Traffic signals are installed at every s = 1.00 km on a long straight road. A signal remains red for

Q: Why can't the travel time be an even multiple of $\tau$?

If the travel time is an even multiple of $\tau$ (like $2\tau$), the next signal will have flipped an even number of times. Since it started in the opposite state (Red), an even number of flips means it will be Red again when you arrive!

Q: Does it matter exactly when we cross the first green light?

Surprisingly, no! As long as you cross the first signal while it is green, traveling at these 'magic speeds' ensures you will hit the exact same relative point in the green phase of every subsequent signal.

Q: What if the car accelerates?

The question specifically asks for *constant* speeds. If the car accelerates, its trajectory on the space-time graph would be a curve, making it much harder to guarantee it stays within the green bands indefinitely.

Q: Why do we use $(2n+1)$ to represent odd numbers?

In mathematics, an even number is always divisible by 2, so it can be written as $2n$. An odd number is always one more (or one less) than an even number, hence $(2n+1)$ where $n$ is an integer.

Q: Are there infinitely many possible speeds?

Yes, mathematically there are infinitely many solutions as $n$ increases. However, physically, as $n$ gets very large, the speed $v$ becomes impractically slow (e.g., crawling at 1 km/h).

Question

Traffic signals are installed at every $s = 1.00 	ext{ km}$ on a long straight road. A signal remains red for $	au = 30 	ext{ s}$ and green for next $	au = 30 	ext{ s}$. The signals are synchronized in such a way that at a time, alternate signals remain red and the other remain green. The scheme is shown in the following figure.
Suggest possible constant speeds at which a vehicle can run on this road without a stop.

Sigma Percentile · Accepted Answer

### The Setup: A Synchronized Dance of Lights

Imagine driving down a long, perfectly straight highway. Every $s = 1.00 	ext{ km}$, there is a traffic signal. These aren't just random lights; they follow a strict, synchronized rhythm. Each signal stays green for $	au = 30 	ext{ s}$, then switches to red for the next $	au = 30 	ext{ s}$.

But here is the brilliant catch: adjacent signals are perfectly out of phase. When signal 1 is green, signal 2 is red. When signal 2 turns green, signal 1 turns red. It is a continuous, alternating dance of lights. Our mission is to find the "magic speeds" that allow a vehicle to cruise down this highway indefinitely without ever touching the brakes.

### The Space-Time Graph: Visualizing the Journey

To truly understand this problem, we must step out of the driver's seat and look at the journey from a bird's-eye view using a **space-time ($x-t$) graph**.

If we plot distance on the horizontal axis and time on the vertical axis, the traffic signals appear as vertical bands of alternating green and red. A vehicle moving at a constant speed $v$ traces a straight, diagonal line on this graph. The slope of this line is determined by the speed. For a non-stop journey, this straight line must thread its way exclusively through the green bands, never once intersecting a red band.

### The Condition for a Non-Stop Drive

Let's assume we cross the very first signal right in the middle of its green phase. To avoid stopping at the second signal (which is $1 	ext{ km}$ away), we must arrive exactly when it is green.

However, remember that the second signal started in the opposite state (red). Since the signals flip every $	au = 30 	ext{ s}$, the second signal will only be green after an **odd number of flips**. If our travel time between signals is an even multiple of $	au$ (like $60 	ext{ s}$), the signal will have flipped twice, returning to its original red state, and we would be forced to stop.

Therefore, the time $\Delta t$ taken to travel the distance $s$ between two consecutive signals must be an odd multiple of $	au$. Mathematically, we write this as:

$$\Delta t = (2n + 1)	au$$

where $n$ is any non-negative integer ($n = 0, 1, 2, \dots$).

### Calculating the Magic Speeds

We know from basic kinematics that the travel time is simply distance divided by speed, so $\Delta t = \frac{s}{v}$. Equating our two expressions for $\Delta t$, we get our master equation:

$$\frac{s}{v} = (2n + 1)	au$$

Rearranging to solve for our constant speed $v$:

$$v = \frac{s}{(2n + 1)	au}$$

Now, let's plug in the given values. The distance is $s = 1 	ext{ km}$, and the time interval is $	au = 30 	ext{ s}$. To get our speed in standard units of $	ext{km/h}$, we convert $	au$ into hours: $	au = \frac{30}{3600} 	ext{ h} = \frac{1}{120} 	ext{ h}$.

Let's test different values of $n$:

**For $n = 0$:**
$$v_1 = \frac{1}{1 	imes (1/120)} = 120 	ext{ km/h}$$
This is the fastest possible speed to surf the green wave. You reach the next signal just as it completes its first flip to green.

**For $n = 1$:**
$$v_2 = \frac{1}{3 	imes (1/120)} = 40 	ext{ km/h}$$
Here, you drive slower. You let the next signal flip three times (red $ightarrow$ green $ightarrow$ red $ightarrow$ green) before you arrive.

**For $n = 2$:**
$$v_3 = \frac{1}{5 	imes (1/120)} = 24 	ext{ km/h}$$

### Conclusion

By continuing this pattern, we generate an infinite sequence of possible constant speeds: **$120 	ext{ km/h}, 40 	ext{ km/h}, 24 	ext{ km/h}, \dots$**

As long as you lock your cruise control to any of these exact speeds, the alternating traffic lights will perfectly align with your arrival, allowing you to drive forever without a single stop.

Visualized Solution

Visualizing the Space-Time Graph

The Condition for a Non-Stop Journey

Analyzing the First Interval

The Phase Shift Logic

Formulating the Master Equation

Solving for Velocity

Calculating the First Speed ( $n = 0$ )

Calculating the Second Speed ( $n = 1$ )

The Sequence of Magic Speeds

The Way Forward

The Sigma Insight: Motion Graphs

The Setup

The Space-Time Graph

The Condition for a Non-Stop Drive

Calculating the Magic Speeds

Conclusion

Similar Questions

A car accelerates from rest at a constant rate $α$ for some time after which it decelerates at a constant rate $β$ to come to rest. If the total time elapsed is $t$ seconds, the total distance travelled is

A car starting from rest, accelerates at the rate $f$ through a distance $s$ , then continues at constant speed for time $t$ and then decelerates at the rate $f /2$ to come to rest. If the total distance travelled is $15 s$ , then

Comprehension Passage

What is the speed $v_{1}$ of the cars on the road?

What is the speed $v_{2}$ of the cars on the bridge?

What is length of the bridge?

A material particle is chasing another one and both of them are moving on the same straight line. After they pass a particular point, their velocities $v$ vary with time $t$ as shown in the figure. When will the chase end?

The speed versus time graph for a particle is shown in the figure. The distance travelled (in metre) by the particle during the time interval $t = 0 s$ to $t = 5 s$ will be ……… .

The v-t graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 s. The total distance covered by the body in 6 s is

Relation between average velocity $v_{a v}$ of a body and time $t$ is shown in the graph. If during the time interval considered, the body did not change direction of motion, draw a graph between instantaneous velocity of the body and time.

Two particles A and B start from the same point and move in the positive $x$ -direction. In a time interval of $2.00 s$ after they start, their velocities $v$ vary with time $t$ as shown in the following figures. What is the maximum separation between the particles during this time interval?