In this section, an equation of motion will be written down, from which some results can be derived about the performance of a car.  Before we begin, let us define the following quantities to describe the dimensions of the car under consideration:

            Width of the car:                                                        w (m)

            Height of the car:                                                       h (m)

            Radius of the wheel and tyre:                                   d (m)

            Drag coefficient:                                                        C_D (-)

            Mass of the car:                                                         M (kg)

            Fractional power loss from engine to wheel:           f (-)

            Other frictional forces:                                               F (newtons)

            Velocity of the car:                                                     v (m/sec)

            Gear ratio:                                                                  g (-)

The quantities in brackets are the units in which we specify these quantities.

To write down the equation of motion, all that is needed is to write down the sum of the forces acting on the car, and equate this to the mass ´ acceleration, according to Newton’s second law, given in (3).  There are three forces acting on the car.  The first is the driving force of the engine.  If the torque produced by the engine is t, then the torque at the driving wheels is

taking account of the losses in propagating from the engine and through the clutch, gearbox and final drive, and the torque multiplication that arises as a result of the gearing.  To calculate the force at the wheel, we note that the wheel has a radius d, and so from the definition of torque, the driving force F₁ is simply

As the velocity of the car increases, it will be subjected to aerodynamic drag.  One of the simplest theories of aerodynamic drag gives the resisting force F_D as

To see how this expression comes about, consider the diagram below:

This shows an object moving at velocity v, so that in one second, the volume of air moved is Av, and hence the mass of air moved in one second is rAv, where r  is the density of air.  As a consequence, the air moved in one second acquires a momentum of rAv².  From Newton’s second law, this requires a force equal to rAv².  Now, according to Newton’s third law, to every action, there is an equal and opposite reaction.  That is, the air that is being moved by the object exerts an equal force back on the object.  In other words, the moving object must experience a resisting force equal to rAv².  The factor of ½ arises from a more exact analysis of the problem, and the drag coefficient allows for the fact that cars (and indeed any other object) are not perfectly blunt objects.

We can therefore see that the aerodynamic drag increases by a factor of 4 if the velocity doubles.  Bearing in mind that we need to take account of other frictional forces F, the equation of motion is simply given by:

where the aerodynamic drag and frictional forces are subtracted from the engine output because they resist the motion of the car.

In order to undertake realistic analyses, it should be noted that the engine torque varies with engine speed.  Let R_M be the engine speed in revolutions per minute, so that we can write the torque in the form

.

It is now necessary to express R_M in terms of the velocity v of the car.

First it is convenient to rewrite the engine speed in terms of revolutions per second:

The rate of rotation of the wheels is then equal to R_S/g rotations per second.  Since the radius of the wheel is d, the circumference is 2pd, and so the velocity of the vehicle is given by

The equation of motion can then be written in the following form:

where the brackets in the first term indicate that the torque is a function of the quantity inside the brackets.  The solution of this equation gives the velocity profile of the car as a function of time.  In arriving at this expression, we have made a number of approximations:

1.                  1.  We have assumed that torque and power losses from engine to transmission is represented by a constant factor f (in particular, that it doesn’t vary with engine speed);

2.         2.  We have assumed that the aerodynamic drag is representable in terms of a drag coefficient and a term proportional to velocity squared;

3.         3.  We have assumed that other frictional losses can be represented by a constant term, rather than varying with the speed of the car.

In spite of these assumptions, the above equation should give us a good representation of the behaviour of the car under acceleration.

To solve this equation, let us assume that the initial velocity of the car at time t=0 is v₀.  The solution then takes the general form:

This is not a particularly friendly expression, and to gain some insight into the behaviour of the solution, it is easier to consider some simplified special cases.

On to Solutions Neglecting Friction