A traffic officer receives complaints on frequent traffic jams at a traffic signal on the main stree

Q: Why do we only analyze one cycle of the traffic light?

Because the motion is periodic! The pattern of moving and stopping repeats identically. Therefore, the average speed over one single cycle is exactly the same as the average speed over a hundred cycles.

Q: Does the length of the traffic queue matter in this problem?

No, the problem simplifies the situation by assuming identical vehicles moving at identical speeds. This effectively treats the traffic as a continuous, uniform stream, allowing us to track a single representative vehicle.

Q: What if the cars take time to accelerate to their moving speed?

The problem explicitly states that the officer 'made reasonable assumptions... to simplify calculations'. One of these assumptions is that vehicles instantly reach their moving speed $v_0$ and instantly stop, ignoring the brief periods of acceleration and deceleration.

Q: Why isn't the initial average speed just the moving speed $v_0$?

Because the cars aren't moving all the time! They spend half of their time completely stopped at the red light. Therefore, their average speed over time is exactly half of their moving speed.

Q: Can we just use the formula $v_{avg} = (v_{initial} + v_{final}) / 2$?

Absolutely not! That formula is strictly for motion with **constant acceleration**. In this problem, the velocity is piecewise constant (it jumps between $0$ and $v_0$), so we must stick to the fundamental definition: Total Distance / Total Time.

Question

A traffic officer receives complaints on frequent traffic jams at a traffic signal on the main street of a busy market. He studied the traffic pattern and to simplify calculations made a reasonable assumptions that all the vehicles are identical in size and move with identical speeds. At present, durations of red and the green signals are equal and average speed of traffic advancement is $v_1 = 1.5 	ext{ m/s}$. For improvement in the situation, if he orders to make duration of green signals $\eta = 2$ times and to leave that of red signals unchanged, what would be average speed $v_2$ of the traffic advancement?

Sigma Percentile · Accepted Answer

### The Real-World Physics of Traffic Jams

Have you ever been stuck at a red light, wondering how much it's slowing down your journey? This problem takes a frustrating real-world scenario and turns it into a beautiful application of kinematics.

We are dealing with a traffic officer who wants to optimize the flow of vehicles. To do this, we model the traffic as a continuous stream of identical cars moving at a constant speed $v_0$ when the light is green, and coming to a complete halt when the light is red.

### The Master Equation: Average Speed

The core concept here is **average speed**. In physics, average speed over a time interval is simply the total distance covered divided by the total time taken.

$$v_{avg} = \frac{	ext{Total Distance}}{	ext{Total Time}}$$

Because the traffic light operates in a repeating cycle (red, then green, then red again), we only need to analyze a single complete cycle to find the overall average speed of the traffic.

### Analyzing the Baseline: Case 1

Initially, the officer observes that the red and green signals have equal durations. Let's call this duration $T$.

During the red light (time $T$), the cars are stationary, so the distance covered is zero. During the green light (also time $T$), the cars move at their constant speed $v_0$. The distance covered during this green phase is simply $v_0 	imes T$.

The total time for one complete cycle is $T + T = 2T$. Therefore, the average speed $v_1$ is:

$$v_1 = \frac{v_0 \cdot T}{2T} = \frac{v_0}{2}$$

We are given that this initial average speed $v_1$ is $1.5 	ext{ m/s}$. This allows us to calculate the actual moving speed of the vehicles:

$$v_0 = 2 	imes 1.5 = 3.0 	ext{ m/s}$$

This means when the cars are actually moving, they are cruising at $3.0 	ext{ m/s}$.

### The Officer's Intervention: Case 2

To improve the situation, the officer decides to double the duration of the green signal ($\eta = 2$), while keeping the red signal duration exactly the same.

Now, the red signal still lasts for time $T$, but the green signal lasts for time $2T$.

During this extended green phase, the cars cover a much larger distance: $v_0 	imes 2T$. The total time for this new cycle is $T + 2T = 3T$.

### The Final Calculation

Let's plug these new values into our average speed formula to find the new average speed $v_2$:

$$v_2 = \frac{v_0 \cdot 2T}{3T} = \frac{2}{3} v_0$$

We already know that the actual moving speed $v_0$ is $3.0 	ext{ m/s}$. Substituting this value in:

$$v_2 = \frac{2}{3} 	imes 3.0 = 2.0 	ext{ m/s}$$

By simply doubling the green light duration, the officer successfully increased the average traffic speed from $1.5 	ext{ m/s}$ to $2.0 	ext{ m/s}$.

### The General Takeaway

This problem beautifully illustrates how macroscopic properties (like average traffic flow) depend on microscopic parameters (like signal timings). The general formula we derived, $v_{avg} = v_0 \frac{T_G}{T_R + T_G}$, is a powerful tool for urban planners to optimize city traffic!

Visualized Solution

$V e l oc i t y - T im e G r a p h S e t u p$

$A v er a g e S p ee d F or m u l a$

$C a se 1 : E q u a l D u r a t i o n s$

$F in d in g M o v in g S p ee d v_{0}$

$C a se 2 : E x t e n d e d G r ee n S i g na l$

$N e w A v er a g e S p ee d S e t u p$

$C a l c u l a t in g v_{2}$

$F ina l A n s w er$

The Sigma Insight: Motion in a Straight Line

The Real-World Physics of Traffic Jams

The Master Equation

Analyzing the Baseline

The Officer's Intervention

The Final Calculation

The General Takeaway

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