When a car passes mark-A, driver applies brakes. Thereafter reducing speed uniformly from 160 km/h a

Q: Why isn't the speed at mark B just the average of 160 and 40, which is 100?

That's the most common trap! The speed decreases uniformly with *time*, not with *distance*. Because the car is moving faster in the first half of the distance, it spends less time there. Therefore, it loses less speed in the first half compared to the second half.

Q: Can I solve this by finding the time taken?

Yes, but it's much longer. You would need to set up equations for distance in terms of time. Using the third equation of motion $v^2 = u^2 + 2as$ completely bypasses the need to find time, making it the most efficient tool here.

Q: What does the negative sign in $ad = -6000$ mean?

The negative sign simply indicates deceleration. The car is applying brakes, so the acceleration vector is opposite to the velocity vector, causing the speed to decrease.

Q: How do I estimate $\sqrt{13600}$ without a calculator?

You don't need an exact value! You know $100^2 = 10000$ and $120^2 = 14400$. Since 13600 is between 10000 and 14400, the square root must be between 100 and 120. That's enough to know it's greater than 100.

Q: Would the answer change if the car was accelerating instead of braking?

Absolutely. If it were accelerating, it would spend less time in the second half, meaning it would gain less speed there. The velocity at the midpoint would be less than the simple average.

Question

When a car passes mark-A, driver applies brakes. Thereafter reducing speed uniformly from $160	ext{ km/h}$ at A, the car passes mark C with a speed $40	ext{ km/h}$. The marks are at equal distances on the road as shown below.
Where on the road was the car moving with a speed $100	ext{ km/h}$? Neglect the size of the car as compared to the distances involved.

Options:

(A) At mark-B

(B) Between mark-A and mark-B

(C) Between mark-B and mark-C

(D) Information is insufficient to decide.

Sigma Percentile · Accepted Answer

The problem of the braking car is a classic kinematics trap that tests your intuition versus mathematical reality. When we see a car slowing down from $160 	ext{ km/h}$ to $40 	ext{ km/h}$ over a certain distance, our brain immediately wants to find the average. The midpoint of $160$ and $40$ is $100$. So, the car must be traveling at $100 	ext{ km/h}$ exactly at the midpoint of the distance, right?

**Wrong.** This is the illusion of the midpoint, and it's exactly what the examiner wants you to fall for. Let's break down the physics and see why our intuition fails us here.

### The Master Equation

We are told the car reduces its speed uniformly. This is the golden keyword. "Uniformly" means the acceleration (or in this case, deceleration) is constant. Whenever we have constant acceleration and we are dealing with velocities and distances—without any mention of time—the third equation of motion is our best friend:

$$v^2 = u^2 + 2as$$

This equation tells us a profound truth: velocity squared ($v^2$) changes linearly with distance ($s$), not the velocity ($v$) itself. This non-linear relationship is the key to unlocking the problem.

### Analyzing the Full Journey

Let's look at the entire journey from Mark A to Mark C. Let the distance between Mark A and Mark B be $d$. Since the marks are at equal distances, the distance between Mark B and Mark C is also $d$. Therefore, the total distance from A to C is $2d$.

We know the initial velocity at A is $v_A = 160 	ext{ km/h}$ and the final velocity at C is $v_C = 40 	ext{ km/h}$. Let's plug these into our master equation for the journey from A to C:

$$v_C^2 = v_A^2 + 2a(2d)$$

$$(40)^2 = (160)^2 + 4ad$$

Now, let's do the arithmetic.

$$1600 = 25600 + 4ad$$

$$4ad = 1600 - 25600 = -24000$$

$$ad = -6000$$

We don't need to find the exact values of $a$ or $d$ individually. The product $ad = -6000$ is the mathematical bridge we need. The negative sign perfectly aligns with our physical reality: the car is braking, so the acceleration is acting opposite to the motion.

### The Midpoint Reveal

Now, let's find out what is actually happening at Mark B, the exact midpoint of the distance. We apply the same equation, but this time only for the journey from A to B. The distance is $d$, and the final velocity is $v_B$.

$$v_B^2 = v_A^2 + 2ad$$

We already know $v_A = 160$ and we just found $ad = -6000$. Let's substitute these in:

$$v_B^2 = (160)^2 + 2(-6000)$$

$$v_B^2 = 25600 - 12000$$

$$v_B^2 = 13600$$

To find $v_B$, we take the square root of $13600$. You don't need a calculator for this. We know that $100^2 = 10000$ and $120^2 = 14400$. Since $13600$ is much closer to $14400$, $v_B$ must be around $116 	ext{ km/h}$. (Exactly, it's $\approx 116.6 	ext{ km/h}$).

### The Final Verdict

Let's step back and look at the timeline of the car's speed:
- At Mark A: $160 	ext{ km/h}$
- At Mark B: $\approx 116.6 	ext{ km/h}$
- At Mark C: $40 	ext{ km/h}$

The question asks where the car was moving at **$100 	ext{ km/h}$**.

Since the speed at Mark B is still $116.6 	ext{ km/h}$, the car hasn't slowed down to $100 	ext{ km/h}$ yet. It must continue past Mark B, braking further, to finally reach $100 	ext{ km/h}$ before it hits $40 	ext{ km/h}$ at Mark C.

Therefore, the speed of $100 	ext{ km/h}$ is achieved **between Mark B and Mark C**.

This happens because the car travels much faster in the first half of the distance (A to B) than in the second half (B to C). Because it's moving faster, it spends less time in the first half. Less time spent braking means less speed lost! It loses only about $43.4 	ext{ km/h}$ in the first half, but loses a massive $76.6 	ext{ km/h}$ in the second half. Physics is beautiful when you look past the illusions!

Select Answer:

Visualized Solution

Visualizing the Journey

Equation of Motion

Applying from A to C

Calculating $a d$

Velocity at Mark B

Computing $v_{B}$

Locating $100$ km/h

Conclusion

The Sigma Insight: Equations of Kinematics

The Master Equation

Analyzing the Full Journey

The Midpoint Reveal

The Final Verdict

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