On a road going out of a city, lights of the last traffic signal glow green for 1.0 min and red for

Q: Why do we only care about the fastest and slowest vehicles?

The fastest vehicle of the second group is at the very front of its pack, and the slowest vehicle of the first group is at the very back of its pack. The gap between the groups is exactly the distance between these two vehicles. Once the fastest catches the slowest, the gap is zero, and the groups merge.

Q: What happens to the vehicles in between the fastest and slowest?

They are safely bounded within the 'zone' of their respective groups. As long as the front of the second group hasn't caught the back of the first group, the zones are distinct and do not overlap.

Q: Why did we convert speeds to km/min?

The time intervals given for the traffic lights are in minutes (1.0 min and 2.0 min). Converting speeds from km/h to km/min keeps all our units consistent and makes the algebra much cleaner, avoiding division by 60 later on.

Q: Does the number of vehicles affect the merging distance?

No! As long as there is at least one vehicle traveling at the maximum speed and one at the minimum speed, the boundaries of the groups remain exactly the same. The density of cars inside the group doesn't change when the boundaries intersect.

Q: What if the red light duration was longer?

A longer red light means the second group starts even later. The fastest car of the second group would need more time and distance to catch up to the first group, so the merging distance would increase.

Question

On a road going out of a city, lights of the last traffic signal glow green for $1.0	ext{ min}$ and red for $2.0	ext{ min}$. After the traffic signal, the road is straight and vehicles run at their constant speeds that lie in the range $60	ext{ km/h}$ to $80	ext{ km/h}$. This mode of operation of traffic lights causes the vehicles to come out of the city in groups. Some distance away from the traffic signal, these groups merge and become indistinguishable. Assuming that the vehicles acquire their constant speeds after the traffic signal in negligible time, which of the following conclusions can you make?

Options:

(A) The groups become indistinguishable $4.0	ext{ km}$ after the traffic signal.

(B) The groups become indistinguishable $8.0	ext{ km}$ after the traffic signal.

(C) Wider is the range of speed of the vehicles; greater is the distance where the groups become indistinguishable.

(D) Larger is the number of fast moving vehicles, shorter is the distance where the groups become indistinguishable.

Sigma Percentile · Accepted Answer

### The Pulse of City Traffic

Imagine standing by a highway just outside a bustling city. The traffic is regulated by a signal that operates on a strict cycle: it glows green for $1.0	ext{ min}$, allowing a cluster of vehicles to escape, and then turns red for $2.0	ext{ min}$, halting the flow.

Because the vehicles travel at different constant speeds—ranging from a leisurely $60	ext{ km/h}$ to a brisk $80	ext{ km/h}$—the cluster of cars that leaves during a single green light doesn't stay tightly packed. Instead, it stretches out over time. The fastest cars pull ahead, while the slowest cars lag behind, creating a distinct "group" or "platoon" of traffic.

### Visualizing the Flow

To truly understand when these distinct groups merge and become a continuous, indistinguishable stream of traffic, we need to visualize their motion. A position-time ($x-t$) graph is the perfect tool for this.

Let's define the boundaries of the **first group**. It starts emerging at $t = 0$. The absolute front of this group is dictated by the fastest vehicle, which travels at $v_{\max} = 80	ext{ km/h}$. The absolute rear of this group is dictated by the slowest vehicle, which leaves the signal just as it turns red at $t = T_G = 1.0	ext{ min}$, traveling at $v_{\min} = 60	ext{ km/h}$.

Now, consider the **second group**. It is held back by the $2.0	ext{ min}$ red light and only starts emerging at $t = T_G + T_R = 3.0	ext{ min}$. Its front is led by a new fast vehicle traveling at $v_{\max}$, and its rear is anchored by a slow vehicle leaving at $t = 4.0	ext{ min}$.

### The Inevitable Catch-Up

The core of the problem lies in a simple realization: the front of the second group is moving faster than the rear of the first group.

The groups will remain distinct as long as there is a physical gap between them. This gap is exactly the distance between the slowest car of the first group and the fastest car of the second group. The moment these two specific vehicles are side-by-side, the gap vanishes, and the groups merge into one continuous stream.

### Setting Up the Mathematics

Let's translate this physical intuition into mathematical equations. We need the position of the slowest car of Group 1 ($x_{1,	ext{slow}}$) to equal the position of the fastest car of Group 2 ($x_{2,	ext{fast}}$).

For the slowest car of Group 1, which started at $t = T_G$:
$$x_{1,	ext{slow}} = v_{\min} (t - T_G)$$

For the fastest car of Group 2, which started at $t = T_G + T_R$:
$$x_{2,	ext{fast}} = v_{\max} (t - (T_G + T_R))$$

Equating the two gives us our master equation:
$$v_{\min} (t - T_G) = v_{\max} (t - (T_G + T_R))$$

### The Final Calculation

To avoid messy fractions, let's convert our speeds from $	ext{km/h}$ to $	ext{km/min}$. 
- $v_{\min} = 60	ext{ km/h} = 1	ext{ km/min}$
- $v_{\max} = 80	ext{ km/h} = \frac{4}{3}	ext{ km/min}$

Substituting these values, along with $T_G = 1$ and $T_R = 2$, we get:
$$1 \cdot (t - 1) = \frac{4}{3} \cdot (t - 3)$$

Multiplying the entire equation by 3 to clear the denominator:
$$3t - 3 = 4t - 12$$

Solving for $t$, we find:
$$t = 9	ext{ min}$$

This tells us that exactly 9 minutes after the first light turned green, the groups merge. But the question asks for the **distance**. We can substitute $t = 9$ back into either position equation. Let's use the simpler one:
$$x = 1	ext{ km/min} \cdot (9	ext{ min} - 1	ext{ min}) = 8.0	ext{ km}$$

Thus, the groups become indistinguishable exactly **$8.0	ext{ km}$** down the road. This elegant interplay of relative speeds and time delays perfectly matches option (b).

Select Answer:

Visualized Solution

$T_{G} = 1.0 min, T_{R} = 2.0 min$

\text{Group 1 Dynamics}

\text{Group 2 Dynamics}

\text{Condition for Merging}

\text{Position Equations}

\text{Substituting Values}

\text{Solving for Time } t

\text{Final Distance Calculation}

The Sigma Insight: Motion in a Straight Line

The Pulse of City Traffic

Visualizing the Flow

The Inevitable Catch-Up

Setting Up the Mathematics

The Final Calculation

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