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 4You 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 = v/t

= (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 = v/t

= (-23.5 – 22 m/s)/

9.5s

= - 4.8 m/s2

Revisiting E.02.04, Problem 4, con’t.Assuming that widely-spaced 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 widely-spaced 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 = v/t

= (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 xn 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 xn 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.