The acceleration you describe 'mg' is the
acceleration due to the pull of the earth on the
object. This is at right angles to the movement of
the object. How could this force slow down the
object?
Dick
Helping teachers who facilitate, motivating
students who learn.
Dick Heckathorn 14665 Pawnee Trail Middleburg
Hts, OH 44130 440-826-0834
www.cvcaroyals.org/~rheckathorn/
Adjunct Physics Teacher - Baldwin Wallace College
Physics is learning how to communicate with ones
environment so that it will talk back.
-----Original Message-----
From: tap-l-owner@lists.ncsu.edu
[mailto:tap-l-owner@lists.ncsu.edu] On Behalf Of
Anthony Lapinski
Sent: Friday, January 05, 2007 8:50 AM
To: tap-l@lists.ncsu.edu; tap-l@lists.ncsu.edu;
phys-l@carnot.physics.buffalo.edu
Subject: [tap-l] stopping distance
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?