Oregon State University
Physics 201
,
Fall 2009
HW12 (due Oct. 12 at 5:00 p.m.)
Page
1
Oregon State University
Physics 201
Fall Term,
l
2009
HW12 Solutions
In these solutions, I try to “think out loud” for you, so I use more words and explanation than you need to put in your own so
lutions.
The points awarded (shown only for the problems scored this week)
are noted near the “vital steps” that you should
have in some recognizable form; these add up to 10 points.
Note that diagrams are often part of the points for setup.
Note
also that I try to avoid putting in most numbers as long as I can—only about 1 point in 10 is awarded for the correct answer,
including units and significant figures.
If you put numbers in earlier, you should keep MORE than 3 significant digits in your
intermediate results.
(If you donʼt, then youʼll propagate error by using a rounded intermediate result.)
Youʼll see that when
I do use intermediate results (because sometime it is more convenient), I keep a couple of extra digits.
Finally, keep in mind
that with kinematics,
there is always more than valid path to the solution
.
I have chosen one I thought was convenient.
1.
Youʼre driving at a speed of 28.0 m/s eastward along a straight, level road at night.
Suddenly
you see the shining eyes of a deer standing in the roadway ahead.
You slam on the brakes and
manage to stop your car just 6.40 m west of the terrified deer.
If your reaction time was 0.365
s, and you first saw the deer from a distance of 120 m, what was the average acceleration ap
plied by your brakes?
You MUST break the analysis up into two parts—because the carʼs acceleration
changes
.
There
are two intervals in which the acceleration is constant, but that constant value is different in the first
interval than in the second.
The first interval is the time it takes you to react to seeing the deer—just
get your foot to the brake pedal.
During this time, the car maintains its previous velocity; the ac
celeration is zero.
During the second interval, the brakes are actually applied, and the car is slowing.
The braking acceleration is constant (assumed here—an average); this is what you want to calculate.
Interval 1
x
i
= 0
(Traveling at velocity v
i
, you see the deer here.)
x
f
= ?
(Your foot hits the brake here.)
(Deer is here at x = 120.)
x
Interval 2: x
i
= ?
(You brake steadily from an initial velocity v
i
, starting there...
... until you stop here (x
(
f
x
= 113.6
).
First, do a quick and easy analysis of Interval 1, measuring time,
Δ
t
(and displacement,
Δ
x
) from the
moment you see the deer (120 m away) until your foot hits the brake pedal—your reaction distance.
a
= 0 in this interval, so
Δ
x
=
v
i
(
Δ
t
) + (
1
/
2
)
a
(
Δ
t
)
2
reduces to
Δ
x
=
v
i
(
Δ
t
)
(“distance = rate x time”).
In this interval,
v
i
= 28.0, and
Δ
t
= 0.365, so
Δ
x
= (28.0)(0.365) = 10.220 m.
Thus (using the
x
coordinate system youʼve defined), the car starts braking at
x
i
= 10.22 and comes
to a complete halt at
x
f
= 113.6 (thatʼs 6.40 m short of hitting the deer at
x
= 120).
The displacement
during braking is therefore
Δ
x
=
x
f
–
x
i
= 113.6 – 10.22 = 103.38 m.
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 Fall '08
 Staff
 Physics, Acceleration, Velocity, ΔT

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