Car Physics: some problems

Cars provide examples for several areas in physics. This page uses the car pictured for a few simple examples to illustrate the chapters Constant acceleration, Weight and contact forces and Energy and power. It also has example problems posed to us about cars. We'll add more later.

UNSW's solar racer sunswift IV on its way to winning the silicon division of the transcontinental World Solar Challenge.

• Approximate drag example
  
• Force and power
  
• Stopping example
  
• Air resistance
  
• Drag coefficient
  
• Example using the drag coefficient
• How fast can a car go?
• Wheel stands ('wheelies') and hard braking: discussed on a separate page
• Car collisions: discussed on a separate page

Acceleration, speed, time, distance

First, a warning. The first two examples below assume constant acceleration – which is usually a chapter in introductory physics books. First, the forwards acceleration of a car is rarely constant. Motors rarely deliver force or torque that is independent of speed and drag is a strong function of speed – turbulent drag is approximately proportional to v². Deceleration can often be rather closer to constant, especially in hard braking and at speeds low enough that the drag may be neglected in comparison with the braking force (strictly, the horizontal force exerted on the tires by the road) or the rolling resistance (which is not stronly dependent on speed). Often, however, a rough estimate is sufficient answer, and that's what the first examples below provide. A better approximation is given in a further calculation below. For precise answers, however, still more detail is required, as we indicate.

Second, a note about units. Discussing cars, speeds are usually given in kilometres per hour, for obvious convenience in everyday life. However, accelerations are usually given in m.s⁻² and, in any case, scientists and engineers usually work in SI units. A kilometer is 1000 m and an hour is 3600 s, so one k.p.h. is 1000 m/3600 s, which is (1/3.6) m.s⁻¹. So, to convert k.p.h. to  m.s⁻¹, divide by 3.6.

Approximate drag example (using the constant acceleration approximation)

A coasting trial. On a training day with negligible wind, sunswift IV entered the flat straight of the race track at 70 k.p.h. The driver then switched the motor off and coasted, using no brakes. After covering a marked kilometer, she was travelling at 50 k.p.h. What is her acceleration? (For this approximate estimate, treat the acceleration as constant, but we'll do a better calculation later.) In the chapter Constant acceleration, we considered motion in a straight line and derived an equation relating the acceleration a to the (final) speed v, the initial speed v0 and the distance covered Δx: v2 − v02  =  2aΔx ,    which we rearrange to give a  =  (v2 − v02)/2Δx. Substituting, we obtain a = − 0.093 m.s−2. The minus sign tells us that it is a deceleration (she is slowing down). To put this in context, compare it with the gravitational acceleration: a is about 1% of g. (A solar car has limited power – about one kilowatt. So drag is very important and that's why sunswift IV has this shape.)

Force and power

Force. From Newton's second law, we write F  =  ma. Including a driver (and ballast, to make the effective driver's mass 80 kg), sunswift's total mass is 244 kg. Substitution gives a force of − 23 N.

As we mentioned above, the drag force is not independent of speed, so neither is the acceleration. Nevertheless, we can say that this is greater than the force at 50 k.p.h. and less than that at 70 k.p.h. For now, let's guess that this is roughly the stopping force at 60 k.p.h. and return to analyse this problem properly later.

If she were travelling at a constant speed of roughly 60 k.p.h. in a straight line, the acceleration would be zero and so would the total horizontal force. So the drive wheel would have to apply a force of magnitude 23 N to the ground. (Only the single back wheel drives, and the motor is in the wheel to minimise losses in the drive train.) This doesn't sound like much: it is about the weight of a 2.3 kg mass, which you can comfortably apply with one finger – to an object at rest. Applying this force at a speed of 60 k.p.h. is less easy. Let's see why.

Power. In the chapter Energy and Power, we saw that the power P applied by a force F applied at angle θ the direction of motion of a point moving at speed v is

P  =  Fv cos θ

Here θ is zero for a horizontal force, so substitution gives P = 380 W. A very fit human can supply 380 W for sustained periods, and rather more for short periods. (sunswift IV can supply rather more than this, too, especially if it's a clear day and the sun is high in the sky. More on this below.)

 



UNSW's solar racer sunswift IV on the Hidden Valley racetrack in Darwin, before the race.

Stopping example

We mentioned before that constant acceleration can be a better approximation for stopping. The solar cars must do stopping and handling tests before qualifying to race. sunswift IV stopped from 35 k.p.h. to rest in 4.5 m. As before, we write
v² − v₀²  =  2aΔx ,    which we rearrange to give
a  =  (v² − v₀²)/2Δx.

Substituting, we obtain a = − 10.5 m.s⁻¹. So her braking acceleration is a little greater than g. With mass 244 kg, this gives a total stopping force of magnitude 2.6 kN

Brakes and friction. Compared with 2.6 kN, the 0.023 kN we calculated above for the coasting trial (at higher speed) may be neglected. So the stopping force is almost entirely that applied by the road to the wheels, which in this case includes regenerative conversion: the wheel motor becomes a generator and turns some of the car's kinetic energy back to electrical energy, which is stored in her battery. (There is also a rolling resistance, but this also is small (rather less than 1% of the braking force.)

As we saw in Weight and Contact Forces, this requires a limiting coefficient of static friction greater than unity. Values as high as 1.2 or 1.3 can be achieved on a clean road in dry conditions, but don't count on being able to stop this rapidly in general!

Air resistance

In many cases, the turbulent drag is proportional to v², to a good approximation. Roughly speaking, the car accelerates a mass of air near it, doing work in the process. In its turbulent wake, the kinetic energy of that air is dissipated as heat. Suppose the car travels dx in time dt. Suppose that, in doing so, the applies a force F to the air, and the magnitude of F equals that of the air resistance F_drag (the force the air applies to the car – Newton's third law). This force F accelerates a volume of air to the speed v of the car, which has cross section A. We'll now look in turn at the questions: What is the mass dm of that air? How much work is done moving it? What power is required? What is the drag force?



How much air is moved? First, imagine a non-aerodynamic vehicle, like a bus. One might expect that, as the bus moves forward dx, it accelerates a volume dV. Let's start with a crude approximation: let's say the volume accelerated is dV = Adx. If the density of the air is ρ, then the mass of air is ρAdx.

How much work is done moving it? By hypothesis, all of that air is accelerated to v, so its kinetic energy is ½mv²  =  ½(ρAdx)v².

Of course not all of that air is accelerated to the speed of the bus: some slips around it. Further, for carefully designed shapes, only a modest fraction might be accelerated – the rest is pushed aside. So let's combine those two effects (fraction of air in front of the vehicle accelerated and the fraction of v to which it is accelerated) in a single constant, C_D, called the drag coefficient. It is a pure number, which we expect to be about one for a bus and rather smaller for an aerodynamic shape like that of the solar racing car. So the work done accelerating the air is ½C_D(ρAdx)v²

What power is required? To get the power required to do this work in time dt, we divide by dt. Putting dx/dt = v, we get:

Power  =  (½C_DρA)v³
Aha, that's why it's so hard to go fast: the power required to overcome turbulent drag goes as the cube of the speed. (If you need to, see the chapter on Energy and Power)

Drag coefficient

What is the drag force? Let's again use the equation P  =  Fv cos θ, where again θ is zero for horizontal forces and velocity. So, dividing both sides by v, we have

F_drag  =  ½C_DρAv²
Which looks fine... except that we don't know C_D: we said above it was 'a number that is about one for a bus and rather smaller for an aerodynamic shape'. In other words, it's what we call a fudge factor. This particular fudge factor is called the drag coefficient. It is defined by the equation just derived, i.e. C_D  =  F_drag/(½ρAv²). However, although it looks like a tautology, it's still a useful equation, because C_D is fairly constant – it depends only weakly on v. So, once we know C_D from data on one or more speeds, we can predict the drag force for other speeds.

An example using the drag coefficient

Let's go back to the coasting example from above. We know the initial and final velocities andwe know the distance covered. Can we work out the drag coefficient? Assume that drag is the only horizontal force. (For a car as slippery as this one, this approximation is only good at high speed. When the drag is low enough, the rolling resistance is non-negligible. Let's put that aside for now.) So, as the car slows, we now know that the drag force and therefore acceleration are not constant. In fact, we have
F_drag  =  ½ρC_DAv²
Now this is the decelerating force, so we can write it as −ma, so
ma  =  −½ρC_DAv²
To make it look a bit simpler, we'll gather the constants together and put   ½ρC_DA/m  =  k, where k is just a constant of convenience, defined by this equation. While we're at it, let's write a as dv/dt: this will give an equation that has v on both sides.
dv/dt  =  − kv²
How does this relate to the distance travelled Δx? We could get Δx by integrating vdt. So let's rearrange the preceding equation to put vdt on one side. We'll need to multiply both sides by dt, and divide by kv. That gives us
dx  =  v.dt  =  − (1/k)dv/v

Notice that dx is a length, but dv/v is a pure number. That tells us that 1/k has the dimensions of length: it is the characteristic length in this problem. (You might already guess the meaning of that length.) In physics, we usually use the symbol λ for a characteristic length, so let's write λ  =  1/k  =  2m/ρC_DA, and so

dx  =  v.dt  =  − λdv/v

Now the differential dv/v (or dx/x, or dt/t etc) appears often and is very important. You need to know that

the integral of 1/x  is  ln x ,   so   dx/x  =  d(ln x).
(Go to Introduction to Calculus if you need some revision.) So now we substitute for dv/v:
dx  =  − λd(ln v)
We integrate both sides to get Δx on the left, so
Δx  =  − λ ln v + constant of integration.

The constant of integration is easy: when Δx  =  0,   v  =  v₀,   so

Δx  =  − λ ln v + λ ln v₀.

This is our answer, but we can make it rather neater. Remembering that   ln a − ln b  =  ln (a/b) , we have

Δx  =  − λ ln (v/v₀)   and, taking antilogs, this gives
v  =  v₀ e^− Δx/λ.
Of course, if you've read the page on Dimensions and Units, you will now want to check that λ really is a length.
λ  =  1/k  =  2m/ρC_DA.
m is in kg, ρ is in kg.m⁻³, A is in m² and C_D is a number, so yes, our equation is dimensionally correct.

Looking at v  =  v₀ e^− Δx/λ , what can we say? Well, it's no surprise that the speed decreases exponentially (though notice that we have an exponential in distance, not time). The characteristic length λ is the distance it takes for the speed to fall from v₀ to v₀/e ; after 2λ, the speed has fallen to v₀/e² , etc. Looking at the expression

λ  =  2m/(ρC_DA),

we see that a car will coast further if its mass is greater: a more massive car starts off with more kinetic energy so the air must do more work to stop it. We also see that we can increase the coasting distance by lowering the density (e.g. doing the experiment at high altitude) , which of course reduces the air resistance. We can also increase the coasting distance by reducing the cross sectional area A, and by reducing the drag coefficient C_D. Which explains why sunswift IV looks like the photos shown above. Let's now be quantitative, using the data in the problem above, for which v  =  50 k.p.h. , v₀  =  70 k.p.h. , and Δx   =  1 km. Above, we had the equation

Δx  =  − λ ln (v/v₀)
which we can rearrrange to give
λ  =  Δx /ln (v₀/v)  =  (1 km)/ln (70 k.p.h./50 k.p.h.)  =  3 km.

So, if the car started at 70 k.p.h. and coasted for 3 km, we predict its speed to be (70 k.p.h.)/e  =  30  k.p.h. , after 6 km, we predict its speed to be (70 k.p.h.)/e²  =  10 k.p.h. , and so on... until when? Here's a puzzle: Our equation Δx  =  − λ ln (v/v₀) predicts that the car would never actually stop! What has gone wrong?

We have assumed that the only stopping force acting on the car is turbulent drag. Because this force goes as the square of the speed, it dominates at high speeds so this is a reasonable (but not excellent) approximation. There is also rolling drag (and, at extremely low speeds, we might even consider viscous drag from the air). The friction in bearings can actually increase at low speeds, as we show in this example from the chaper on Oscillations.

How fast can a car go?

Suppose that a power P is transmitted to the wheels and that that power is used only to overcome aerodynamic drag, so, using the equations above for power P, drag force F_drag and characteristic length λ,

P  =  F_dragv  =  ½C_DρAv²v  =  (m/λ)v³  or

v  =  (Pλ/m)^1/3

For solar cars, the power available depends on whether one uses batteries or not. However, for the solar car land speed record, batteries are not allowed: the power used must be supplied, in real time, only by the sun. If we take P = 1200 W and m = 240 kg, then this last equation predicts a speed of 25 m/s or 89 kph. This is approximately the speed that Sunswift achieved to set the Land Speed Record for solar cars. Yes, 89 kph on the power of a hair-dryer.

Other car physics problems on Physclips
• Wheel stands ('wheelies') and hard braking: discussed on a separate page
• Car collisions: discussed on a separate page