Speeds of two identical cars are u and 4u at a specific instant. The ratio of the respective distanc

Q: Why do we assume the retardation is the same for both cars?

The problem states the cars are 'identical'. This implies they have the same mass and the same tires. Since they are on the same road, the coefficient of friction $\mu$ is the same. Retardation due to friction is $a = \mu g$, which is independent of mass and velocity!

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

We are given initial velocities ($u$ and $4u$), final velocities ($0$, since they stop), and we need to find the ratio of distances ($s$). Time ($t$) is neither given nor asked for. The third equation, $v^2 = u^2 + 2as$, perfectly links these three variables.

Q: Does the mass of the car affect its stopping distance?

Surprisingly, no! The frictional force is $f = \mu mg$. Using Newton's second law, $ma = \mu mg$, so the mass $m$ cancels out, leaving $a = \mu g$. A heavier car has more momentum, but also experiences more friction, perfectly balancing out the stopping distance.

Q: What if the second car was moving at $3u$ instead of $4u$?

Because stopping distance is proportional to the square of the initial velocity ($s \propto u^2$), if the speed was $3u$, the stopping distance would be $3^2 = 9$ times longer. The ratio would be $1:9$.

Q: I got $1:4$ as my answer. What did I do wrong?

You likely assumed that stopping distance is directly proportional to velocity ($s \propto u$). Remember, kinetic energy ($\frac{1}{2}mu^2$) is being depleted by work done ($F \times s$). Since energy depends on $u^2$, the distance $s$ must also depend on $u^2$!

Question

Speeds of two identical cars are $u$ and $4u$ at a specific instant. The ratio of the respective distances at which the two cars are stopped from that instant is

Options:

(A) 1:1

(B) 1:4

(C) 1:8

(D) 1:16

Sigma Percentile · Accepted Answer

### The Setup: Two Cars, One Road

Imagine you are standing by a long, straight highway. Two perfectly identical cars zoom past you. The first car is cruising at a comfortable speed of $u$. The second car, however, is tearing down the road at a blistering speed of $4u$—exactly four times as fast as the first one.

Suddenly, both drivers slam on their brakes at the exact same instant. The tires screech, and both cars eventually come to a complete halt. Our mission is to find the ratio of the distances they cover before stopping, which we call their **stopping distances** ($s_1$ and $s_2$).

### The Physics of Stopping

Before we dive into the math, let's think about what makes a car stop. When brakes are applied, the friction between the tires and the road provides a retarding force. Because the problem explicitly states that the cars are **identical**, they have the same mass and the same tires. Since they are on the same road, the coefficient of friction $\mu$ is identical for both.

According to Newton's Second Law, the frictional force is $F = \mu mg$. The retardation (negative acceleration) is therefore $a = \frac{F}{m} = \mu g$. Notice how the mass $m$ completely cancels out! This means both cars experience the exact same constant retardation $a$, regardless of how fast they are going.

### The Mathematical Execution

We know the initial velocities ($u$ and $4u$), the final velocities ($0$, because they stop), and we want to find the distances ($s_1$ and $s_2$). The perfect tool for this job is the **third equation of motion**, which links these variables without needing the time $t$:

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

Let's set up the equation for the **first car**. Its final velocity $v$ is $0$, and its acceleration is $-a$ (since it's slowing down):

$$0^2 = u^2 - 2as_1$$

Rearranging this to solve for $s_1$, we get:

$$s_1 = \frac{u^2}{2a}$$

Now, let's do the exact same thing for the **second car**. Its initial velocity is $4u$:

$$0^2 = (4u)^2 - 2as_2$$

Squaring the initial velocity gives us $16u^2$. Rearranging for $s_2$ yields:

$$s_2 = \frac{16u^2}{2a}$$

### The Golden Rule of Road Safety

Now, we just need to find the ratio $\frac{s_1}{s_2}$. If we look closely at our expression for $s_2$, we can factor out the $16$:

$$s_2 = 16 \left( \frac{u^2}{2a} \right)$$

Since the term inside the parentheses is exactly $s_1$, we can write:

$$s_2 = 16 s_1$$

Finally, taking the ratio gives us:

$$\frac{s_1}{s_2} = \frac{s_1}{16 s_1} = \frac{1}{16}$$

This result reveals a profound and life-saving principle of physics: **Stopping distance is directly proportional to the square of the initial velocity** ($s \propto u^2$). If you double your speed, your stopping distance quadruples. In our case, because the second car was going four times as fast, it required a massive $4^2 = 16$ times more distance to stop! This is the fundamental reason why speed limits exist and why high-speed driving leaves so little margin for error.

Animated Solution for Physics - Kinematics: Speeds of two identical cars are $u$ and $4 u$ at a specific instant. The ratio of the respective distances at which the two cars are stopped from that instant is

Select Answer:

Visualized Solution

$v_{i} = u, 4 u$

$v^{2} = u^{2} + 2 a s$

\text{Car 1 Setup}

$s_{1} = \frac{u ^{2}}{2 a}$

\text{Car 2 Setup}

$s_{2} = \frac{16 u ^{2}}{2 a}$

\frac{s_1}{s_2} = \frac{1}{16}

$s \propto u^{2}$

The Sigma Insight: Equations of Kinematics

The Setup

The Physics of Stopping

The Mathematical Execution

The Golden Rule of Road Safety

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