SEPTEMBER 17, 2013
Announcements & Reminders
• To the teacher: Turn on the recording!
•
• To students:
• Off-campus students: Raise your hand if your received your tape
measure.
• On-campus students: Raise your hand if you picked up your tape
measure from the DEEP office.
• Use Moodle to answer questions as well as ask them.
• Planning for the week: L105 is due Friday. The lab requires quite a
bit of analysis using Logger Pro. Get an early start. I’ll be available
on WebEx to answer questions about the analysis from 8 – 9 PM
Thursday night. This is an optional session. You’ll get an automated
email notice; however, log in only if you have a question.
More stuff about the quiz
• Be sure to use a computer with working Flash and Java.
There will be a PencilPad problem and some IWP
animations. From my experience, Java does not work with
Chrome.
• The quiz will appear in your WebAssign assignments at
6:00 AM. The item code will be Q01.
• The first item on the quiz is an academic honesty check
off box. To be specific, here’s what academic honesty for
a quiz means:
Academic Honesty on Quizzes and Tests
• The only thing you may have open on your computer is the
•
•
•
•
•
WebAssign quiz itself and any animations linked to quiz questions.
The only other electronic device you may use is a hand calculator.
If you print out the approved AP Physics B equation list in advance,
you may use that.
The only person you may communicate with is your proctor (or with
me in the event that your proctor needs to call me with a tech
issue).
The only person you may get the password from is your proctor.
After the quiz, keep mum until I’ve announced that I’ve released the
key. Besides the fact that some students take the quiz throughout
the day, some students will occasionally take quizzes early or late.
Revisiting E.02.04, Problem 4
You were to
determine the
acceleration from a
position vs. time
graph. Here’s a
snapshot at t = 0.
Note that the run of
the blue tangent line
is always 2.0 s. The
corresponding rise
is (12 – (-32))m =
44 m. Therefore,
the velocity at t = 0
is (44 m)/(2.0 s) =
22 m/s.
Revisiting E.02.04, Problem 4, con’t.
It’s convenient to use
the point at which the
slope, hence the
velocity, is 0. The
average acceleration
calculated using this
point and t = 0 is:
aav = Dv/Dt
= (0 – 22 m/s)/4.6s
= - 4.8 m/s2
Revisiting E.02.04, Problem 4, con’t.
At t = 9.5 s, the velocity
is (-(41) – 6))m/s/2.0s =
-23.5 m/s. The average
acceleration calculated
using t = 0 as the other
point is:
aav = Dv/Dt
= (-23.5 – 22 m/s)/
9.5s
= - 4.8 m/s2
Revisiting E.02.04, Problem 4, con’t.
Assuming that widelyspaced points are used
to increase accuracy
and the graph is read
correctly, why should
one expect to get the
same value of
acceleration for
different pairs of points
selected for the
calculation?
Revisiting E.02.04, Problem 4, con’t.
Assuming that widelyspaced points are used
to increase accuracy
and the graph is read
correctly, why should
one expect to get the
same value of
acceleration for
different pairs of points
selected for the
calculation?
The acceleration is
given to be uniform.
Methods of solving the hare and tortoise problem
(P105)
This was the strategy described in the problem statement:
Equate the positions (xh = xt) to obtain
voht = xot + vott + at2/2
Solve for t using the quadratic formula.
t = {-(vot – voh) +/- sqrt[(vot – voh)2 – 4(at/2)xot]} / [2(at/2)]
The discriminant must be >= 0.
(vot – voh)2 – 4(at/2)xot >= 0
Solving for at,
at <= (vot – voh)2 / (2xot)
Alternative method (or variation) used by 3 students
When the acceleration
of the tortoise is just
enough to catch the
hare, the animals have
the same velocity at the
meeting point.
vh = vt
We know this because
the hare’s x-t graph is
tangent to that of the
tortoise at the meeting
point.
Alternate method, con’t.
The displacements of the hare and the tortoise in time t
are:
Hare: xh = xoh + vav,ht = voht
Tortoise: xt = xot + vav,tt
= xot + [(vot + vt)/2]t
Now we substitute vt = vh = voh.
xt = xot + (vot + voh)t/2
The next step is to equate the displacements and solve for
t.
Alternate method, con’t.
xh = xt
voht = xot + (vot + voh)t/2
t(voh – vot/2 – voh/2) = xot
t(voh/2 – vot/2) = xot
t = 2xot/(voh – vot)
= 2(100. m)/(10.0 – 1.0) m/s
= 22.2 (or 22) s
Alternate method, con’t.
The displacement is simply:
xh = voht = (10.0 m/s)(22.2 s) = 222 m
The acceleration can be found from the definition of
acceleration.
a = Dv/Dt
= (v – vo)/t
at = (vt – vot)/t
Now we substitute vt = voh and t = 2xot/(voh – vot).
at = (voh – vot)/[2xot/(voh – vot)]
= (voh – vot)2/(2xot)
= [(10.0 – 1.0)(m/s)]2/(200. m)
= 0.405 m/s2
Time for questions about the L109 prelab
• A question from Moodle: How do we calculate the Dxn for
the last point on our picture since there is no xn+1?
• Another question: Do we need to include a list of
apparatus in our lab write up?
Time for questions about the L109 prelab
• A question from Moodle: How do we calculate the Dxn for
the last point on our picture since there is no xn+1?
You don’t. The same is true for the first point.
• Another question: Do we need to include a list of
apparatus in our lab write up?
no
Complete the 2nd question on the prelab now
• I’ll reset WebAssign to extend the due time to 10:00 PM
and to let you see whether your answers are correct.
• I recommend printing a paper table and writing you
answers there in case you encounter connection issues
with WebAssign.
• I’ll be online if you have questions. Raise your hand if you
need to get my attention.