Animated Solution for Physics - Laws of Motion: Consider a car moving on a straight road with a speed of $100{\text{ ms}}^{-1}$. The distance at which car can be stopped, is $[{\mu }_k=0.5]$

Imagine you are cruising down a straight highway at a blistering speed of $100\text{ m/s}$ (which is a whopping $360\text{ km/h}$!). Suddenly, you spot a hazard and slam on the brakes. The car skids, the tires screech, and eventually, you come to a complete halt. The burning question is: How much distance did the car cover before stopping?

This classic physics problem is all about understanding the interplay between kinetic energy, friction, and kinematics. Let's break it down step-by-step.

When the brakes lock the wheels, the car slides. The only horizontal force acting on the car is the kinetic friction between the tires and the road. This frictional force acts in the direction opposite to the car's motion, causing it to decelerate (or retard).

According to the laws of friction, the kinetic frictional force $f_k$ is given by:

Since the car is on a flat horizontal road, the normal reaction $N$ perfectly balances the weight of the car, so $N=mg$. Therefore:

By Newton's Second Law ($F=ma$), the retardation $a$ produced by this frictional force is:

Notice something beautiful here? The mass $m$ of the car cancels out! This means the stopping distance depends only on the speed and the road conditions, not on whether you are driving a lightweight sports car or a heavy truck.

Since the frictional force is constant, the retardation is also constant. This allows us to use the equations of uniformly accelerated motion. The most suitable equation here is the third equation of motion, which relates initial velocity $u$, final velocity $v$, acceleration $a$, and displacement $s$ without involving time:

We use a negative sign because the acceleration is acting opposite to the velocity. We know the car eventually stops, so the final velocity $v=0$. Substituting our retardation $a={\mu }_{k}g$ into the equation:

Rearranging this to solve for the stopping distance $s$, we get a very famous and highly useful standard result:

Now, it's just a matter of plugging in the numbers given in the problem: - Initial velocity, $u=100\text{ m/s}$ - Coefficient of kinetic friction, ${\mu }_k=0.5$ - Acceleration due to gravity, $g=10{\text{ m/s}}^2$

Substituting these values into our master equation:

Let's simplify the denominator first. $2\times 0.5$ is exactly $1$, leaving us with just $10$ in the denominator.

So, the car travels a massive $1000\text{ meters}$ (or $1\text{ kilometer}$) before coming to a complete stop!

Look closely at the formula $s=\frac{u^2}{2{\mu }_{k}g}$. It tells us that the stopping distance is directly proportional to the square of the initial speed ($s\propto u^2$).

This is a profound realization for road safety. If you double your speed, your stopping distance doesn't just double—it quadruples! If you triple your speed, you need nine times the distance to stop. Physics isn't just for exams; it's the rulebook of reality.