AP Physics C Summer Work

AP Physics C Derivatives and Integrals-Acceleration and Velocity

You probably already know some of what follows from Honors Precal. Velocity is the time derivative of position, and acceleration is tl1e time derivative of velocity. Also, you must integrate an acceleration function to determine the velocity function, and integrate velocity to determine position (it's like going backward from differentiation, which is why inteb>rals are antiderivatives ). Some basic mies for derivatives and integrals follow. They are excerpted from tl1e Red Knight Physics website. Other helpful

websites include: Paul's online math notes, http://tutorial.math.lamar.edu/ and

https://www.mathsisfun.com/calculus/integration-introduction.html.

Basic Differentiation In algebra, you were taught how to find the slope of a straight line, both by interpretation of a linear function (for example, we know that y = 2x + 3 has a slope of 2 because the slope is the coefficient of the xterm) and with the slope formula, (y2 - y 1)/(x2 - x1). However, for non-linear functions, the first method cannot be used, and the second can only provide and average slope over an interval. Calculus provides us with a way to find the "slope" at any point along any function, whether it is linear or not. This is done by measuring the slope of a line tangent to the curve at that point. An example of a few tangent lines is shown below. /~--

,,/ \

Each of the red lines is tangent to the graph of fix) at the corresponding blue point. Notice that the lines seem to follow where the function would be if it continued in either direction with a constant slope. In calculus, we use a method known as differentiation to allow us to find the slopes of these tangent lines. Before we begin learning how to differentiate, it is important to understand a few things. Looking at a linear function, slope is a rate of change. More specifically, it is the rate of change of y with respect to x. In simpler term s, the slope of a line tells us how much they value changes for each increment that x changes. For linear functions , this value remains constant. However, the rate of change of a non-linear function can't be a constant, leaving only one possibility: it is another function . This brings us to differentiation. Differentiation is the process of using a given function to find the function representing its rate of change, called its derivative. For a function fix), the derivative of/{x) is denotedF(x). This may sound complicated, but it's actually a simple process carried out by using a set of simple rules based on the structure of the function. Here, we will only cover simple polynomials, as they are the only types of functions on the AP Physics C exam.

Rule #l: The Constant Rule lf/(x) = c, where c is any constant,f ' (x) = 0 Example: /(x) = I, sof'(x) = 0 Example: _/(x) = -IO, sof'(x) = 0 Example:.f{x) = 3p, sof'(x) = 0 Remember that p is a constant, and so any constant multiplied by p is also a constant. Example:.f{x) = 4(27 + 32), sof'(x) = 0 Note that what 4(27 + 32) evaluates to is irrelevant in finding the derivative because it is still a constant.

Rule #2: The Power Rule If/(x) = x", thenf'(x) = nx"· ' Example:.f{x) = x3

, soF(x) = 3x2

Example:.f{x) = x2, sof'(x) = 2x

Note tliat when tl1ere is no exponent written on x, it is to the 1st power. Example:_/(x) = x, sof'(x) = x0 = I Note tl1at any number to the power of 0 is I, so x0 = 1

Rule #3: The Constant Multiple Rule If/(x) = cg(x), where c is some constant, thenf'(x) = cg'(x) Example:.f{x) = 2x2

, sof'(x) = 2*2x = 4x Example:_/(x) = 5x, sof'(x) = 5 Remember that the derivative ofx by the power mle is 1, and I *5 = 5

Rule #4: The Sum and Difference Rules Jf_/(x) = g(x) + h(x), thenf'(x) = g'(x) + h'(x) If/(x) = g(x) · h(x), thenf'(x) = i((x) - h'(x) Example: _/(x) = x3 + x2

, sof'(x) = 3x2 + 2x Example:.f{x) = 2x2

- 3x, sof'{x) = 4x - 3 Example:.f{x) = 5x4 + 2x3

- x2 - 3x + 5, soF(x) = 20x3 + 6x2

- 2x - 3 Remember that the derivative ofa constant is 0, so any constant terms in.f{x) become 0 inf'(x)

Using just these four simple rules, it is possible to take the derivative of any simple polynomial containing only one variable, like those in the examples for rule #4. Also note that if tbe function has parenthesis, it can be turned into a simple polynomial by distributing. The most basic way to apply this to phys ics is through kinematics. In kinematics, we cover position and displacement, velocity, and acceleration . Upon analyzing these three properties of motion, you may notice that velocity is the rate of change of position over time, and acceleration is the rate of change of velocity over time. This means that velocity is the derivative of position with respect to time, and acceleration is the derivative of velocity with respect to time. Knowing this information , we see that for any position function x(t), x' (t) = v(t), the velocity function , and v' (t) = a(t) , the acceleration function. Below, we will look at the fourth Big 5 equation, which actually takes the form of a position function. Note that for any problem using this equation, a, v;, and X; are all constants , only I is variable.

X = x(t) = 1/,or + V;I + x1

x '(t) = v(t) = 2 * 1/ 2al + v, =at + V;

v'(t) = a(t) = a

Numbers can be plugged into either of the first two equations to find a missing value, depending which equation is appropriate to the situation. In an actual calculus cl ass, you will learn many more rules for differentiating more complex functions, but these are all the rules you should need to know for the AP Physics C exam. But you're not done yet, derivatives aren't the only thing you need to know about calculus.

Basic Indefinite Integration Differentiation is a useful tool for physics, however it cannot solve all of our problems. For example, what if you want to go backwards, such as from a velocity function to its position function? Differentiation cannot go backwards like this, however, calculus has another tool for this task: integration. Integration is simply going backward to put the variables back in (and sometimes

finding what the constant was). There are two broad types of integration: indefinite and definite. Here, we will start with indefinite integration and its applications, then continue to definite integration and its applications. While the derivative is the slope of the tangent to the curve, integration is the area under the curve (it can also be called "antidifferentiation"). Like differentiation, indefinite integration is a process that follows certain rules. Integration also has its own special notation, which looks something like the following: /J(x)dx. The symbol int; tells you that you need to integrate,j(x) is the function to be integrated, and dx tells you that you are integrating with respect to x. As long as the variable after d matches the variable in the function and there are no other variables in the function , the following rules apply exactly.

Rule #I: Constant Rule oflntegration f kdx = kx + c, where k and c are constants. Example: / 5dx = 5x + c Note the + c that appears after integration. This is called the constant of integration, and must be added at the end of the solution to any indefinite integration problem. The reason for this is because the derivative of any constant is 0, and since you are essentially going from the derivative back to the original function, it is possible for any constant to appear at the end of the function. So for the above integral, the actual solution could be 5x, 5x + 3, 5x - I 0, etc., as all these functions have a derivative of 5. Example:/ -7dx = -7x + c Example:/ 4pdx = 4px + c Remember, p is a constant, so this rule applies to it. Example:/ dx = f ldx = x + c

Rule #2: Power Rule of Integration / x"dx = (x"• 1)/(n+ 1) + c Example:/ x2dx = x3/3 + c Example: f x' dx = x4/4 + c Example: f xdx = f x'dx = x2l2 + c Note that this rule does not work for integrating x·' = 1/x, as it would result in x0/0, which is undefined . There is a separate rule for this, but it should not be necessary for AP Physics.

Rule #3: Zero Rule of Integration /Odx =c

Rule #4: Constant Multiple Rule of Integration / k/{x)dx = kf!{x)dx Example: / 2xdx = 2/ xdx = 2x2/2 + c = x2 + c Example: f 5x2dx = 5/ x2dx = 5x3/3 + c Example:/ 2l3x4dx = '!J x'dx = 21,(x'l5) + c = 2x'll5 + c

Rule #5: Sum and Difference Rules oflntegration /.!{x) + g(x)dx = /J(x)dx + / g(x)dx /j{x) - g(_x)dx = /.!{x)dx - / g(_x)dx Example: f 3x2 + 5dx = 3x3/3 + 5x + c = x3 + 5x + c Example: f 2x3 + 4x2 + 3x + ldx = 2x4/4 + 4x3/3 + 3x2/2 + x + c = x4/2 + 4x3/3 + 3x2/2 + x + c

Using these rules, it should be possible to integrate any function used in the AP C curriculum. However, there are two questions that remain: "what do we do with the + c?" and "How do we apply this to physics?"

The first question is answered using an initial condition. For example, the integration problem might be to integratej'(x) = 3x + 2, and the initial condition might be/{!) = 3. So in order to solve this, we first integratef'(x) indefinitely, as follows:

f'(x)=3x+2 /{x) = f 3x + 2dx /{x) = 3x2/2 + 2x + c

Now we plug the initial condition,/{!)= 3, into the function , replacing x with 1 andj(x) with 3, and solve for c:

3 = (3*1 2)/2 + 2*( + C

3=3/2+2+c 3=7/2+c c = 3-7/2 = - l/2

Now that we know what c equals, we simply plug it back into /{x): /{x) = 3x2/2 + 2x - 1/2

Now we know exactly what./(x) is, with only one variable. Now, we must apply this to physics. The most basic application is again in kinematics. Remember that the derivative of position is velocity, and the derivative of velocity is acceleration. Now, notice that integration is, essentially, the reverse of differentiation, so that means that the integral of accel eration is velocity, and the integral of velocity is pos ition. See the appropriate section's practice problems for more details . Now that we have covered basic differentiation and indefinite integration, we have only one concept remaining to explain: definite integration and its physical applications.

Basic Definite Integration Now that we have learned indefinite integration, we can begin learning definite integration. Definite integration uses the same set of rules as indefinite integration, but the notation and method have some additions. Also, the answer to a definite integral is not a function, but a number. We will talk more about the meaning of this number later. First, take a look at the notation for a definite integral : rx2

JxJ /{x)dx Notice that the only change in notation is the addition of x, and x2 at the top and bottom of the integral symbol. These are called the limits of integration, and define the interval on which you are performing the definite integration . x, is the lower limit, and thus is usually a lower number, and x2 is the upper limit, and thus usually a hi gher number. This idea will become more clear as we practice a few problems, but in order to do this, we must look at arguably the most important theorem in calculus:

The Fundamental Theorem of Calculus

f,7' f'(x)dx = ./(x2)-/{x1)

7 Example: f 2 3dx / 3dx = 3x + c (3*7 + c) - (3*2 + c) 21 + c-6- c 21 -6 = 15 Note that the c from each equation cancels out because of the distribution of the negative sign to the second equation. This will happen in every problem, so the+ c can be ignored in definite integration.

Example: f { -3x + 4dx J -3x + 4dx = -3x2/2 + 4x + c

(-3*52/2 + 4*5) - (-3*22/2 + 4*2) -3*25/2 + 20 + 3*4/2 - 8 -75/2 + 18 -75/2 + 36/2 = -39/2 = -19.5 Note that in calculus, you would normally leave the answer as a simplified improper fraction, like -39/2. In physics, we generally use decimal answers, like -19.5. Also, be very carefol to distribute the negative to all tenns of the second equation. Failing to distribute this negative properly can result in radically incorrect answers.

Now that we have seen how to use definite integration on basic polynomials, we need to understand what it means and when it is applicable to physics. The more relevant explanation of what a definite integral means would be the total change in./(x) from x1 to x2 . So for example, if you are given a velocity function , v(t), and took the definite integral from 11 to tr, then since the integral of velocity is position, you would be finding the change in position, or displacement. If you did the same to an acceleration function , a(t), then you would be finding the change in velocity over that time interval. This should be all the calculus you need to know for the AP Physics C exam.

Practice Problems

AP Physics Basic Derivative Practice Evaluate the derivat ive of each function.

1. y=3 2. f(x)=x2+4 3.f(x)=x2+2x-3 4.f(x)=2x3-x2+3x 5.f(x)=x'-.!..cosx 2

7.y=_l_-3sinx 8.g(t)=1tcos t X

4 13.y= ;=, 14.f(x)=x3-3x-2x'

9.f(x)= -1-

3x3 10.f(x)=~

3x 2

15.f(x)=x2-3x-3x2 16.g(t)=t2- ~

t

6.f(x)=S+sinx

I 17.f(x)=x+-;;,

x3 -3x2 +4 18.f(x)= ,

2x 2 -3x +l 19.h(x) 20. y=x(x2 + 1) 21.f(x)=½ + 1/x

x-

± 23 .f(x)=4✓x +3cosx 22.h(s)=s 5

27.f(x)=2x+sec x

x 2 +2x 32.f(x)=-

x

28.f(x)= _I__ 3sin x X

33. y=x +cotx

X

24.f(x)=2sinx+3cosx 4x 2

25.f(x)=-5- 26. f(x)= tan x

2

29.f(x)=(4x+ 1)' 30.f(x)=4(3-x)' 31. f(x)=(x2+2x)(x+ 1)

34.y= '-✓x +4secx 35.f(x)=Ssec x + tan x

Basic Integral Practice

I. f 4xdx 2. f x3dx

3. f 3x4dx 4. f (3x+4)dx

5. f 5dx 6. f -7dx 7. f 4pdx

8. f dx 9. f x2dx

10.f x3dx

11.f xdx 12.f 2xdx 13.f 5x2dx 14.f 2/3 x4dx 15.f 3x2+5dx 16.f (2x3 + 4x2 + 3x + l)dx

Kinematics w/ Graphs 1. An object's position during a IO second time interval is shown by the graph below:

:/~ ·.I ~,-,_,

1 :. 3 s e 1 a s rn . J

·2

a.) Determine the object's total distance traveled and displacement. b.) What is the object's velocity at the following times: I = I, I= 3, and I = 6. c.) Detennine the object's average velocity and average speed from I = 0 to I = I 0. d.) What is tlte object's acceleration at I= 5?

2. An object's velocity during a 10 second time interval is shown by the graph below: V~lcc1ly(m!s)

a.) Determine the object's total distance traveled and displacement. b.) At t = 0, the object's position is x = 2 m. Find the object's position at I = 2, I = 4, I= 7, and I = 10. c.) What is the object's acceleration at the following times: I = I, I = 3, and I = 6. d.) Sketch the corresponding acceleration vs. time graph from I= 0 to I= JO.

3. An object's position during a given time interval is shown by the graph below: Pot~>Qn lmtl•ri;

;• '\,.,!' f,e_,c,_MI

I

a.) At which of the marked points is the object's velocity the greatest? The least? b.) rs the object's acceleration positive or negative between points A and B? c.) Suppose this curve can be modeled by the function x(t) = t3

- 9.5t2 + 23t- 9. Find the object's

velocity and acceleration at t = 1, t = 3, and t = 5. d.) Using tl1e function from part c, detennine the object's maximum and minimum positions and velocities within the interval from t = 1 tot= 6.

Velocity and Acceleration Practice (Using Derivatives and Integrals)

I . An object is moving in one dimension according to the formula x(t) = 2t'+t2-4. Find its velocity at t=2 s.

2. Find the position and velocity functions of an object for arbitrary times if it has an acceleration of a(t)=2t-4, an initial velocity of +4 mis, and an initial position x=0 at time t=0.

3. An object moves vertically according to y(t)= l2-4t+2t'. At t=3 s, what is its acceleration?

4. A particle's motion can be described with the following equation: x=7 .8+9.2t-2. lt'. What is its velocity at t=3.5 s?

5. A particle's position on the x axis is described by: x=4-27t+t3 Detennine the velocity and acceleration equations. ls there ever a time when velocity is zero?

6. An object's velocity is given by v=[J0-(t-5)2]. Find the expression for acceleration.

7. An object moving along the x axis has its velocity described by tl1e function v=2t2 mis, where t is in seconds. Its initial position is x(0)=lm at time zero t(0)=0 s. at t=l s, what are the particle's (a)

position, (b) velocity, and (c) acceleration? Hint: when you integrate to get position, there will be a constant. You find the constant by using the initial position of I m at time zero.

Practice Exercises 1. A ball is thrown straight up. Ignoring air resistance, which of the following is true at the ball's

highest point? A. velocity is zero. B. acceleration is zero.C. Acceleration is down. A) A only B) B only C) Conly D) A and B E) A and C

2. Objects R and S start at the origin. Their v-t graphs are shown. After 2 s, what can you say about the two objects? A) they are 25 m apart B) they are at the same position C) they are 5 m apart D) they are 20 m apart E) they are moving at constant speed

• ,(sec)

3. After 4 s, A) R has gone 35 m B) R is at rest for an instant C) S has returned to the original position D) A and B E) A, B, and C

4. A car with a positive initial velocity slows to a stop with constant acceleration. Which graph could represent the position-time graph?

5. Which (using the same graphs) could represent the velocity-time graph?

6. A ball is thrown straight up near the edge of a 25 m cliff with a speed of 20 mis. If it misses the cliff edge on the way down, it will hit the ground at about how many seconds?

A) 2 B) 3 C) 4 0) 5 E) 7

7. The velocity-time graph of a ball is shown. Which position-time graph might go with the v-t graph?

r-. l fl) I I (CJ

, r

()

8. A vertically-moving object has the following position function: y(t)=12-4t+2t3• At 3 s, what Is its

acceleration in mis2? A) 4 B) 50 C) 54 0)36 E)12

9. A car moving at 10 mis accelerates nonuniformly and has the following acceleration function: a(t)=3t2• What is its velocity in mis at 2 s? A)22 8)18 C)3 0)8 E)12

10. An object moves in a manner such that its position is proportional to 15"'. Therefore, velocity squared will be proportional to: A)t312 B)t712 C) t3 E)t

Problem-Solving 1. A two stage rocket leaves the launch pad with an acceleration of 4 mis2 upward . At 10 s

into the launch, the first stage of the rocket is released. It no longer has fuel to force it upward and is in "free fall." The second stage is now accelerating at 6 mis2

• (a) how high is the rocket when stage 1 separates? (b) how fast is the rocket moving at separation? ( c ) what is the maximum height of the first stage after separation? (d) what is the distance between the two states 2 s after separation?

2. A go cart has the following velocity-time graph. (a) what are the time intervals of increasing velocity? (b) what time intervals show decreasing velocity? ( c) what is the displacement between 2 and 5 s? (d) sketch what the corresponding acceleration-time graph would look like.

5'11/2016 Phfsicsl.AB: Chase Problems IK2.

..t .. Physics ""Vonfine LAB

Worksheet Chase Problems #2

Spursuer = 0

gap" + S1eader

v0 t + ½at2 j number j v0 t + ½at2

vt I I vt

Refer to the following information for the next three questions.

Two balls are rolling in tlie same direction. Wl1en first obser;ed, tile slower ball is 1.5 meters in front of faster moving ball . The slower ball is traveling at 30 cm/sec while !lie faster ball is moving at 50 cm/sec.

C) Sketch (using general shapes) and then describe the pos ition-time and velocity-time graphs for this

chase problem. C ircle on your position-time graph where the two balls collide.

position (rn) velocity (cmlsec)

time (sec) time (sec)

How long will it take the fast ball to "rear-€nd" the slower ball?

How far did the slower-moving ball travel before it was struck by the faster-mov ing ball?

Refer to the following information for the next eight questions.

Interpret this graph showing two cars traveling norU1 on 1-95 in adjacent lanes. One car has just joined t11e flow of traffic after fix ing a flat tire.

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velocity (m/sec)

PhysicslAB: Chase Problems #2

B

time o--------~1~0---------(sec)

Althoug~ t_h.e_c_§r§ _are_side:bY:Side at time t = 0, w hich C§r 1'Ji ll_ini tia_lly_!a_ke the lead? Support your answer.

What acceleration does car B experience?

At what time w ill the cars be traveling at the same speed?

At what t ime w rll they once agarn be sIde-by-srde?

W hat s_peed w ill each c_cir_be 1T1oying at the exact f!:l_Oment t~al_! h.§y pa_s_s each_ other?

How far did each vehicle travel during the chase?

If both cars maintain their veloci ties in part (d) after they pass each other, who would remain the leader for the

rest of the trip?

Sketcl1 (using general shapes) and tt1en describe ! lie posit ion-time graph of this problem.

position (m)

time (sec)

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5111/2016 PhysicsLAB: SVA Relationships #4

~ -.... Physics ...,~nfine LAB

Worksheet SVA Relationships #4

s-t v-t a-t

instantaneous y-coordinate of point position

.. - - - -

displacement difference in two y- area between graph and - - -coordinates x-axis

instantaneous velocity slope of tangent (to the y-coordinate of point - - -graph) at specified t ime

change in velocity - -- difference in two y- area between graph and coordinates x-axis

instantaneous - - - slope of graph at y-coordinate of point acceleration specified time

Refer to the following information for the next two questions.

Group #1 : Use the given posItion-tIme, velocity-time. and acceler'at;on-t1rne waphs to deter111n1e the miss inq values for position, P, and tl·:e final velocity, v These graphs show an object gaininq speed while traveling in a negative direction.

position Cm )

225

C) v

C) P

p D I I 2 9

"elocity Cm/sec)

2 9

-13

-~ - - - I V


acceleration Cm/sec>)

2 9

I I

-3 - L

Group #2: Use the given position-time, velocity-time, and acceleration-time graphs to determine the missing values for position, P, and the final velocity, v. These graphs show an object gaining speed while traveling in a positive direction.

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5111/2016

position Cm)

p

3

p

7: 2 8

"elocity Cm/sec)

PhysicslAB: SVA Relationships #4

acceleration Cm/sec•)

5

2 8


2 8

Group #3: Use the given position-time, velocity-time, and acceleration-time qraphs to determine the rn:ssing values for position. P, and the final velocity, v . These graphs show an object losing speed wni!e traveling in a negative direction .

position Cm)

250-~

,-L~ I

1

V

p

PhysicslAB Copyright© 1997-2016 Catharine H. Colwell Al rights reserved. Application Programmer

Mark Acton

7

"elocity Cm/sec)

1 7 I

V •_:--7 -35 ✓

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acceleration Cm/sec•)

4

7

212

5111/2016 PhysicsLAB: Freefall #1

..t ... Physics .... Vantine LAB

Refer to the following information for the next five questions.

Scenario # 1: A rock dropped from a 20 meter bridge falls into the river below.

Which kinematics variables are stated in this problem?

O va D vr D a O s D t initial velocity final velocity acceleration displacement time interval

At what velocity does it strike the water? ,

Which kinematics equation did you use to solve this problem?

O v}=v;+2as

Worksheet Freefall #1

Based on the ORIGINAL givens, which equation would you now use to determine how much time it takes for the rock to reach the water?

0 v; =v; + 2as

How much time did the rock take to reach the water?


0 s =v t +].at2

• 2

Scenario /1!2 : rho rock is now tossed upwards at 6 rnisec from a 20 meter bridge falls into t110 river below.

At what velocity does it strike the water?



Scenario ii3: The rock is now thrown downward at 6 rnisec from a 20 meter bridge falls into the river below.

At wl1at veloc ity does ,t strike the water?



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5111/2016 PhysicsLAB: Freefall #1

Scenario #4: A rock is thrown straight up into the air and returns to its original release pos ition after 3 seconds.


O va O v1 O a O s O t initial velocity final velocity acceleration displacement time interval

If you want to determine how fast the rock was released, which kinemati cs equation should you use?

0 a= v, -v, t

How fast was it released?

O v; =v;+2as

Which equation(s) could you now use to determine how high it rose up into the air?

How high did it rise?

PhysicsLAB Copyright © 1997-2016 Catharine H. Colwell Al rights reserved. Application Programmer

Mark Acton

[ . 1 J s=-(v+v1 )t 2 • D v; =v;+2as

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5/11/2016 PhysicsLAB: Freefall #3(Honors)

A .... Physics ~~nfine LAB


Worksheet Freefall #3 (Honors)

Wi1ile holding !,is, rifle ,it shoulder-level. a 1,8 meter- tall hunter accidentally discharges :t straight up into the air,

If the bullet exits the barrel of the rifle at 200 m/sec how many seconds does the hunter have to "step aside"

to avoid being hit by the descending bullet?

How high did the bullet rise in the air before it starting fall ing back down to earth?

I fhEl d<les 11?1 rr_iove fast enough, at what velocity would the descending bullet strike his shoulder?

Refer to the following information for the next two qm,stions.

Wrien rising to spike a ball in a volleyball parne, a player jumps vertically 1,5 meters off the floor,

How much total t ime does he spend in the air, assuming he lands in the same position frorn which he left the ground.

At what velocity did he hit the ground at the end of the jump?


Two students are tossing a set of keys from one to the other. The fi rst student (who initially has tl1e keys) is 1,8 meters tall and is standing on the ground 4 meters below the second student who is on a catwalk .

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5111/2016 PhysicsLAB: Freefall #3 (Honors)

4.0moters

Umotors

ground

The student on the ground tosses the keys upward, releasing thern exactly as his hand reaches the top of l1is head, with just the right velocity so that thei r apex coincides w ith the second student's outreached hand.


O v0 D v1 D a O s O t initial velocity final velocity acceleration displacement time interval

How fast were the keys tossed?

How much time did the keys spend in the air?

Refer to the following informatiorr for the next three questions.

Suppose that tho student on U1e iJalcony was distracted and failed to catch tho keys and they fall back d<)llJn to the oround.

W ith what velocity will they strike the grass at the feet of the first student?

True or False. The keys spent the same amount of t ime falling to the ground as they spent ris ing towards the balcony.

0 True O False

True or False. The keys struck the ground at the sarne speed as they were originally tossed upwards by the first student.

0 True 0 False

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5111/2016 PhysicslAB: Freefall #3 (Honors)

Refer to the following i nformation for the next four questions.

Three students are standing side-by-side next io the railinq on a fifth floor balcony. Simultaneously. the three students release their pE:nnies .

One student proceeds to drop a penny to li1e ground below. rhe second student tosses his penny straight dowmva;ds at 15 m/sec, v1i'lr !e The third student tosses his penny straiqht upwards at 15 1111sec.

Which penny or pennies strike(s) the ground first?

0 the penny that was dropped

0 the penny that was tossed upwards

0 the penny that was tossed downwards

Which penny or pennies strike(s) the ground last?

D the penny that was dropped

[ J the penny that was tossed upwards

D the penny that was tossed downwards

Which penny or pennies strike(s) the ground with the greatest final velocity?

f.] the penny that was dropped

W the penny that was tossed upwards

D the penny that was tossed downwards

(I) As the pennies are falling, a person on the 3rd floor times one of the pennies as it passes her 1.5-meter

tall bedroom window. If the penny took 0.15 seconds to clear her window, how fast was the penny traveling just as it entered the top of her window frame? ,


A student, while packing up his book bag that islocated near the edge of a lab table 95-cm tall, accidentally drops his pencil on the floor.


D v0 Llv r D a O s O t initial velocity final velocity acceleration displacement time interval

With what velocity with ttie pencil hit the floor?

How much time does it take the pencil to fall and strike the floor?

(I) If the room temperature is 25 °C. how much total time passed between when he lost his grip on the

pencil and when he hears the sound of it hitting the floor?

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5111/2016 PhysicslAB: Freefall #3 (Honors)

(I)• Modeling this situation, how deep is a well if t11e sound of a penny striking the water is heard 4 seconds

after the penny is released? You may assume that the temperature remains 25 °C.

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5111/2016

A .. Physics ""~nfine LAB

PhysicsLAB: Positioll-Time Graph "Story" Combinations

Worksheet Position-Time Graph "Story" Combinations

Refer to the foll owing information for the next four q ttesfions.

For each of the graphs shown below. choose the description that would correctly outline the behavior of the cart in each "colored" section.

0 a cart originally at rest, uniformly gains speed in a positive direction , maintains that speed, and then uniformly loses its speed finally coming to a stop.

0 while traveling in a positive direction, a rapidly moving cart uniformly loses speed, comes to a stop and rests, and then uniformly regains its original speed.

0 a cart originally at rest, uniformly gains speed in a posit ive direction, maintains that speed, uniformly loses its speed coming to a stop, rests. uniformly gains speed in a negative direction, maintains that negative velocity, then uniformly loses speed and comes to a final rest.

0 while traveling in a negative direction, a rapidly moving cart uniformly loses speed, comes to a stop and rests, and then uniformly regains its original speed but in a positive direction.

0 a cart originally at rest. uniformly gains speed in a negative direction, maintains that speed, and then uniformly loses its speed finally coming to a stop.

0 none of the above

0 a cart originally at rest. uniformly gains speed in a positive direction, maintains that speed, and then uniformly loses its speed finally coming to a stop.

0 while traveling in a positive direction, a rapidly moving cart uniformly loses speed, comes to a stop and rests, and then (1niformly regains its original speed.

0 a cart originally at rest, uniformly gains speed in a posit ive direct ion, maintains that speed, uniformly loses its speed coming to a stop, rests , uniformly gains speed in a negative direction, maintains that negative velocity, then uniformly loses speed and comes to a final rest.


0 a cart originally at rest, unifom1ly gains speed in a negative direction, maintains that speed, and then uniformly loses its speed f inally coming to a stop.

0 none of the above

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5111/2016 PhysicsLAB: Positi~ Time Graph "Story" Combinations

0 a cart originally at rest, unifom1ly gains speed in a positive direction, maintains that speed, and then uniformly loses its speed finally coming to a stop.

0 while traveling in a positive direction, a rapidly moving cart uniformly loses speed, comes to a stop and rests , and then uniformly regains its original speed.

0 a cart originally at rest, uniformly gains speed in a positive direction, maintains that speed, uniformly loses its speed coming to a stop, rests, uniformly gains speed in a negative direction, maintains that negative velocity, then uniformly loses speed and comes to a final rest.

0 while traveling in a negative direction, a rapidly moving cart uniformly loses speed, comes to a stop and rests , and then uniformly regains its original speed but in a positive direction.

0 a cart originally at rest, uniformly gains speed in a negative direction, maintains that speed, and then uniformly loses its speed finally coming to a stop.

0 none of the above

0 a cart originally at rest. uniformly gains speed in a posit ive direction, maintains that speed, and then uniformly loses its speed finally coming to a stop.

0 while traveling in a positive direction, a rapidly moving cart uniformly loses speed, comes to a stop ancl rests, and then uniformly regains its original speed.

0 a cart originally at rest, uniform ly gains speed in a positive direction, maintains that speed, uniformly loses its speed coming to a stop. rests. uniformly gains speed in a negative direction, maintains t11at negative velocity, then uniformly loses speed and comes to a final rest.


0 a cart originally at rest, uniformly gains speed in a negative direction, maintains that speed, and then uniformly loses its speed finally coming to a stop.

0 none of the above

Refer to the following information for the next question.

On your own paper draw a posit ion-time graph illustrating the following story.

A person who walks two blocks at a moderate speed, waits at an intersection for a short t ime until the

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5/11/2016 PhysicslAB: Position-Time Graph •story" Combinations

~ "walk" light turns "green," then walks the next block more s lowly, and finally runs the fi nal two blocks very ~ rapidly.

block 5 ,--+-----1t-----1t-----1t-----1----1---+-

block 4 1---1----11-----1~---1----t----t---+-

block 3 ■---+-----1~---1----1---+---+---+-

block 2 ■---+-----1----1----t---+---+---+-

block 1 1---t----+----+---t---+---+---t-

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5 10 15 20 25 30 35 time (min)

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5/11/2016 PhysicslAB: Derivatives: Instantaneous vs Average Velocities

~..._ Physics "'Vantine LAB

Resource Lesson Derivatives: Instantaneous vs Average Velocities

As we have already discussed, there are distinctions that need to be remembered when using the terms distance traveled, net displacement, average speed, average velocity, instantaneous speed, and instantaneous velocity. There are times when certain terms are synonymous and other times when they are absolutely and completely different.

A few examples to refresh your memories:

• While traveling in one direction along a straight line path, total distance traveled and net displacement would be equivalent HOWEVER, if the path changed direction at any time, those two terms would then be completely different with distance representing Hie lengtl1 of the path traveled and displacement represent ing the length of the straight-line vector that begins at the starting point and ends at the finishing point, independent of the path taken.

A--------------- B

C~D

In both of these examples 1 the displacements are identical whereas the d istance t raveled between points C and D is much greater than the distance traveled between points A and B since the direction switches back and forth causing the path length to be much longer.

• While traveling at a constant speed, the calculations of an object's constant speed, average speed, and instantaneous speed would all yield equivalent values HOWEVER, if the speed was varied throughout the internal then the average speed during that interval would only occasionally equal its instantaneous speed at any one moment in the interval. Obviously, the notion of a constant speed would no longer be discussed under those conditions.

6 .---~-~---.-

2

Both o f these graph represent position vs time graphs for objects traveling along straig h t -l ine paths. On the

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5/11/2016 PhysicslAB: Derivatives: Instantaneous vs AverFYJS Velocities

left, the first object t ravels at a constant speed beginning at x = 5 and end ing at x = -6 taking approximately 5.5 seconds to complete t he trip. On the right graph , a second object is travel ing at a variable rate beg inning at x = 1 traveling down to x = 0 then up to x = 6 and finally all the way back to x = -8 taking app roximately 5 seconds to comp lete its trip.

■ Average speed and average velocity during an interval also take on their own unique properties. Average speed represents the rat io of total distance traveled divided by t11e total time taken to travel that distance while average velocity represents the ratio of the object's net displacement divided by the total time required to acl1ieve that displacement. Whenever total distance traveled and net displacement differ, average speed and average velocity will be different. Since average speed is a scalar it is always reported as a positive (or zero) value; while average velocity (being a vector quantity) can have a negative, zero, or positive value depending on how you assign positive/negative directions in setting up the problem.

On t he left graph d isplayed above the first object's average speed (11 units/ 5.5 seconds) equals t he magnitude of its average velocity (11 units/ 5.5 seconds, down}. While on the right graph the second object's average velocity (9 units/ 5 seconds, down) is much smaller than its average speed (15 units/ 5 seconds).

■ On the other hand, an object's instantaneous speed will always equal the magnitude of its instantaneous velocity.

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5111/2016 PhysicsLAB: Average Velocity-A Calculus Approach

~ --- Physics ....,~nfine LAB

Resource Lesson Average Velocity - A Calculus Approach

Motion of a particle in one dimension when given a position (displacement) function

Suppose a particle is moving back and forth along the x-ax is with a pcsition function (the coordinate giving the location of the particle on the x-axis) given by x(t) = t2- 21 - 3.

Assume that tis in seconds, t < 0 is not possible, and the coordinates on the x-ax is are in meters. We already know the following facts:

1. The initial position is x(O) = -3 meters which is 3 meters to the left of the origin 2. The coordinate on the x-axis giving location when t = 2 is x(2) = 4 - 4 - 3 = - 3 meters. 3. The particle is at the origin when x(t} = 0 = (t - 3)(t + 1) or when t = 3 and t = -1

(we exclude the case where t = - 1 s ince we exclude t < 0). 4. This position function x(t) is really a vector function called the "displacement" of the particle from the

origin. x(O) = - 3 means the particle is a distance (magnitude of the vector x) of 3 to the left (direction) of the origin.

Let's use these facts to answer the next two questions.

@ What is the average velocity between times t = 2 and t = 5?

@ What is the ins tantaneous velocity at t = 2?

Motion of a particle in one dimension when given a velocity or acceleration function

Suppose we know that the velocity of a particle moving back and forth on the x-axis is given by v(t) = 21 - 2 and the initial position x(O) = -3. Notice that this is the same particle as in Part I above.

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5111/2016 PhysicslAB: Average Velocity -A Calculus Approach

Some basic facts:

1. v(t) = 21 - 2 is the derivative of the position function x(t) = t2- 2t - 3 . 2. a(t) = 2 is the derivative of v(t) and is the uniform (or constant) acceleration of the particle. 3. v(t) = 0 when t = 1 which invites us to investigate the sign of our velocity function 4. v(t) < 0 when t < 1 and v(t) > 0 when t > 1.

To detenmine an object's position given its velocity function and an initial position we will use a basic fact from calculus:

Given the rate of change of a quantity and a time interval, the "net change" is the integral of the rate of change over the given time period.

Let's use these fact to look at three final questions.

@ Wl1at is the position of the particle after the first 5 seconds of motion?

@ What is the total distance traveled by the particle in the first 5 seconds?

@ Is the particle speeding up or slowing down when t = 4?

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5/11/2016 PhysicslAB: Accelerated Motion: Practice with Data Analysis

~ .. Physics '"Vontine LAB

Worksheet Accelerated Motion: Practice with Data Analysis


The following table represents a group's data for the motion of their ball down a hill.

t position

sec m 0 I 0

0.5 0.25

1 1 1.5 I 2.25

2 4

2.5 I 6.25 3 9

3.5 12.25 4 16

4.5 20.25

5 25

Cl) What was the ball's acceleration?


The follo,vlng table represents a 9roup's data for the motion of their cart ciown ti1e 11allway.

t cositlon

sec m 2 5

2.8 9.8

3.6 16.2

4.4 I 24.2

5.2 33.8 6 I 45

8.8 I 57.8

7.6 72.2

8.4 I 88.2

B.2 105.8 10 125

(;), What acceleration did the cart experience?


The following table represents a group's data for the motion of box sliding down a slide.

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5/11/2016 PhysicsLAB: Accelerated Motion: Practice with Data Analysis

t I position

sec cm 0 I 0 1 0.8

2 3.2

3 7.2

4 13,8

5 I 23 6 32.8

7 46.2

8 I 60.2 9 75.9

10 I 92

Cl) What acceleration did the box experience? I

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(Jf)d / tJ r Cal cul vlho-n s

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: ''

Graphical Methods-Summary

A graph is one of the most effective representations of the relationship between two variables. The independent variable (one controlled by the experimenter) is usually placed on the x-axis . The dependent variable (one that responds to changes in the independent variable) is usually placed on the y-axis. It is important for you to be able interpret a graphical relationship and express it in a.written statement and by means of an algebraic expression.

Modification Algebraic Graph shape Written relationship required to ITneadze representation

graph ..

- H - .. . . ~ . -~ . . . -· .. _:;:-, .. . ' ..

y As x increases, y rem ains None y = b , or the same. There is no y · rs constant relationship between the variables .

... -X

.h As x increases, y increases None y-= mx + b y I/_ proportionally. Y is directly proportional to x.

- "'-X

a 1 (I\

~ As x increases, y

Graph y decreases. y vs - , or y z= m- + b

X \x) Y is inversely proportional -1 y VS X to X .

... -X

· ,

jl

J _ x2 y = mx 2 + b y Y is proportional to the Graph y VS

square of x.

'

-X

a Graph y2 vs X y 2 = mx + b y The square of y is

proportional to x .

..._ .... X

When you state the relationship, tell how y depends on x ( e.g., as x increases, y ... ).

AP Physics C Summer Work

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