# Total Stopping Distance at 80mph: How do you calculate it?

• i.am.lost
In summary: The braking distance, however, is not linear due to varying deceleration with speed. So when doubling the speed, the thinking distance only doubles, but the braking distance quadruples. Therefore, the total stopping distance at 80mph is approximately 144m.In summary, the conversation discusses how to calculate the total stopping distance for a car traveling at 80mph, given a table of stopping distances at various speeds. Multiple methods are proposed, including using an approximation formula, a mathematical forecast formula, finding the average increase between known distances, using the time and velocity formulas, and using the kinetic energy formula. The individual arrives at the solution by realizing that the concept of doubling the velocity quadruples the kinetic energy applies to the braking distance, not
i.am.lost
How do i work out the total stopping distance for a car at 80mph given that:
SPEED ... TOTAL STOPPING DISTANCE
20mph ... 12m
30mph ... 23m
40mph ... 36m
50mph ... 53m
60mph ... 73m
70mph ... 96m
Note that mph denotes miles per hour and m metres

The above figures are widely agreed on (as stated in the Highway Code of Great Britain), but it stops at 70mph and I must now work out the total stopping ditance at 80mph. Please give formula(e), as I am totally stumped.
I have tried the following methods, so please tell me if any are correct (d = total stopping distance):

1) Using the APPROXIMATION formula:
d = (v^2 / 20) + v
where v is velocity

2) Using the mathematical forecast formula which I want to avoid:
d (80mph as x in this formula ONLY) = a + bx
where a = AVERAGEy - (b * AVERAGEx)
and b = (SUM((x - AVERAGEx)*(y - AVERAGEy))) / (SUM((x - AVERAGEx)^2))
using the different speeds in mph as x and the known stopping distances as y
by SUM i mean sigma - so each value of x and y have to be subbed in

3) Finding the average increase between each known stopping sitance, and adding it to 70mph

4) Using t = (v - u) / a
where t is the time taken to stop
v is the final velocity (always 0 in this case)
u is the starting velocity
and a is the acceleration

Then using AVERAGEv = (v + u) / 2
where v is the final velocity (always 0 in this case)
and u is the starting velocity

And substituting the value of AVERAGEv and t in d = AVERAGEv * t

5) Using KE = 0.5 * m * v^2
where m is the mass of the vehicle
and v is the starting velocity (i think it should be starting)

And then substituting the value for kinetic energy in d = WD / f
where WD is the work done (equal to kinetic energy i think)
and f is the force applied to the brakes

The best ones i think are 4 and 5 (esp. 4), but the problem is i have so many unknowns, and when i try to work my way back to get acceleration, for example, i get different values for every known stopping distance. So now i am totally stumped.

Hmmm... I've just had a flash of lightning but it might just be me. Using KE = 0.5 * m * v^2 it is obvious that doubling the velocity quadruples the kinetic energy. Hence, the braking distance is quadrupled.
This is proven in my figures as the stopping distance at 20mph is 12m, and at 40mph it is 36m (36 / 12 = 4).
It is also kind of disproven in the figures as the stopping distance at 60mph is not quadruple 30mph (73m and 23m, respectively).
Can anyone explain why this is the case for the 60/30mph? And is my theory right?

Final Word:
Woohoo! I have now finally solved it. The concept introduced in the addendum is actually correct, but it applies to the BRAKING distance not the total stopping distance - which includes the thinking distance. So by multiplying the braking distance at 40mph by 4, and then finding the trend for the thinking distance (which is simple - t(n) = 3n). By taking the sum of the braking and thinking distances, i have now arrived at the total stopping distance at 80mph!

Last edited:
Fitting the data in the table to a quadratic:

d = .015714 v^2 + .26286 v + .60000

or if v is in feet / second and d in feet

29.333 39.37
44.000 75.46
58.667 118.11
73.333 173.88
88.000 239.50
102.667 314.96

d = .023968 x^2 + .5878 x + 1.9740

1.9740 feet = probabaly due to ronding in the table.

.5878 is reaction time

braking decleration = 20.86 ft / sec^2 = .648g

My guess is that the table assumed braking deceleration of .65 g and .6 second reaction time and rounded to the nearest meter. Most of the simplified braking distance charts assume braking deceleration of .7g and vary on reaction time. In the real world, aerydynamic drag is a factor, so deceleration isn't a constant at higher speeds, and an integral would be required. Also some cars decelerate stop much better than .7g's especially with ABS.

Last edited:
Final Word:
Woohoo! I have now finally solved it. The concept introduced in the addendum is actually correct, but it applies to the BRAKING distance not the total stopping distance - which includes the thinking distance. So by multiplying the braking distance at 40mph by 4, and then finding the trend for the thinking distance (which is simple - t(n) = 3n). By taking the sum of the braking and thinking distances, i have now arrived at the total stopping distance at 80mph!

Yes - the thinking time is assumed constant, so the thinking distance is linear.

## What is stopping distance for a car?

Stopping distance for a car is the distance that a car travels from the moment the brakes are applied to the moment the car comes to a complete stop. It includes both the distance traveled while the driver reacts to the need to stop and the distance traveled while the car is braking.

## What factors affect the stopping distance of a car?

The stopping distance of a car is affected by several factors, including the speed of the car, the condition of the road surface, the condition of the tires, and the condition of the brakes. In addition, factors such as weather conditions and the weight of the car can also impact stopping distance.

## How can the stopping distance of a car be calculated?

The stopping distance of a car can be calculated by adding the driver reaction distance and the braking distance. The driver reaction distance is the distance the car travels while the driver reacts to the need to stop, and it is equal to the speed of the car divided by 2. The braking distance is the distance the car travels while braking, and it is equal to the square of the speed of the car divided by the coefficient of friction multiplied by 2.

## What is the coefficient of friction and how does it impact stopping distance?

The coefficient of friction is a measure of how well two surfaces grip each other. In the context of stopping distance for a car, it refers to the grip between the tires and the road surface. A higher coefficient of friction means that the tires have a better grip on the road, which can result in a shorter stopping distance. Factors such as road conditions, tire pressure, and tire tread depth can impact the coefficient of friction.

## What are some tips for reducing stopping distance while driving?

Some tips for reducing stopping distance while driving include maintaining a safe speed, keeping a safe distance from other cars, regularly checking and maintaining tire pressure and tread depth, and keeping brakes in good condition. In addition, practicing defensive driving techniques and staying alert and focused while driving can also help reduce stopping distance.

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