Example 6-5 How Far to Stop?
The U.S. National Highway Traffic Safety Administration lists the minimum braking distance for a car traveling at
40.0 mi>h to be 101 ft. If the braking force is the same at all speeds, what is the minimum braking distance for a car
traveling at 65.0 mi>h?
Set Up
Three forces act on the car—the gravitational
force, the normal force exerted by the road, and
the braking force—but if the road is level, only
the braking force does work on the car. (The other
two forces are perpendicular to the displacement,
so Fd cos 90° = 0.) We’ll use Equations 6-2, 6-8,
and 6-9 to find the relationship among the initial
speed vi, the braking force, and the distance the car
must travel to have the final speed be vf = 0.
Solve
Work done by a constant force,
straight-line displacement:
(6-2)
W = Fd cos u
Kinetic energy:
1
(6-8)
K = mv 2
2
Work-energy theorem:
(6-9)
Wnet = Kf 2 Ki
n
d
Fbraking
mg
The car ends up at rest, so its final kinetic energy is
zero. So the work-energy theorem tells us that the net
work done on the car (equal to the work done on it
by the braking force) equals the negative of the initial
kinetic energy.
Work-energy theorem with
final kinetic energy equal to
zero:
1
Wnet = 0 - K i = - mv 2i
2
The braking force of magnitude Fbraking is directed
opposite to the displacement, so u = 180° in the
­expression for the work Wbraking done by this force.
Net work = work done by the braking force:
Wnet = Wbraking
= Fbrakingd cos 180° = 2Fbrakingd
Substitute Wnet = 2Fbrakingd into the work-energy
theorem and solve for the braking distance d. The
result tells us that d is proportional to the square of
the initial speed vi.
Work-energy theorem becomes
1
-Fbraking d = - mv 2i
2
Solve for d:
mv 2i
d =
2Fbraking
We aren’t given the mass m of the car or the magnitude Fbraking of the braking force, but we can
set up a ratio between the values of distance for
vi1 = 40.0 mi>h and vi2 = 65.0 mi>h.
Use this expression to set up a ratio for the value of d at two
different initial speeds. The mass of the car and the braking
force are the same for both initial speeds:
d
vi
v=0
v 2i2
d2
= 2
d1
v i1
Solve for the stopping distance d2 corresponding to
vi2 = 65.0 mi>h:
d2 = d1
v 2i2
v 2i1
= 267 ft
Reflect
= 1101 ft2 a
65.0 mi>h
40.0 mi>h
b
2
This example demonstrates what every driver training course stresses: The faster you travel, the more distance you
should leave between you and the car ahead in case an emergency stop is needed. Increasing your speed by 62.5%
(from 40.0 to 65.0 mi>h) increases your stopping distance by 164% (from 101 ft to 267 ft). The reason is that stopping
distance is proportional to kinetic energy, which is proportional to the square of the speed.