# Re: Repeal of the National Speed Limit Law

```In article <3ten51\$3d4@curly.cc.emory.edu>, lparker@curly.cc.emory.edu (Lloyd R.
Parker) says:
>
>michaelj@relay.relay.com wrote:
>
>: First of all, KE is not exponentially increasing. Energy is work per unit
>: time, and work is force times distance, which means that KE has to be
>: expressed in units of force times velocity, not mass times velocity squared.
>
>Sorry, but an object in motion possesses an amount of KE equal to 1/2
>times its mass times its velocity squared.  Basic physics.  To take the
>object from 10 mph to 0 mph requires removing a certain amount of KE
>equal to 1/2 times the mass times 100 (1/2m10^2 - 1/2m0^2).  To take the
>object from 20 mph to 10 mph requires removing KE equal to 1/2 times the
>mass times 300 (1/2m20^2 - 1/2m10^2).
>
>The amount of force you have to use is not a constant -- it will depend
>upon how much KE you have to remove.

Hmm.. we really need to do something about these lloyderers. Care to
cite your reference Lloyd? Unfortunately, I don't have a physics text
here, but in this case that hardly matters.

Allow me to reiterate. When attempting to make calculations in
physics (and chemistry) it is important to remember that if the units
don't come out right, you've made a mistake or some invalid assumptions.

W = F * D                                                      (1)

Work is force times distance. Pretty basic comic book physics.

E = W / T                                                      (2)

Energy is work over time. Again, basic comic book physics. If we assign
units to the values on the right (F in Newtons, D in Meters, T in
seconds) we have:

Energy = Newtons * Meters / Second                             (3)

What this means is that for any change in velocity the energy used to
effect that change is directly proportional to the force applied to the
object and is directly proportional to the change in velocity. If you
apply a small amount of force, the energy used will be small (and
incidentally, the velocity change will take a long time). If you apply a
large amount of force, the energy used will be higher and the elapsed time
will be less for the same velocity change.

This is why it takes more energy to accelerate quickly from 0-60 mph than it
does to accelerate slowly from 0-60 mph. When you finally get to 60 mph your
car has the same kinetic energy relative to the Earth, but the energy you
expended accelerating can vary.

To talk about kinetic energy as though it were an absolute quantity is
just plain wrong. Kinetic energy is always measured relative to a frame
of reference. For example, suppose I am driving 75 mph and am overtaking
a car that is moving at 65 mph. I have a given kinetic energy with
respect to that other car. If I slow to match speeds with that other
car, I now have kinetic energy equal zero with respect to that other car.

Or another example, the Solar system (and Earth) are moving through space at
some fixed velocity. With respect to some hypothetical observer, the kinetic
energy of a car stationary on Earth is going to be extremely large. Now,
suppose that the Solar system hit some (relatively) dense interstellar gas
and slowed down by a few 100 km/hr. Does this mean that suddenly cars on the
planet Earth would be able to stop faster? Somehow I don't think so.

We can express this in another way. Let's take a look at the basic equation
for acceleration (or deceleration, which is negative acceleration):

F = MA

Force equals mass times acceleration. It is valid to average the acceleration
if you know the beginning velocity, the ending velocity, and the time involved.
So we're decelerating from X mph to X-5 mph. Let's assume that we always use
enough force to do this in 3 seconds. To make the units match, we can convert
some mph figures to feet/second:

75 mph = 110 ft/sec
65 mph =  95 ft/sec

We can also assume that the mass of the car is constant. Let's assume that it
is 1000 Kg (a convenient number). So we have a velocity change (Vd) of 15
ft/sec over a 3 second interval, giving us an acceleration of 5 ft/sec/sec.
We can plug in those numbers, giving us:

F = 1000 Kg * 5 ft/sec/sec

Or

F = 5000 units of force

You can do the same math using different starting and ending velocities and
the result will be the same. The important thing is not how fast you are going
when you start, it's the change in velocity.

We can rearrange the F = MA equation to express it in terms of elapsed time,
in which case:

T = M * Vd / F

Or, elapsed time equals Mass times (Velocity change) over Force. What this says
is that for a constant mass, a constant force, and a fixed change in velocity,
the elapsed time is always the same. If you increase the force, the time
decreases. If you increase the mass or the velocity change, the time
increases.

The energy used to effect this velocity change can be calculated using:

E = F * Vd

If we plug in the numbers:

E = 5000 kg ft/sec/sec * 15 ft/sec

Which gives us:
2     3
E = 75000 kg ft / sec

Or to change units:
2
E = 75000 kg (ft/sec) / sec

Hm.. there's your velocity squared. But wait a second, how did that
extra time value get in there?  Let's do a little reasoned analysis of
this equation. A reasonable interpretation of this would be that for a given
mass and a given velocity change, that the energy used would be directly
proportional to the mass, directly proportional to the square of the velocity
CHANGE and inversely proportional to the time it takes to effect that change.
We already know that energy is inversely proportional to time, from the
equation E = W / T so that would mean that the work done in slowing a car by a
given velocity would be given by the mass of the car times the square of the
velocity change. Always.

So obviously something is missing in the kinetic energy picture. I'm not
going to try to guess what it is, I'd rather go dig out my physics text,
which is at home. It may be that Jeff and Lloyd are applying the KE equation
in ways that don't apply (gee, do you think so?). It may be that they don't
have the right equation. I can't verify that at the moment.

Also, please note that when you increase your velocity relative to a frame
of reference, your kinetic energy always increases. When you decrease your
velocity relative to a frame of reference, your kinetic energy always
decreases. But note also that it always takes a POSITIVE amount of energy
to effect a velocity change, either positive or negative. That would suggest
that what you are doing when you press the brake pedal is not absorbing the
kinetic energy of the vehicle. So where does the kinetic energy go? Actually,
it is dissipated as heat and noise. That's why the brakes heat up and sometimes
squeal.

Michael Johnson, Relay Technology, Inc.
michael@maine.maine.edu, michaelj@relay.relay.com

```

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