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 The Physics Behind Stopping a Car

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Braking Distance | Stopping Distance Equation | Stopping Distance Tables
Calculator | Common Misconceptions | Conclusion
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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?

  1. 10 feet.
  2. 20 feet.
  3. 40 feet.
  4. 80 feet.

The answer, which surprises nearly everyone, is (d) 80 feet (on dry, level pavement and neglecting driver reaction time). 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:
  • $d$ = Total stopping distance (reaction + braking), meters.
  • $v$ = Vehicle speed, kilometers/hour.
  • $r$ = Driver reaction time, seconds.
  • $b$ = Braking coefficient factor.
Notes:
  • 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 time (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 Mass

For 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 Area

At 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 Distance

Large 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

The "take-home": Virtually no one realizes that a car's braking distance increases as the square of its speed (disregarding reaction time). Ordinarily, not knowing physics and math is only inconvenient — but for car stopping problems it can get you killed.

Reader Feedback

Vehicle stopping distance in inclement conditions

Thank you for providing such clear explanation on stopping distance. It will definitely inform my writing.

Are aware of common additional factors that would be included in the case of rain or snow while driving? Certainly there are a large number of variables that cannot be easily tabulated outside of test conditions. I'm trying to discover if there is a general maxim that could be suggested about stopping in certain conditions.

Example - if it takes approximately 200 feet to stop an average car on clear, flat, dry pavement using average braking power, could we establish a corollary that generally describes the stopping distance for other conditions like "Because of the XYZ variables, driving in wet conditions requires 1.8 times the stopping distance as in dry"?

One cannot reliably do this. Consider the variables:

  • The notorious combination of a pea-gravel surface and anti-lock brakes, the latter of which will glide over the gravel and apply almost no braking force, wrongly calculating that traction has been lost. This combination of factors must be experienced to be believed.
  • The first season's rainfall over pavement coated with a full season of oil buildup from past traffic.
  • New snow on top of a layer of old snow. When this happens in steep terrain, it leads to avalanches. When it happens on roadways, it leads to a false sense of security because the top snow layer looks fresh and pliable, but it hides a slick surface beneath.
  • Black ice, very dangerous and often appearing when the air temperature is well above freezing because the pavement radiates its heat directly into space, disregarding the intervening air's temperature (in physics, radiation is much more efficient than convection).
  • Irregular surfaces with patches of water and hydroplaning effects.

No, these conditions and others mean one cannot say with any certainty what a stopping distance will be on a surface other than dry, level pavement.

I hope this helps.

Braking distance on a slope

Thank you for the information on the mechanics of stopping distances for an average vehicle on a "level", dry surface, tires of average condition etc. But...how does the math/physics change if the surface isn't level but has a slope? Let say 10%. The mass of the vehicle isn't changed. How about the frictional forces? First, for a slope s expressed as a percentage, the angle in degrees is equal to tan-1(s / 100), so for a 10% slope it's 5.71 degrees — call this θ.

The vertical component of the mass (that bears down on the tires and road surface) is on average m cos(θ) (m = vehicle mass), so for the 10% slope case, the effective frictional mass is 99.5% of the level mass. But the vehicle's inertial mass (working to prevent a change in velocity) remains the same. Therefore we already have a factor in the vertical dimension that works against an efficient stop.

Added to this is the effect of the slope. A force proportional to m sin(θ) is added to — or subtracted from — the forces acting on the vehicle and its tires. For 5.71 degrees this is roughly equal to 0.1 m. So for a downhill direction, the effective stopping distance from this factor alone is increased by 10%. I emphasize that this factor can't be evaluated independently of the prior factor ("vertical component"), which has the effect of reducing the vehicle's effective braking mass but without changing its inertial mass.

More formally, for intermediate angles between zero and 90 degrees, the math becomes very tricky because it also depends on the behavior of the car's suspension and its center of mass. The above equations are only applicable — and only approximately — for angles near zero.

All the above becomes practically unworkable if we try to calculate the specific effect on the four individual tires for a vehicle with a high center of mass (the tires nearer the center of mass get greater loading, those farther from the center of mass get less).
Going to the extreme, if a vehicle is in free fall (in a vacuum), there are no fictional forces so at that point it is removed from the equation completely, right? Yes. At that point it's a classic falling object on a ballistic trajectory, no braking force any more. Interestingly, in an environment with less gravitational acceleration like the moon, masses can be lifted against gravity more easily, but they have the same inertia, so getting an object moving (applying an acceleration) on a level frictionless surface requires the same amount of force as on earth. The Apollo astronauts found it surprisingly difficult to adjust to having much less gravitational mass but the same inertial mass — some just fell over. So how does the physics change for a 10% slope? As above. In summary, a simple answer is not available. After computing the above test case, this isn't something I would dream of making a definitive statement about. Consider a top-heavy vehicle or a vehicle that's leaning to the side as well as uphill or downhill — that would prevent any realistic advance estimate of stopping distance.

I hope this helps.

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