OPENING QUESTIONS:

PLEASE GRAB YOUR ASSIGNED CHROMIE!

Consider the following equation showing the position of a particle over time:

1) x = t² + 2t + 4

What are two methods you could employ to determine when the velocity of that particle is zero?

2) How do you find an equation for velocity vs. time for that particle?

3) Try using your calculator to graph the equation.

4) What does that graph tell you without doing any mathematics at all?

5) Any thoughts on how you might use a table of data to find the equation of a curve using your calculator?

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APC LEARNING TARGETS:

B) Students should understand the special case of motion with constant acceleration, so they can:

(1) Write down expressions for velocity and position as functions of time, and identify or sketch graphs of these quantities. (B & C 1.3)

(2) Use the equations below and to solve problems involving one-dimensional motion with constant acceleration. (B & C 1.4)

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FORMULAE OBJECTUS:

0) vt=x (we won't EVER use this for vertical motion--- why not?)

1) v_f = v_i +at

2) v_avg = (v_i + v_f)/2

3*) x_f = x_i + v_it + 1/2at²

4*) v_f² - v_i² = 2a∆x

 

WORK O' TODAY:

Please review the frozen pond problem

Let's review the LOGIC for solving yesterdays vertical motion problems (Did you get my email???) starting from when our person first throws the ball up in the air

1) Evaluating Motion at Point B:

• Recognize this is the first part of vertical motion--- "What goes up..."
• the ball began motion at v_max and rises straight up until v_final = 0. The object is at its max height at that time.
• Use v_f² - v_i² = 2a∆y to find ∆y (which is really h_max)
• Now use v_f = v_i +at to find time to reach h_max

2) Evaluating Motion at Point C:

• Recognize this is the second part of vertical motion--- "Must come down..."
• The height of the ball at point C is the same as the height for the ball at A. Therefore the time to go from point A to point B is THE SAME as the time for the ball to go from B to C... Make sure you understand that!
• We begin analyzing the motion of the ball from point B to point C when the velocity of the ball is zero...so v_i = 0.
• We can find the velocity of the ball at point c using v_f² - v_i² = 2a∆y

3) Evaluating Motion at Point D:

• I rewrote the problem to say that we KNOW the time after launch to point D is 5.0 seconds
• We subtract the time for the ball to go up to point B (2.04 seconds) which means the ball has been falling for 2.96 seconds from point B.
• Use v_f = v_i +at to find v_f at point D.
• Use y_f = y_i + v_it + 1/2at² to find how far down from D the ball has fallen (hint: it is VERY helpful to define y_i = 0 here. You can use h_max as well, but that just means more math!

4) Evaluating Motion at Point E:

• The ball began falling at hmax where vi = 0. The TOTAL distance the ball falls is the height of the building plus the height the ball flew initially to get to hmax.
• Use v_f² - v_i² = 2a∆y to solve for vf.
• Use v_f = v_i +at to find total flight time from point B to E.

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One of the key skills I'm going to teach you this year is the importance of avoiding what I call "pattern matching" when you solve problems in physics-- That means don't immediately start gathering formulas and chugging in numbers and hoping for the best.

Take a listen to this on WHY it is sooooo important to avoid pattern matching (the problem looks like this, I have formulas for that so I'll plug and chug even though I'm not entirely sure what's going on) or read the story here

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Please make a NICE BIG NOTE TO SELF in your notes for today. Please remind yourself that our equations of motion are for CONSTANT ACCELERATION only.

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Now let's take a look at example 2.8 on page 39.

A very, very good way to learn this material is to take an example and work through it step by step.

Let's do that together...

Step #1: Read the problem through once to get a 'flavor' of the situation described

Step #2: After we identify this problem as a motion problem, look for key works, phrases or implications that identify whether this is a one-dimensional, two-dimension or three-dimensional (VERY UNLIKELY) motion problem.

Step #3: Look for key "weasel" words that stand out or seem a little bit out of place... Ignoring those terms is very, very perilous.

• Which term or phrase seems a little bit out of place?

Step #4: You might be tempted to jump right in and practice 'pattern matching'. Which is to say the problem looks like something that fits with some formula so you write down a formula, plug in values, plug and chug and hope for the best. NO NO NO NO

(You might not like this part, but trust me, as you're learning this can be very, very important)

• Write down ALL of your initial conditions
• Be sure to include relevant missing information
• Make some sort of note so you are CERTAIN you know what the problem is asking you to find.
• NOW right yourself a little qualitative statement that clarifies the physics of the situation at all relevant points

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My solution:

Initial conditions and assumptions

V_car,i = 45.0 m/s (constant)

V_m,i = 0.0 m/s

V_m,f = ?? m/s

A_m = 3.00 m/s² (constant)

Find the time for the motorcycle to catch up to the car.

I'M NOT GONNA PLUG AND CHUG 'CAUSE MR W SAID THAT's A NO-NO!

I know the car is moving at a constant rate, and I know the motorcycle is accelerating uniform ally in order to catch up.

The problem seems to indicate that both the car and the motorcycle are moving in the same direction and only in 1 direction. There is nothing in the problem to contradict that, so I MUST assume this is motion in one direction and I can use my formulas for that.

I also know that I have to find a key to what ties the motion of the motor cycle to the motion of the car. My options seem to be:

a) To find out when both objects are going the same velocity

b) To find when both objects have gone the same distance

c) To find when both objects have been moving the same length of time relative to the origin.

There is also something going on with that whole 1 second later thing that is kinda fishy... I'll come back to that later.

NOW Close your books.

Work with your groupies to solve the problem using my qualitative analysis approach first.

THEN go back and compare what you did with what the author suggests... DO NOT assume you are wrong just because your approach is different... you may be completely accurate or very, very close.

Rationale:

a) Won't work because both objects can easily be going the same velocity but not be in the same place

b) Looks promising because if both cars are at the same position relative to the origin then the motorcycle has caught the car.

c) Doesn't look good, because both car and motorcycle are always traveling the same amount of time since the origin so that won't help.

Let's take a look at b.

I have an equation of motion for an object moving in one dimension at a constant velocity. So let's examine that:

One of the things that can drive you nuts is the proliferation of formulas. YOU DO NOT NEED TO MEMORIZE THEM ALL. The book suggests using the following formula (2.7 or something):

x_f,car = x_i,car +vt

BUT I HATE memorizing silly partial formulas, so I'm gonna use one of my big guns:

x_i,car + (v_i,car )(t) + 1/2(a_car)t²

 

The motorcycle is NOT moving at a constant velocity but it is moving at a constant acceleration so I can use this:

x_i,motorcycle + (v_i,motorcycle )(t) + 1/2(a_motorcycle)t²

However, this option works for us only when the distance traveled for BOTH vehicles is the same, so x_f,car will equal x_f,motorcycle

or

x_i,car + (v_i,car )(t) + 1/2(a_car)t² = x_i,motorcycle + (v_i,motorcycle )(t) + 1/2(a_motorcycle)t²

However, there are several initial values that are 0, so I'm going to take those out and revaluate

x_i,car + (v_i,car )(t) + 0 = 0 + 0 + 1/2(a_motorcycle)t²

Which is pretty much where the big ends up...

Follow the book from there