A car moving with a speed of 50 km/h, can be stopped by brakes after atleast 6 m. If the same car is

Q: Why do we use the third equation of motion instead of the first or second?

The problem gives us initial velocity, final velocity (since the car stops), and distance. It asks for another distance. Time is neither given nor asked for. The third equation of motion ($v^2 = u^2 + 2as$) is the only one that relates these variables without requiring time.

Q: Do I absolutely have to convert km/h to m/s?

If you are calculating the exact value of acceleration, yes, you must convert to standard SI units (m/s). However, in this specific problem, because we are comparing two cases and taking a ratio, the conversion factor ($\frac{5}{18}$) cancels out completely. Using the proportionality shortcut ($s \propto u^2$) bypasses the need for conversion entirely.

Q: Why is the acceleration negative?

Acceleration is a vector. If we take the direction of the car's motion as positive, the brakes apply a force in the opposite direction to slow it down. This opposing force creates an acceleration in the negative direction, which we commonly call retardation or deceleration.

Q: Can this problem be solved using the Work-Energy Theorem?

Absolutely! The Work-Energy Theorem states that the work done by friction equals the change in kinetic energy: $-f_k \cdot s = 0 - \frac{1}{2}mu^2$. Since frictional force $f_k$ is constant, we get $s = \frac{mu^2}{2f_k}$. This again proves that $s \propto u^2$.

Q: What if the mass of the car was different in the second case?

Interestingly, the stopping distance is independent of the car's mass! The frictional force is $f_k = \mu mg$, so retardation is $a = \frac{f_k}{m} = \mu g$. Since mass cancels out, the retardation depends only on the coefficient of friction $\mu$ and gravity $g$. So, a heavier car and a lighter car will have the same stopping distance if they have the same tires and initial speed.

Question

A car moving with a speed of $50 	ext{ km/h}$, can be stopped by brakes after atleast $6 	ext{ m}$. If the same car is moving at a speed of $100 	ext{ km/h}$, the minimum stopping distance is

Options:

(A) $12 	ext{ m}$

(B) $18 	ext{ m}$

(C) $24 	ext{ m}$

(D) $6 	ext{ m}$

Sigma Percentile · Accepted Answer

### The Physics of Braking

Imagine you are cruising down a highway. Suddenly, you spot an obstacle and slam on the brakes. Your car doesn't stop instantaneously; it skids forward, covering a certain distance before coming to a complete halt. This distance is known as the **stopping distance**.

In this problem, we are exploring how this stopping distance changes when you alter your initial speed, assuming the braking capability of your car remains exactly the same.

### Setting Up the Equations

To analyze this mathematically, we turn to the equations of kinematics. We know the initial velocity $u$, the final velocity $v$ (which is $0$ since the car stops), the constant retardation $a$ provided by the brakes, and the stopping distance $s$. The perfect tool that connects all these variables without involving time is the **third equation of motion**:

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

Since the car comes to rest, $v = 0$. The equation simplifies to:

$$0 = u^2 + 2as$$

$$2as = -u^2$$

### Analyzing the First Case

In our first scenario, the car is moving at $u_1 = 50 	ext{ km/h}$. It stops after covering a distance $s_1 = 6 	ext{ m}$. Let's plug these values into our simplified equation.

*(Note: While it's a good habit to convert km/h to m/s by multiplying with $\frac{5}{18}$, you will soon see that in ratio-based problems, these conversion factors often cancel out beautifully!)*

$$2a(6) = -\left(50 	imes \frac{5}{18}ight)^2$$

From this, we can extract an expression for the constant retardation term $2a$:

$$2a = -\frac{\left(50 	imes \frac{5}{18}ight)^2}{6}$$

We will hold onto this expression. There is no need to calculate the exact numerical value of $a$ just yet.

### The Second Case: Doubling the Speed

Now, the driver is moving twice as fast, at $u_2 = 100 	ext{ km/h}$. The brakes are applied with the exact same force, meaning the retardation $a$ is identical. We need to find the new stopping distance, $s_2$.

Using our equation again:

$$2as_2 = -\left(100 	imes \frac{5}{18}ight)^2$$

Now, we substitute the expression for $2a$ that we found from the first case:

$$\left[ -\frac{\left(50 	imes \frac{5}{18}ight)^2}{6} ight] s_2 = -\left(100 	imes \frac{5}{18}ight)^2$$

Notice how the negative signs and the $\frac{5}{18}$ conversion factors cancel out perfectly on both sides. Rearranging the terms, we get:

$$\frac{s_2}{6} = \left( \frac{100}{50} ight)^2$$

$$\frac{s_2}{6} = 2^2 = 4$$

$$s_2 = 6 	imes 4 = 24 	ext{ m}$$

### The Proportionality Shortcut (The Pro-Move)

While the algebraic substitution is rigorous, there is a much faster, more elegant way to solve this—a method highly favored in competitive exams like JEE.

Let's look back at our core equation: $0 = u^2 + 2as$. Rearranging it for stopping distance $s$, we get:

$$s = \frac{-u^2}{2a}$$

Since the retardation $a$ is constant for a given car and road surface, we can clearly see that the stopping distance is **directly proportional to the square of the initial velocity**:

$$s \propto u^2$$

This is a powerful insight! It means that if you change your speed by a factor of $n$, your stopping distance changes by a factor of $n^2$.

In our problem, the speed increased from $50 	ext{ km/h}$ to $100 	ext{ km/h}$. The speed was **doubled** ($n = 2$). Therefore, the stopping distance must become $2^2 = 4$ times the original distance.

$$s_{new} = 4 	imes s_{old} = 4 	imes 6 	ext{ m} = 24 	ext{ m}$$

This shortcut not only saves precious time but also builds a deeper physical intuition about how dangerous speeding can be. Doubling your speed doesn't just double your braking distance; it quadruples it!

Animated Solution for Physics - Kinematics: A car moving with a speed of $50 km/h$ , can be stopped by brakes after atleast $6 m$ . If the same car is moving at a speed of $100 km/h$ , the minimum stopping distance is

Select Answer:

Visualized Solution

$Visualizing the Braking Process$

$Equation of Motion$

$Analyzing Case 1$

$Extracting the Retardation Relation$

$Analyzing Case 2$

$Calculating New Stopping Distance$

$The Proportionality Shortcut$

The Sigma Insight: Equations of Kinematics

The Physics of Braking

Setting Up the Equations

Analyzing the First Case

The Second Case

The Proportionality Shortcut (The Pro-Move)

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