Tail of Grp 1 (starts t_g) Head of Grp 2 (starts t_g + t_r) Merge Point t=1 min t=3 min

2. Mathematical Formulation

Let’s define the timeline:

  • $t=0$: Start of Green light for Group 1.
  • $t=1.0$ min: End of Green light. Tail of Group 1 leaves. Speed = $v_{min}$.
  • $t=3.0$ min ($1.0$ min Green + $2.0$ min Red): Start of Green light for Group 2. Head of Group 2 leaves. Speed = $v_{max}$.

Let $x$ be the distance from the traffic light where they merge.

Time taken by Tail of Group 1: $\Delta t_1 = \frac{x}{v_{min}}$

Time taken by Head of Group 2: $\Delta t_2 = \frac{x}{v_{max}}$

The Head of Group 2 starts $2.0$ minutes ($t_r$) after the Tail of Group 1. For them to meet at the same time:

$$ \text{Travel Time}_{slow} = \text{Travel Time}_{fast} + \text{Delay} $$ $$ \frac{x}{v_{min}} = \frac{x}{v_{max}} + t_{gap} $$

Note: The “gap” in start times between the Tail of Group 1 and Head of Group 2 is exactly the Red Light duration $t_r = 2.0$ min.

3. Calculation

Given values:

  • $v_{min} = 60 \, \text{km/h}$
  • $v_{max} = 80 \, \text{km/h}$
  • $t_r = 2.0 \, \text{min} = \frac{2}{60} \, \text{h}$

Solving for $x$:

$$ x \left( \frac{1}{60} – \frac{1}{80} \right) = \frac{2}{60} $$ $$ x \left( \frac{80 – 60}{60 \times 80} \right) = \frac{2}{60} $$ $$ x \left( \frac{20}{4800} \right) = \frac{1}{30} $$ $$ x \left( \frac{1}{240} \right) = \frac{1}{30} $$ $$ x = \frac{240}{30} = 8.0 \, \text{km} $$

Conclusion: The groups become indistinguishable at a distance of 8.0 km.

Correct Answer: (b)