Can a Motorcycle Brake Faster Than a Car?

There are two schools of thought on whether a motorcycle can stop faster than a car. On the one hand, a motorcycle is much, much lighter than a car so wouldn’t it stop faster than a heavy car? Everyone expects a truck or train to stop slower than a car due to their mass, so the same probably applies for a car against a bike? On the other hand, cars have four brakes, four larger tires to grip the road with and weight remains on the rear wheels even in a heavy braking situation, so perhaps they stop quicker?

Interestingly, if you do a Google search you’ll find that a straight forward answer isn’t forthcoming. Many swear black and blue that a bike will beat a car every time, yet there’s also a lot of anecdotal evidence to suggest the opposite. So what is the answer? Well, it depends…

The most common view is that because a bike is lighter, not only can it accelerate faster, it can also decelerate quicker too. But this is actually completely wrong as you’re dealing with different physics in each instance. For acceleration, a bikes power to weight ratio is often far superior to that of a car and that’s the reason it accelerates quicker. But power to weight has no bearing on stopping.

Let’s break (get it?) down the physics. I’m relying heavily on a wonderful post on the

Firstly, the force on a vehicle when stopping is:

F = ma

Where F is the force on the vehicle, m is its mass and a is the acceleration (negative in this case).

Next, that force is applied to the tires via traction as follows:

F = μW

Where W is the weight of the vehicle, and μ is the coefficient of friction.

The weight of the vehicle is its mass, m, times the gravitational force, g, so:

F = μmg

The maximum stopping force that can be applied is the maximum frictional force that the tires can cope with, so:

ma = μmg

What all this shows is that we can cancel the use of the variable m as it appears on both sides of the equation. In simple terms, it means that mass has no bearing whatsoever on stopping distances. The equation to thus show the maximum possible deceleration is

a = μg

So (negative) acceleration equals the coefficient of friction of the tires, multiplied by the gravitational force.

So, that light bike you have? When it comes to braking, it’s irrelevant. It’s down to the tires. And because a car has four of them instead of two (and they’re wider), it points towards a car being able to stop quicker. So, cars stop faster than tires. Case closed. Well, not quite…

The reason it’s not so straightforward is that an average bike has far higher performance characteristics than an average car. In general, an average sports bike will have higher quality brakes and better tires than an average hatchback or a family saloon.  So your Yamaha R6 or Triumph Street Triple will probably out brake Miss Daisy in her Toyota Camry. But as soon as you get a car that’s remotely sporty, say a Golf GTI, the car will begin to stop in shorter distance than a bike. Physics will win and with those four tires, the equation of a = μg will see the car pull up quicker.

As our video shows, an average bike will outperform an average car. In the future though we’ll put the same bike up against a hot hatch or similar to show how the reverse becomes true. Regardless of what bike you ride or what cars are around, remember that you are your own crumple zone. Always leave plenty of room for an emergency stop as what would be a small fender bender in a car can potentially become a serious injury (or worse) if you run up the back of someone.


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  • Janne Juslin

    The physics of friction is actually not that simple. In heavy braking all of the bikes weight is on the front wheel, which deforms heavily as it’s pressed aqainst the tarmac. The actual area of contact increases with pressure, caused by “asperities” (roughness). This increases the effect of electrostatic and/or Van der Waals and hydrogen bonding force. The same weight shifting happens with the car far less. Less pressure, less “sticktion”. I quess which has less deceleration is strongly a question of surface roughness and material. Would be really intresting to see real G-values for superbike on street tyres vs. sports hatch on different tarmacs, dry and wet and so on. Would anybody have the resources to drive those tests? The whole motoring world would be most greatful:)

    • Unfortunately we don’t have those resources 🙂 Next time MotoGP is on – i believe they sometimes have a G-meter reading, would be interesting to see what they pull under braking.

  • Chris Barker

    You’re missing the fact that your equation, a=ug, has no component of surface area. So your explanation is actually wrong. Were friction a function of pressure, which is a function of mass and surface area, then you’d be correct. But it’s not. Thus we should expect a bike and a car to brake exactly the same, ASSUMING the same u (tyre compound). In truth, as you do rightly point out, an R6 is likely to have better tyres than a camry, but perhaps less so that a GTI. In my experience, all superbikes have very soft tyres; tyre cost is SO not a consideration when you’re hurtling at 200km/h, and 2 bike tyres are actually quite cheap vs 4 decent car tyres. Having said all that, why then do bikes so often end up in awful crashes due to an inability to brake. Well I put it down to crazy speeds that other drivers fail to calculate correctly. That, plus a lack of ABS in most bikes, means a long ass slide when trying to brake nicely.

  • Kan Tutan

    The average motorbike on the street has:
    -no high performance sport tires
    -no high performance disk brakes
    -no abs (the average motorcycle dealer doesn’t even know what is an ABS)
    -only one wheel in front

    The average motorcyclist on the street has:
    -no proper education on applying brakes in panic stops
    -no regular training on panic stops

    This makes the average motorcyclist a good candidate for brake related accidents.
    It also means an average motorcyclist cannot safely stop his/her motorcycle in panic stops.
    This is why I sent all of my motorcycle drivers to driving schools once they are qualified to work in our company.

    • Kevin Farr

      You are obviously not in Australia. Where it is compulsory to have training to ride a Motorcycle. Part of the test is Emergency braking! Cars drivers do NOT do this kind of training at all. My bike (as do sooooo many these days) has ABS standard (soon it is being legislated as Compulsory). And I personally have never had anything less than the best tyres money can buy (my life depend on it). And I don’t know any motorcycle riders that don’t do the same. I have been riding for over 40 years, on all styles of Road bikes. Yes I have come off but only because of the Mistakes of Car drivers doing stupid stuff (not looking or checking mirrors)

      • Kan Tutan

        Here in the Philippines, I have never encountered a single motorcycle driver that had undergone a motorcycle driver training education.
        On top of the subsidized education, they have to be paid to attend a driver education.

  • normanzb

    based on above theory, as the weight has nothing to do with braking distance, a fully loaded lorry with 8 tires and huge contact patch gonna out brake all the cars? don’t be ridiculous. I don’t understand how this article still here online confuses more people in believing that their stupid car out braking anything else in the world, no matter it is bike (based on above theory in the article) or lorry (based on the real life experiences that they got tailgated all the times by the lorries);.

    learn the physic properly here:
    you will need more force to stop a heavier object than a lighter object, so that is all come down to mass / stopping force ratio. so there isn’t a straight answer to that who will win the match when bike compared with a car, bike is lighter but also has smaller stopping power.

  • normanzb

    what you proofed above is not that “braking distance has nothing to do with mass. ”

    As you used using F = μW in the equation, what it gives you is that the how much force you need to break the tyre’s kinect friction force barrier to make it sliding.

    so what you proofed by using those equations in question is actually “while braking is happening, the mass of the vehicle has nothing to do with the tyre’s ability of holding the vehicle without sliding it”, another words, as long as you brake with a fixed force N and with same set of tyres, the load of the car (heavy or light) will not affect the tyre be locking up or not.

  • normanzb

    For tires, μ decreases as weight increases, so the overall friction increased by weight increase isn’t large enough to compensate the kinetic energy increase caused by gaining weight.

    And also for the same reason, you need a large size tire to increase the grip.

    If weight doesn’t really play a role in stopping distance, then, let me ask, should there be no difference in stopping distance when the same car is fully loaded ?

  • freemba

    Yes, this does not sound correct. If mass had no bearing on the braking distance, you would have the same braking distance for a car without passengers and luggage and the same car which is much lighter and has only a driver. The same logic will then lead you to stating that a lorry would have the same braking distance as a passenger car, if we put it on 4 passenger car tires.. I think you need to take into consideration the effectiveness of brakes, abs or no abs, mastery of the driver, stability of the vehicle while braking, tire specific things, etc. But I am pretty sure that all things equal the braking distance of the bike will be substantially shorter, as the brakes have to deal with a much lower mass / impulse / inertion.