From a town, cars start at regular intervals of 30 s and run towards another town with constant spee

Q: Why doesn't the distance between the cars decrease when they slow down?

Because they slow down *at the exact same instant*. If Car A slowed down first, Car B would catch up a bit, reducing the distance. But since they both hit the brakes simultaneously, their relative velocity remains zero during the deceleration, keeping their separation constant.

Q: What if the problem said they enter a 'bad weather zone' one by one?

That's a completely different scenario! If they enter a zone one by one, the first car slows down while the second is still going fast, so the distance between them shrinks. However, the *time interval* between them passing any point in the zone would remain exactly 30 seconds.

Q: Do we need to convert km/h to m/s to solve this?

You can, but it's unnecessary work. Since we are taking a ratio of speeds ($v_1/v_2$), the units cancel out perfectly. It's always smart to look for ratios before blindly converting units!

Q: Does the total distance to the second town matter?

Not at all. As long as the cars have established their new constant speed and constant separation before reaching the town, the arrival time interval depends only on that separation and the new speed.

Q: Is the time interval always inversely proportional to the speed?

In this specific case where the *spatial separation* is locked in and constant, yes. Time = Distance / Speed. Since Distance is constant, Time is inversely proportional to Speed.

Question

From a town, cars start at regular intervals of $30	ext{ s}$ and run towards another town with constant speed of $60	ext{ km/h}$. At some point of time, all the cars simultaneously have to reduce speed to $40	ext{ km/h}$ due to bad weather condition. What will be the time interval between arrivals of the cars at the second town during the bad weather?

Options:

(A) $20	ext{ s}$

(B) $30	ext{ s}$

(C) $40	ext{ s}$

(D) $45	ext{ s}$

Sigma Percentile · Accepted Answer

Have you ever been on a long road trip, watching the cars ahead of you and behind you? There is a certain rhythm to highway driving. This problem captures that rhythm perfectly, but it introduces a twist that tests your fundamental understanding of kinematics: the difference between **temporal separation** (time gap) and **spatial separation** (physical distance).

Let's break down this beautiful problem step by step.

### The Setup: A Convoy of Cars

Imagine a straight highway connecting two towns. Cars are leaving the first town one after another. We are given two crucial pieces of information about the initial state of this convoy:
1. The speed of every car is $v_1 = 60 	ext{ km/h}$.
2. The time interval between consecutive cars departing is $\Delta t_1 = 30 	ext{ s}$.

Because the cars are moving at a constant speed, this time gap translates directly into a physical distance gap. Think about it: when Car B starts its journey, Car A has already been traveling for exactly $30 	ext{ s}$.

Therefore, the physical distance $d$ between any two consecutive cars is simply the distance the first car covered in that time:

$$d = v_1 	imes \Delta t_1$$

As long as the cars maintain their speed of $60 	ext{ km/h}$, this physical distance $d$ remains constant. They are like beads on a string, moving together.

### The Crucial Word: "Simultaneously"

Now comes the twist. Bad weather strikes. The problem states: *"all the cars simultaneously have to reduce speed to $40 	ext{ km/h}$"*.

Read that word again: **simultaneously**. This is the linchpin of the entire problem.

What happens to the physical distance between the cars when they slow down at the exact same instant?

Imagine Car A and Car B. At $t = 0$, they are separated by distance $d$. At $t = 0.001 	ext{ s}$, they both hit the brakes. Because they decelerate identically and at the exact same time, their relative velocity is zero. Car A does not pull away from Car B, and Car B does not catch up to Car A.

Consequently, the physical distance $d$ between them **does not change**. It is locked in.

### The Final Leg: Arriving at Town 2

After the simultaneous deceleration, the entire convoy is now moving at a new, slower speed:

$$v_2 = 40 	ext{ km/h}$$

However, as we just established, the physical distance between them is still the original $d$.

Now, imagine standing at the entrance of the second town. You see Car A arrive. How long will you have to wait for Car B? You have to wait for Car B to cover that physical distance $d$ at its new speed $v_2$.

This gives us our new time interval, $\Delta t_2$:

$$\Delta t_2 = \frac{d}{v_2}$$

### The Elegant Calculation

We know that $d = v_1 	imes \Delta t_1$. Let's substitute this into our equation for $\Delta t_2$:

$$\Delta t_2 = \frac{v_1 	imes \Delta t_1}{v_2}$$

Notice how elegant this is? We don't even need to convert units like $	ext{km/h}$ to $	ext{m/s}$ because we are dealing with a ratio of speeds ($v_1 / v_2$), and the units will perfectly cancel out!

Let's plug in the numbers:

$$\Delta t_2 = \left( \frac{60}{40} ight) 	imes 30 	ext{ s}$$

$$\Delta t_2 = 1.5 	imes 30 	ext{ s} = 45 	ext{ s}$$

The cars will now arrive at the second town with a time gap of **$45 	ext{ s}$**.

### A Trap to Avoid

Before we wrap up, let's consider a classic variation of this problem that traps many students. What if the problem said, *"The cars enter a bad weather zone one by one, where their speed drops to $40 	ext{ km/h}$"*?

In that scenario, Car A would slow down while Car B is still moving fast. Car B would catch up slightly, meaning the **physical distance $d$ would decrease**. However, because they enter the zone at $30 	ext{ s}$ intervals, they would also pass any point inside the zone at $30 	ext{ s}$ intervals. The time gap would remain unchanged!

Always pay close attention to whether a change happens **simultaneously** across space, or **sequentially** as objects pass a specific point. It makes all the difference in kinematics.

Select Answer:

Visualized Solution

$v_{1} = 60 km/h, Δ t_{1} = 30 s$

$d = v_{1} \times Δ t_{1}$

$Simultaneous Deceleration$

$Δ t_{2} = \frac{d}{v _{2}}$

$Δ t_{2} = \frac{60}{40} \times 30$

$Δ t_{2} = 45 s$

$Food for Thought$

The Sigma Insight: Motion in a Straight Line

The Setup

The Crucial Word: "Simultaneously"

The Final Leg

The Elegant Calculation

A Trap to Avoid

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Comprehension Passage

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