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Re: [Phys-l] stopping distance




There is an interesting human perception sidelight to this issue.

Since the force is linear in distance but the energy decreases as the
square of the velocity (draw the graphs), the perception is, especially
for braking from higher speeds quickly, that the car will not stop in
time. People then tend to push harder on the brake pedal and go "phew"
when the car comes to a "quicker" stop in the final seconds.

This became an issue for fighter pilots who would think that their
landing airplane would not stop in time and would try to brake harder,
thus causing extra wear on the brakes. They would report, "thank
goodness that the brakes finally started working," probably using more
colorful language.

Cheers,

Rick

Richard E. Swanson, Ph.D.
Dean of Instruction
Physics Professor
Sandhills Community College, Pinehurst, NC 28374
swansonr@sandhills.edu (910) 695-3715

"Anthony Lapinski" <Anthony_Lapinski@pds.org> 1/5/2007 8:50:05 am

Using the Work-Energy Theorem, you can determine how the stopping
distance
is related to the car's initial velocity:

W = KE
fd = 0.5mv2
mmgd = 0.5mv2
mgd = 0.5v2

If we assume the brakes "lock" the wheels to create a skid mark, then
the
above result shows that the stopping distance does NOT depend on the
car's
mass.

However, students tell me and Driver Manuals often state that the
weight
of the vehicle IS a factor (heavier = longer stopping distance). In
the
above calculation, I am dealing with an idealized case, but in
"real-life"
anti-lock brakes change the actual stopping distance depending on the
weight of the vehicle. I've also heard that large trucks have a unique
braking system. Can anyone elaborate on these ideas?

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