It's common knowledge that large, heavy trucks can't brake as sharply as small cars. According to the Federal Motor Carrier Safety Administration:

The average stopping distance for a loaded tractor-trailer traveling at 55 mph (in ideal conditions) is 196 feet, compared with 133 feet for a passenger vehicle

But why should this be? For a mass of $m$ and coefficient of friction $\mu$, the force of friction (between tires and road) is $\mu mg$, so the acceleration $\mu g$ is theoretically independent of mass, which means the stopping distance is also theoretically independent of mass.

Looking around the web for answers to this question, I see a variety of explanations:

  • Often it is claimed it's just because of the greater momentum of the truck. This doesn't make sense, because under ideal conditions the acceleration due to friction would be independent of mass.
  • Partly it's due to air brake lag. Unlike hydraulic brakes in passenger cars, air brakes require significant time to engage. According to this cdl training page, air brake lag at 55 mph adds about 32 feet, only enough to explain half the difference in total stopping distance.
  • I've seen it claimed that not all wheels would have traction during a hard stop. But theoretically, it shouldn't matter how many wheels have traction, should it? As long as at least one braking wheel has traction, the frictional force should be independent of the contact area.
  • Is it because the air brakes aren't able to apply enough force to the wheel or dissipate energy fast enough because the weight-per-wheel ratio on an 18-wheeler is higher than on a 4-wheel passenger car?
  • Is it because the anti-lock brake system isn't as good on a tractor-trailer as on a passenger car, due to the slower response of air brakes? Anti-lock brakes need to detect skidding and rapidly adjust braking force.
  • Is the coefficient of friction lower for tractor-trailer tires? This calculator suggests that might be the case, though I don't see where they are getting their numbers for truck tires vs passenger car tires. It might make sense that a trucking company would choose a harder tire for reduced wear, at the cost of less traction.
  • I've also seen it said that the truck driver is less willing to brake hard because he doesn't want his load to shift or the trailer to jackknife.

What's the best explanation for the difference in stopping distance?

  • $\begingroup$ How heavy is a loaded semi? How many tons per wheel, and how does that compare to a car? Remember that the brakes have to dissipate the energy without overheating. You perhaps have not followed a loaded semi on a long downhill descent in the mountains and smelled the hot brakes. Too hot and the brake fluid boils and the brakes fade or fail. $\endgroup$
    – Jon Custer
    Commented May 24 at 0:20
  • $\begingroup$ The size of the brake pads should be the limiting factor. Brake too quickly, and they burst into flames. $\endgroup$ Commented May 24 at 1:34

2 Answers 2


The stopping distance difference between large trucks and cars is what you'd expect based on the sources cited in the question. If a typical passenger car initially travelling at $55$ mph stops in $133$ feet, and $32$ feet is a typical air brake lag, and the coefficient of friction for large truck tires is $80\%$ that of high performance car tires (from the cited calculator), then the expected stopping distance of a large truck is $$32\,\mathrm{ft}+\frac{133\,\mathrm{ft}}{0.80}=198\,\mathrm{ft}$$ This agrees with the value of $196\,\mathrm{ft}$ given in the question. (The precise agreement is fortuitous. Given the rough inputs, the answer would better be written as $ \sim200\,\mathrm{ft}$.)

This leads to two physics questions:

  • Why do air brakes in trucks have a time lag after the brake pedal is pressed?
  • Why do the tires of large trucks have lower coefficients of friction?

Brake Lag

The brake fluid typically used in a passenger car is almost incompressible, so when the brake pedal is pressed, the pressure is communicated to the brakes at the speed of sound in the fluid. Air, however, is a compressible gas, so when the brake pedal in a large truck is pressed, gas actually has to flow through tubing to build up the pressure needed to activate the brakes. This takes a bit of time.

Tire Coefficient of Friction

An typical American passenger vehicle might weigh 2 tons spread over 4 wheels and travels on average (in the USA) about $12,000$ miles per year. A fully loaded heavy Class-8 semi-trailer truck can weigh 80 tons spread over 18 wheels and travels on average (in the USA) over $60,000$ miles per year. This means truck tires have to support up to $9$ times more weight than a car tire and must be much more robust. According to Chapter 1 of "The Theory of Ground Vehicles"

for heavy truck tires, the high load intensities necessitate the use of tread compounds with high resistance to abrasion, tearing, and crack growth, and with low hysteresis to reduce internal heat generation and rolling resistance. Consequently, natural rubber compounds are widely used for truck tires, although they intrinsically provide lower values of the coefficient of road adhesion, particularly on wet surfaces, than various synthetic rubber compounds universally used for passenger car and racing car tires.

Heavy truck tire friction is also reduced because they are inflated to such high pressures, typically around 100 psi, about three times that of typical passenger car tires. As noted in the answers to "Why do tires with low air pressure experience more friction?", higher air pressure reduces tire friction.


The force of friction of one object sliding over another is given by $F = \mu N$, where $N$ is the force pressing the two objects together.

For a mass, $m$, sliding over a horizontal table, $N = mg$.

$\mu$ is a coefficient for those two objects. Suppose you had two sets of identical objects pressed together with force $N$. You would expect to total friction force to be twice as big as for one object. Or if you glued the objects together, twice the friction force from the new double-sized objects. The point is that $\mu$ depends on the objects.

For cars and trucks, friction is from pressing brake pads against brake rotors. The normal force, $N$ is not $mg$.

If you have 18 wheel trucks with big strong brakes that are very firmly pressed together with high pressure air, you get a lot of friction. Cars have 4 wheels, smaller brakes, and less force. There is less friction.

On the other hand, trucks have a lot more mass than cars. So it takes a bigger force to decelerate them. There is a lot more kinetic energy that has to be converted to heat by the brakes.

You could build brakes big enough to stop a truck in the same distance as a car. But engineers chose not to.

  • $\begingroup$ It's plausible that big enough brake pads are the limiting factor, but where's the confirmation of that? It could just as easily be that stopping is limited by the coefficient of friction between tires and road. For a car, certainly the limit on how fast you can stop is the tire/road friction, not the brake pad friction. We know this because properly functioning brake pads in a car are capable of making the wheels skid (exceeding the maximum amount of static friction between tires and road) if it weren't for the antilock brake system. $\endgroup$
    – causative
    Commented May 24 at 3:11
  • $\begingroup$ Trucks also have antilock brake systems (ABS), which would suggest that truck brake pads are also powerful enough to make the wheels skid without the ABS, which would suggest that tire-on-road friction (and not pad-on-rotor friction) is the limiting factor for trucks the same as it is for cars. (Unless maybe the truck brake pads are only capable of that for a very short time before they overheat). $\endgroup$
    – causative
    Commented May 24 at 3:17

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