• Written By Saif_Ansari

# Stopping Distance Formula: Definition, Derivation and More

If a driver hits the brakes of a car, the car doesn’t come to rest immediately. How far will a vehicle travel when the brake is applied? How much friction force is required between tyres and roads to stop a vehicle within a certain distance? Answers to all the above questions can be given by the Stopping Distance Formula and its application.

When the driver applies the brakes, the distance the car covers before it comes to a stop is known as stopping distance. The value of stopping distance depends on the speed of the car and the coefficient of friction between the wheels and the road, or we can say that it depends on the retarding force acting on the car. In this article, you will learn more about the stopping distance formula with examples. Continue reading to know more.

## Concept of Stopping Distance

When the vehicle is moving at a specific velocity, and suddenly the driver applies brakes, then you can observe that the vehicle comes to rest after covering a certain distance. This distance is called stopping distance. The stopping distance is the distance covered between the time when the brakes are applied to a moving vehicle and the time when the vehicle stops entirely.

Suppose a driver driving a car on the road at a certain speed suddenly sees a Hazard (Men, animals, etc.), and thinks to apply the brake. The time interval between the driver saw the hazard and pressing of the brake paddle is known as reaction time. During this time, the car will travel some distance. The actual braking begins after this. Most of the time, this reaction time is quite small and can be neglected in calculating stopping distance for low-speed cars or vehicles.

The stopping distance depends on the vehicle’s speed, the roughness of the road surface, and reflexes of the car’s driver, and it is denoted by $$s$$. The SI unit for stopping distance is meter (unit of distance).

### Friction

The sole reason for stopping any vehicle after the application of brakes is friction. It is the resistive force that acts when there is a relative motion between two surfaces. It plays a major role in day-to-day life, from walking humans on the ground to stopping the plane on the runway. There are two types of friction, static friction and kinetic friction. Static friction acts between two surfaces when they aren’t moving relative to each other, while kinetic friction acts between two objects or surfaces when they are in relative motion.

The static friction force is given by,
$$f = {\mu _s}N$$
where $${\mu _s}$$ is the static friction coefficient, and $$N$$ is the normal reaction between the objects or surfaces.
The kinetic friction force is given by,
$$f = {\mu _k}N$$
where $${\mu _k}$$ is the static friction coefficient, and $$N$$ is the normal reaction.

### Stopping Distance Formula Derivation

Before going for the derivation of the stopping distance formula, we will have to recall two equations of motion,
$$v = u + at$$ and $$s = ut + \frac{1}{2}a{t^2}$$
where $$v$$ is the final velocity, $$a$$ is the acceleration, $$u$$ is the initial velocity, and $$s$$ is the displacement.
From the above two-equation, we can write as:
$${v^2} = {u^2} + 2\,as$$……(i)
In the case of stopping distance, the final velocity is zero, and acceleration must be negative (Friction force will oppose the motion), or the body must be under retardation motion.
Let the retarding force (Friction, air drag, etc.) be $$F$$ acting on the body of mass $$m,$$
Then the acceleration of the vehicle, $$a = \frac{{ – F}}{m}.$$
Now put the value of final velocity $$\left( {v = 0} \right)$$ and acceleration in equation (i).
$${0^2} = {u^2} + 2\,as$$
$$\Rightarrow s = – \frac{{{u^2}}}{{2a}} = \frac{{m{u^2}}}{{2F}}$$
Thus the stopping distance formula for any retarding force $$F$$ will be,
$$s = \frac{{m{u^2}}}{{2F}}.$$

### Stopping Distance Formula Due to Friction

When friction force acts as retarding force, then we can write as:
$$f = \, – \mu mg = \, – ma$$
$$a = \, – \mu g$$
Now stopping distance formula becomes, $$s = \frac{{{v^2}}}{{2\mu g}}$$
Where,
$$v$$ – The velocity of the vehicle
$$\mu$$ – The coefficient of friction
$$g$$ – The acceleration due to gravity

### Factors Affecting the Stopping Distance

After discussing the stopping distance formula or stopping distance equation, we noticed that there are certain parameters that play a major role in the stopping distance. Let us have a look at the factors affecting the stopping distance.

1. Weather
We can observe that in poor weather conditions such as during the rainy season or snowfall, a car or any other vehicle faces so much trouble with brakes and the total stopping distance is likely to be longer. It happens because of a reduction in the coefficient of friction. For snow or ice, it is ten times than on the paved road.

Another important fact is road conditions, which could be a major factor affecting the stopping distance. A road could be particularly greasy or slippery if it’s been raining for a long time after a period of hot weather or if the oil has been spilt on it.

3. Driver Condition
The driver’s condition is one of the most vital factors because driving requires good eyesight. A driver’s age, how aware they are going to be while driving and if they have taken or consumed any kind of drugs or alcohol can all influence how quickly it takes them to react.

4. Car Condition
Even though now we have advanced technologies to make the car and at the same time many modern cars might be able to stop at considerably shorter distances than the normal car. The condition of the car will also have an important impact on the stopping distance.

5. Velocity
The braking distance increases with velocity. This is explained by the relationship between velocity and kinetic energy. The greater the velocity, the greater the kinetic energy the brakes have to transfer.

### Solved Example on Stopping Distance Formula

Q.1. A car travelling at a speed of $$45\,{\rm{m}}\,{{\rm{s}}^{ – 1}}.$$ The driver puts on the brakes when he sees a stop sign. If the coefficient of friction between the tyres and the road is $$\mu = 0.5,$$ then, what is the stopping distance of the car?
Sol:

Given,
The velocity of the car $$= u = 45\,{\rm{m}}\,{{\rm{s}}^{ – 1}}$$
The coefficient of friction $$= \mu = 0.5$$
The acceleration due to gravity, $$g = 9.8\,{\rm{m}}\,{{\rm{s}}^{ – 2}}$$
Now, we are asked to determine the total stopping distance of the car. We know that the stopping distance of a car is given by the stopping distance formula:
$$s = \frac{{{u^2}}}{{2\mu g}}$$
where,
$$u$$ – The initial velocity of the vehicle
$$\mu$$ – The coefficient of friction
$$g$$ – The acceleration due to gravity
Substituting all the required values in the above equation, we get:
$$\Rightarrow s = \frac{{{v^2}}}{{2\mu g}} = \frac{{{{\left( {45} \right)}^2}}}{{2 \times 0.5 \times 9.8}}$$
$$\Rightarrow s = 206.6\,{\rm{m}} \sim 207\,{\rm{m}}$$
Therefore, the stopping distance of the car is $$207\,{\rm{m}}.$$

Q.2. A car is moving with a velocity of $$40\,{\rm{m}}\,{{\rm{s}}^{ – 1}}$$ and suddenly applies brakes. Determine the acceleration of the body, which covers a distance of $$10\,{\rm{m}}$$ before coming to rest.
Sol:

Given:
Velocity, $$u = 40\,{\rm{m}}\,{{\rm{s}}^{ – 1}}$$
Stopping distance, $$s = 10\,{\rm{m}}$$
The stopping distance formula is given by,
$$s = \frac{{{u^2}}}{{2a}}$$
Then acceleration is,
$$a = \frac{{{u^2}}}{{2s}} = \frac{{{{40}^2}}}{{2 \times 10}} = 80\,{\rm{m}}\,{{\rm{s}}^{ – 2}}$$
The retardation generated by the car brake is $$80\,{\rm{m}}\,{{\rm{s}}^{ – 2}}.$$

### Summary

When the driver applies the brakes, the distance the car covers before it comes to a stop is known as stopping distance. The value of stopping distance depends on the speed of the car and the coefficient of friction between the wheels and the road. For retarding $$F$$ acting on the body of mass $$m,$$ It is given by, $$= \frac{{m{u^2}}}{{2F}}.$$
When retardation force is friction force, then stopping distance is given by, $$s = \frac{{{v^2}}}{{2\,\mu g}}.$$ It depends on various factors like weather conditions, soil or road conditions, the technical standard of the vehicle, the condition of the brakes, and the condition of the driver.

### FAQs Based on the Stopping Distance Formula

Below here we have provided some of the most asked questions related to the stopping distance formula:

Q.1. What is the formula to calculate the stopping distance of a vehicle?
Ans:
Stopping distance formula is given by,
$$s = \frac{{{u^2}}}{{2a}}$$
Where $$u$$ – The initial velocity of the vehicle
$$a$$-is the retardation.

Q.2. How the stopping distance depends on the coefficient of friction?
Ans:
Stopping distance formula can be written as,
$$s = \frac{{{u^2}}}{{2\mu g}}$$
Where $$u$$ – The initial velocity of the vehicle
$$\mu$$ – The coefficient of friction
$$g$$ – The acceleration due to gravity
Thus, we can say that stopping distance is inversely proportional to the coefficient of friction.

Q.3. What is reaction distance, and how it is helpful?
Ans:
The reaction distance is the distance you travel from the point of detecting a hazard until you begin braking or swerving.
The reaction distance can be decreased by
1. Anticipation of hazards.
2. Preparedness.

Q.4. What is braking distance, and what are those factors which affect it?
Ans:
The braking distance is the distance the car travels from the point when you start braking until the car stands still.
The braking distance is affected by
1. The vehicle’s speed (quadratic increase; “raised to the power of $$2$$”):
Weather conditions, Soil or road conditions, the technical standard of the vehicle, condition of the brakes, condition of the driver (e.g., reaction time due to fatigue possibly higher than $$1$$ second)