Home | | Mathematics | | Share This Page |

Braking Distance

Question: if a car going 20 miles per hour (MPH) requires 20 feet to stop, how much distance is required at 40 MPH?

- 10 feet.
- 20 feet.
- 40 feet.
- 80 feet.
The answer, which surprises nearly everyone, is (d) 80 feet (on dry, level pavement and neglecting driver reaction distance). This is because the energy of a moving car is proportional to its mass times the square of its velocity, based on the kinetic energy equation from physics:

\begin{equation} \displaystyle E_k = \frac{1}{2} m v^2 \end{equation} Where:

- $E_k$ = Kinetic energy, joules
- $m$ = Mass, kilograms
- $v$ = Velocity, meters/second
It turns out that a car's braking distance is proportional to its kinetic energy. The energy is dissipated as heat in the brakes, in the tires and on the road surface — more energy requires more braking distance. This explains why

braking distance increases as the square of a car's speed.

Stopping Distance Equation

We can use the kinetic energy idea, and a knowledge of driver reaction times, to write an equation that predicts car stopping distances ("stopping" distance is the sum of reaction and braking distance). Here is the equation's canonical form:

\begin{equation} d = r v \frac{10}{36} + \frac{v^2}{b} \end{equation} Where:Notes:

- $d$ = Total stopping distance (reaction + braking), meters.
- $v$ = Vehicle speed, kilometers/hour.
- $r$ = Driver reaction time, seconds.
- $b$ = Braking coefficient factor.

- The left-hand side of the equation ($r v \frac{10}{36}$) converts the driver's reaction time into distance traveled during that time.
- The right-hand side of the equation ($\frac{v^2}{b}$) computes braking distance by applying a braking coefficient factor ($b$) to the square of the car's velocity. Assuming dry, level pavement, a typical value for $b$ would be 170, but this is an empirical factor — it's derived from field measurements.
- This equation may be rewritten for non-metric measurement units, but it's simpler and more reliable to convert its arguments and results to/from metric units:

- To convert input velocities from miles per hour (MPH) to KPH, multiply by 1.609344.
- To convert output distances from meters to feet, multiply by 3.28084.
- Remember that this equation provides the car's total stopping distance — the sum of driver reaction distance (left-hand side) and braking distance (right-hand side). To compute these values independently, isolate the equation's sides (driver reaction distance = $r v \frac{10}{36}$, braking distance = $\frac{v^2}{b}$).

Stopping Distance Tables

Here are tables of typical values generated using the above equation and that agree closely with data published by public safety organizations.

Metric units: (KPH, meters):

Imperial units (MPH, feet):

These tables assume dry, level pavement and a driver reaction time of 1.5 seconds. It turns out that, within broad limits and because of the physics of tire friction, the size of one's tires and their loading (from vehicle mass) don't significantly change the outcome for most vehicles (details below in the "Common Misconceptions" section), so the above tables provide reasonably accurate stopping-distance predictions — but the equation provided earlier is more flexible and useful than these tables.

Calculator

This calculator provides results for user-entered speeds, driver reaction times and braking coefficients. Choose input and output units and enter values in those units.

Entries Speed: KPH MPH Reaction Time: Seconds Braking Coefficient: Empirical Results Reaction Distance: Meters Feet Braking Distance: Reaction+Braking Distance:

Common Misconceptions

Vehicle MassFor a fixed tire size and within reasonable limits, increasing a vehicle's mass shouldn't increase its braking distance. The reason is that the heavier vehicle's tires apply more force to the road — braking effectiveness results from a combination of surface area and force. The increased inertia of the heavier vehicle is balanced by its increased surface force.

Tire Surface AreaAt first glance, one might think increasing the size and road-contact surface area of a tire should improve its braking performance — after all, there's more rubber contacting the road. But as it turns out, for a given vehicle mass, each square meter of a larger tire's surface presses against the road with less force, and (as explained above) braking effectiveness results from a combination of surface area and force. This is why we don't see gigantic tires on the vehicles of safety-conscious drivers — it just doesn't work.

Moving in the other direction, if we make tires too small, the energy of braking would melt their surfaces, destroying their effectiveness. Also small tires tend to wear out more quickly in normal operation, so there's a lower practical limit to tire size.

Truck Braking DistanceLarge truck operators often claim that a large truck must have more braking distance, because stopping a greater mass requires more distance. This is false, and I'm going to prove it below. Once you've read the proof, you will realize the big-truck braking-distance argument makes no sense. Here we go:

Imagine a sport-utility vehicle (SUV) that weighs four tons and has four tires. Its braking distance can be accurately predicted using the stopping distance equation provided earlier.

Compare the SUV with a big truck that weighs 20 tons and has 20 tires. Can this big, heavy truck — five times more massive than the SUV — stop in the same distance? Yes,

it must be so— read on.

Now imagine five four-ton SUVs driving close together, almost touching. If they all apply their brakes at once, each SUV will stop in the same distance as when separated

^{*}.

Now imagine the five SUVs are connected together by metal bars, so they

become one vehicle— a vehicle that weighs 20 tons and has 20 tires. What has changed? Each driver applies his brakes in the same way, therefore the connected assembly of SUVs stops in the same distance that the individual SUVs do when separated.By being connected, the five individual four-ton SUVs have become a vehicle that weighs 20 tons, has 20 tires, and

stops in the same distance as one SUV.Q.E.D.

^{*}It's true that, in present-day reality, big trucks do require more stopping distance than small cars, but the reason is economics, not physics. In principle, big trucks could be designed to stop in the same distance as small cars, if we wanted to pay for the engineering improvements.

Conclusion

Here are this article's "take-homes":

A car's braking distance increases as the square of its speed(disregarding reaction time). Twice as fast, four times the stopping distance.Heavy vehicles with adequate brakes should stop in the same distance as light vehicles, because the heavy vehicle's tires are either more numerous or are pressing down on the road with more force.Ordinarily, not knowing physics and math is only inconvenient, but for car stopping problems it can get you killed.

Reader Feedback

Truck Stopping Distances

Sloping terrain, wet pavement, leaves on the road — Help!

Home | | Mathematics | | Share This Page |