due the first day of class. kgazq9j - shelbyed.k12.al.us · Morion is all around us, so it's easy...

Physics C (Mechanics)

Welcome to AP Physics C - Mechanics at OMHS! Here is your summer assignment, due the first day of class. First step: Join the Google Classroom (join code r4zno7d) .

AP Physics C is a calculus-based course which requires strong analytical skills. In addition to the science concepts, physics often seems like a course in applied mathematics. The following assignment includes a derivative review, an introduction to integrals, and a review of velocity and acceleration (horizontal and vertical motion). I have attached a summary of velocity and acceleration, but you will need to review these concepts more thoroughly at PhysicsLab (http: //www.physicslab.org/Lessons.aspx and go to Kinematics) or physicsclassroom.com.

This packet contains lessons/review of kinematics so that we can spend more time learning electricity and magnetism later in the year. Be sure to start early on the packet and pace yourself. DO NOT wait until the night before school starts to begin the assignments. If you do a few problems each week, next year will be SO much easier for you, since you will not have had time to forget everything you've learned. Here are a few things that you need to know about AP Physics in general:

1. It's not just "plugging in numbers." AP focuses more on concepts than on plugging in. Almost every problem is new and different, including on the national exam. You'll be taught the principles of physics, and then will apply those principles in various situations.

2. Buy a review book (now ... ). If you didn't take general physics or AP Physics 1, then I'd also buy a "quick study" book that explains the basics.

3. There are a lot of good websites (prettygoodphysics is VERY good, PhysicsLab, physicsclassroom, AP Central, etc.). Also, many universities have good physics websites with practice quizzes. MIT Open Courseware has great lectures, as do Dan Fullerton (APlusPhysics) and Khan Academy. Become familiar with them over the summer.

4. We will do online homework (Quest out of UT Austin). I will give you instructions for registering on the first day of school.

5. You must provide your own calculator ( simple, with trig functions, not programmable) for class, and will be expected to have it with you every day.

Assignments in the Packet *Basic Derivatives and Integrals Practice *Kinematics with Graphs *Velocity and Acceleration Practice (Derivatives and Integrals) *MC and FR Practice Exercises *Worksheets from PhysicsLab site *3 Graphs from PhysicsLab

Have a great summer! Send me an email if you have questions and check your email for updates from me. ([email protected]).

Mrs. Humphrey ©

kgazq9j

CHAPTER

Motion .in.One Dimension

Morion is all around us, so it's easy to understand why the earliest physical thinkers focused on describing motion and determining irs causes. In this chapter and the next, you'll learn how to describe mo tion without considering what is actually causing the motion. This study is called kinematics. Yo u'll need to understand th e nature of the quantities used to precisely describe morion and the relationships these quantities have with each other. With his long incl ines and water clocks, Galileo Galilei was the first to study kinematics experimentally. Just as he did, yo u'll begin with objects constrained to move along a straight line, or onedimensional morion .

Position and Displacement To describe where an objec t is, you must introduce a coordinate system, pick an origin , and choose the posi tive direction for each axis. In one dimens ion yo u need on ly one axis, freq uently labeled the x-axis for horizontal motion or the y-axis for vertical motio n. The position of an object is its value of x . As the object moves, its position changes. The change in position is call ed rhe displacement of rhe object, writ ten as /::,x. The delta symbol, /::,, stands for the change in a quantity, with the change always determined using "final-initial." For displacemem, the equation is

/::, x = x1 - X;

where x1 is the final position of the body and x; is rhe in itial position. As an object moves, its pos ition depends on rime, and this functional dependence is often written as x = x(t). Pos ition and displacement are both directed quantities, which means they carry a sign that determines their direction relative to the coordinate sys tem. (When you generalize later to

more than one dimens ion, you' ll see that such directed quantities are called vecto rs. ) If an object moving in one dimension never changes direction, the magnitude of its displacement will equal rhe distance traveled . But if the object does change direction, you' ll have to break up its ·path carefully into segments to determine the total distance traveled. When you throw a ball straight up in th e air to a height of 7 m, its displacement is O when you catch it, but it has traveled a distance of 14 m.

Velocity -------------------➔

Velocity is the rate at wh ich the pos ition of an object changes. The average velocity over a time interval is the displacement during the interval divided by the duration of the interval:

lx Vav =~

Let's look at the x vs . t graph for the morion (figure I ).

x(m) +--+---+-+--+------I--+-

6 t(sec) figure 1

From the graph you can see that at t = 1 s, the object is at abou t x = 6 m, and art = 4 s, it is at about x = 2 m. During this 3 second interval:

lx 2 - 6 m v,v = T, = ~ = - 1.33 5

If the fun ction x = x(t) changes cons iderab ly, then the average velocity is very sensitive to the interval over which it is averaged. Fo r example, if you had chosen the rime interval fro m I to 3 s, then x would have changed from about 6 to about 9 m, fo r an average velocity of

!J.x 9 - 6 m v,v = T, = 3-=-T = + I. 5 s

You' ll notice that the average velocity involves a change on the vertical axis divided by a change on rhe horizo ntal axis. Th is is a slope formula, and the average velocity is rhe slope of the lin e connecting rhe two points defining th e interval.

I

".

Average velocity isn't the most precise way to describe motion because it depends on the particular interval you've chosen to average over. For more precision, you'll need to use instantaneous velocity. You can probably surmise that as an object moves, it has a well defined velocity at every instant. The problem with defining such a quantity is that at each instant it is frozen like a photograph, so how do you determine the displacement and the time interval? You can use the face chat if the interval is small enough, che function will not change much. Instantaneous velocity at a given time is the average velocity over a tiny interval (imagine it infinitely small) centered at that time:

(/1x) . 6.x dx V = fu small =11'f_'.:;ofu = dt

I Because the average velocity is the slope of the line connecting the two endpoints of the interval, the instantaneous velocity will be the slope of the tangent line drawn to the graph at the given time. Of course, from your work in calculus, you should recognize chat the derivative of a function gives the slope of the tangent line.

li]!fui}1W)~

I By drawing a tangent line to' th~ curve at x j ~ (figure 2), you can see from the slope that the instantaneous velocity is approximately 1.2 mis.

J

I x(m)

iI fl .,

~ ij ffi

101+--+'---+-+L...+-+---t---

,:+-- -+---+

" I. :s ·2 6 t(sec)

j figure2

I I Motion in One Dimension

l

11

• n object is m_oving in one di1;1ension ac~cii-ding to che formula x(t) = 2t3 + i' - 4. Find it,s vdoi:iry at t = 2 s

dx · 2 .v(t) = dt = Gr + 2t

v(2) = 28 7.

Acceleration

Acceleration is the ra te at which velocity changes. Just as you did for ve locity, you can define average acceleration over an interval or instantaneous acceleration at a given time.

6.v a,v =-;;:; D,v) . D.v dv d2x

a = (fu sma ll =J)~6, = --;f; = ---;y:

v(!p-) 10

2 -t---i·--i-----t-----t---i---+--

I /,

,I

J!i

6 t(sec) 1 figure 3

1If you look at a graph of v vs . t (figure 3), the average accele ration over an interval is just the slope of the line connecting the rwo ends of the interval, and the instantaneous acceleration l is the slope of the tangent line drawn to the graph at the given time. I

Chapter 1

~ .

~---There is one ocher useful piece of information you can get from the v vs. t graph . Given the functional form v = v(t), you can integrate as follows:

v(t) = ! => dx = vdt

rdx = fvdt xo 0

X - Xo = rvdt

The lefr-hand side of rhe last equat ion is the displacement over the interval, and the righthand side is the area under the v vs. t graph. Thus by calculating the area under such a graph, we can determine the displacement. This is particularly simple if the graph is a sequence of

connected lines. ·I 0

Find the displacement over ·the interval from Oto 6.s in the rwo graphs below.

'°+- ' I

2+ I I F\-+

2 J 4 l 6 t(scc)

~ '... ,(T)

vs."')\

-i,--t-- --l---..--~ l(~c~)

..:..10·

figure 4

The left graph can be thought of as a ;ectangl~ _aqd a triangle, giving an area of

area = B • 3 + .l(3 • 8) = 36 m. 2 .

In the right graph, you can count the .boxes to det~rmine the area. Each box has a value 5 • 2 = · 10 m. From O to 4· s/you. have 35 bo)(eS. From 4 to 6 s, you have 1 box, but the area is negative since fr lies. below th~ t-axis _and corresponds to

decreasing x-values. Thusyou have a total _of 2.5 boxes or 25 m for the displacement.

Motion in One Dimension

Given the functional form of the acceleration, you can inregrate co find the velocity.

a(t) = ~~ => dv = adt

J"dv = J,'adt VO 0

v - = fadtv0 0

Find the position and velocity of an object for arbitrary times if it has'an ·· : ;' acceleration of a(t) = 2t ~ 4, an initial velocity of+ 4 •irt/s; and' an initial°' '•": position x. = 0 at ·rime t = 0. · · · · . . ·: 1, .

• I: ;..• ... i 1~f":•'.h.\ :j~:_.,

From the preceding equations you can write: .,,,-, ._. ,.;. ",, i!{._-i-; er,:- ""rY.'-•<ifi''>l'i ,. · .:.· ;·:.: .~r·~.;1):•;::.!(~\~~:1 ~ : , .•~t(~·.•~i~f v-4= f(2t-4)dt x~o={<?·-4f·f 4 )dt \ '·: ... ·

. 0 • . ..~ - .. · . 0 · · ' ;,·,. -\~'l)t•i•', 't ..J' : '.

v(t) = ? - 4t + 4• x(t) =-}c3 - 2?+ 4t

Constant Acceleration Equations ----- ---- -----

When acceleration is constanr, you can use some simple equations that relate the various kinematic variables. To understand their conrenr, consider the v vs. t graph (figure 5) .

velocity

v/- --- ------ - =

VO-

V - VO slope= T=o

time

figure 5

Chapter I

Since the acceleration is the slope of the v vs. t graph, constant acceleration means a straight line. If the interval begins at t = 0 and extends to some arbitr;,ry time t, the slope is a = ~v, =~, or you can w rite

.u. t- 0 v = v0 + at

Since the graph is a straight line, the average velocity over the interval is the median of the

'· 1

velocities at the endpoints: vov = ½(v + v0l. Since by definition v,v = t, you can write

!:ix= 2 (v + v0)tf 1

j You can combine these rwo equations algebraically to eliminate certain variables and obtain other equations. The end result is a set of four equations:

Ii 1. v = v0 + at 3. l:i.x = v0t + +a?

lf ~:!:ix= 2 (v + v0lt 4. v2 = vJ+ 2a!:ix g ' n Rather than thinking in terms of the graphs, you can derive these equations from direct integration beginning with the definition of a and using che face chat it is constant. For example: dv~ a = dt ⇒ dv = adt

I lt

J"dv = J'adt •o o

v - = at ⇒ v = v0 + atv0

You can apply these equations to the motion of an object over an interval where the acceleration a is constant. During this interval, the veloci ty changes from v0 to v while a displacement !:ix occurs. In a typical problem, you' ll be given three of the five quantities v, v0, a, l:i.x, and t; you can determine the remaining rwo by using the four equations. You can usually use several different ways to solve such problems, depending on which equations yo u~ decide to use and which quantities you decide to solve for first.

i r

~if

:·1


--~i\:.-'t, ~~ri~_____________________ _ '~t/!!.ft~n automobile moving along a straight road passes an observer who records

a speed of20 m/s. A second observer 100 m from the first records a speed of 30 mis. Find the value of the acceleration, assu_ming it is constant, and the time it took co travel che 100 m.

For chis interval youdiave:

"o = 20 s !:ix= l 00m

V = 30 ~ t =?

a=?

Yo u can use the fourth equation to find the acceleration si nce it does not invo lve t: ·

2 v = v5 + 2ail.x

302 = 202 + 200a

a= 2 )- .!!e.2

. s

You can then use ch e first equation co find t::1 v = v0 + at

:11 30 = 20 + 2.5t

1!,, t = 4s

One of the most important situations where you can apply one-dimensional constant acceleration is the case of an object moving vertically under only the influence of gravity. Gravity causes all masses near che Earth's surface co experience an acceleration of g = 9.8

2 m/s directed downward. On the AP test, you will always be permitted co replace 9.8 m/s2

with 10 m/s2 when you're solving numerical problems.

C) Chapter I

-,---

? pv

I

You can use the third equation to get t and the first equation to get v: : . : EXAMPLE,;.;_ . : :,'' ·/·.. -~l'.•

+; • /J.y = v0t +fa? v = v0 + at

-25 = 2Dt - 5? v = 20 - 10(5)

(t-5)(t+ l)=0 v= -30~

t = 5s

This is just one possible path to the solutions . Depending on the intervals you choose to analyze and the variables you solve for first, yo u could use d ifferent equations to end up with the same results .

·~· /J.y = ? /J.xAverage Velocity t =? Vav = 71t

m -IO~

/J.x dxInstantaneous Velocity v =l im 7tt V = ---;f;t.1-•0

You know two of the five quantities, so yo u can solve for the others. Choosing the first equation, you can find t: /J. vAverage Acceleration

a av = tft

I v = v0 + at

0 = 20 - !0t Instantaneous Acceleration a= lim ./J.v dv a = dtt., - ol!.rt = 2 s

You could then use the second equation to find /J.y: General Velocity from Acceleration v - = J'adtv0 0

6.y = 21

(v0 + v)t General Position from Velocity x - = J'vdtx0

06.y = 21

(0 + 20) (2) = 20 m

Constant Acceleration EquationsThe highest point will be 45 m above the ground. ). V = Vo + tit To find the total time in the air, you could choose an interval beginning with the start of the

1motion and ending just before the ball hits. For this interval you have: 2. 6. x = 2 (v + v0)t

Vo= 20 ~ 3. 6.x = v0t + ½a? 6.y=-25m

v = 4. v2 = v5 + 2aih t;:::?

a=-!Om2 s

'!f G

I Chapter I


11

·

AP Physics C Derivatives and Integrals-Acceleration and Velocity

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

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

https ://www.mathsisfun.com/ca lcu lus/i ntegratio n-i ntroducti on. 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 -y1)/(x2 - x 1). 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.7 - ,-:-/ .

M_i " \ /-·.•. ~ ,,, /

-~---;:

Each of the red lines is tangent to the graph of.f(x) 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 ofy with respect to x. In simpler terms, 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.f(x), the derivative of.f(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 #1: The Constant Rule If./(x) = c, where c is any constant,f'(x) = 0 Example:f(x) = I, sof'(x) = 0 Example:/{x) = -10, sof'(x) = 0 Example:.f(x) = 3p, sof'(x) = 0 Remember that pis a constant, and so any constant multiplied by pis also a constant. Example:./(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:_/{x) = x2, sof'(x) = 2x

Note that when there is no exponent written on x, it is to the Ist power. Example:_/{x)=x,sof"(x)=x0 = I Note that any number to the power of0 is 1, so x0 = I

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

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

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

, sof"(x) = 3x2 + 2x Example: _/{x) = 2x2

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

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

- 2x - 3 Remember tliat the derivative of a constant is 0, so any constant tenns in./(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 the function has parenthesis, it can be turned into a simple polynomial by distributing. The most basic way to apply this to physics 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 ftmction. 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 t is variable.

X =x(t) = 'l,a/2 + V;l + X;

x'(t) = v(t) = 2 * 1/ 2a/ + v1 =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 class, 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 wi ll start with indefi nite integration and its applications, then continue to definite integration and its applications. While the derivat ive 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 fo llows certain rul es. Integration also has its own specia l notation, which looks something like the following: /j{x)dx. The symbol int; tells you that yo u need to integrate,/{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 oth er variabl es in the function , the fo ll owi ng rules apply exactly.

Rule #1: Constant Rule of Integrationf kdx = kx + c, where k and c are constants. Example: / Sdx = Sx + 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 tl1e above integral, the actual solution could be 5x, 5x + 3, 5x - I0, etc., as all tl1ese fu nctions have a derivative of 5. Example: / -7dx = -7x + c Example: f 4pdx = 4px + c Remember, p is a constant, so this rule applies to it. Example: fdx = f Idx = x + c

Rule #2: Power Rule of Integrationf x"dx = (x"" )/(n+ I ) + c Example: f x2dx = x 3/3 + c Example: f x3dx = x'/4 + c Example: f xdx = f x'dx = x2/2 + 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 /0dx = c

Rule #4: Constant Multiple Rule of Integrationf kf(x)dx = k/f(x)dx

2Example: f 2xdx = 2/ xdx = 2x2/2 + c = x + c Example: f 5x2dx = 5/ x2dx = 5x3/3 + c

2Example: f 2/ 3x

4dx = 'Ii/ x'dx = / 3(x5/5) + c = 2x5/ 15 + c

Rule #5: Sum and Difference Rules of Integration /.f(x) + g(_x)dx = //{x)dx + f g(_x)dx /f{x) - g(_x)dx = /.f(x)dx - fg(_x)dx Example: f 3x2 + Sdx = 3x3/3 + 5x + c = x3 + 5x + c Example: f 2x3 + 4x2 + 3x + ldx = 2x4/4 + 4x3/3 + 3x2/2 + x + c= x'/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 integratef"(x) = 3x + 2, and the initia l condition might be/{ J ) = 3. So in order to solve this, we firs t integratef'(x) indefinitely, as follows:

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

Now we plu g the initial condition,/{ ! ) = 3, into the function , replacing x with I and/{x) with 3, and solve for c:

3 = (3*1 2)/2 + 2*1 + 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 on ly 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 acceleration is velocity, and the integral of velocity is position. See the appropriate section's practice problems for more details. Now that we have covered basic differentiation and indefinite integration , we have on ly 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 indefi nite integration , but the notation and method have some additions. Also, the answer to a definite integral is not a function , but a number. We wi ll talk more about the meaning of this number later. First, take a look at the notation for a definite integral :

J~1 f{x)dx Notice that the on ly change in notation is the addi tion of x, and x2 at the top and bottom of the integral symbol. These are called the li mits of integration, and define the interval on which you are performing the definite integration. x, is the lower limit, and th us is usually a lower number, and x2 is the upper limit, and thus usually a higher number. Th is idea will become more clear as we practice a few problems, but in order to do thi s, we must look at arguably the most important theorem in calculus:

The Fundamental Theorem of Calculus

s:;f'(x)dx = .f(x,) -/{x,)

Example: f J3dx f 3dx = 3x + c (3*7 + c) - (3*2 + c) 2 1 + c-6-c 21 -6 = 15 Note tliat 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.

5 Example: f2 -3x + 4dx f -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 careful 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 infix) from x 1 to x2 . So for example, if you are given a velocity function , v(t), and took the definite integral from 11 to 11, 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 derivative 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'-.!._ cos x 6.f(x)=5+sinx2

1 2 I rr 7.y=.!._-3sin x 8.g(t)=rrcost 9.f(x)= x, 10.f(x)= x, 11.f(x)= ( x)' 12.f(x)= ( x)'

X 3 3 3 3

4 I 13.y=-=, 14.f(x)=x3-3x-2x' 15.f(x)=x2-3x-3x·' 16.g(t)=t'-i 17.f(x)=x+-;,

X t

18.f(x) x' -3x 2

+ 4 19.h(x)= 2 x 2

- 3x + I 20.y=x(x2+1) 21.f(x)=½ + 1/xx' X

4x 2

22.h(s)=s ' 23.f(x)=4 ✓x + 3 cos x 24.f(x)=2sinx+3cosx 25.f(x)=-- 26.f(x)= tan x 5 2

27.f(x)=2x+sec x 28.f(x)= .!._ - 3 sin x 29.f(x)=( 4x+1) 2 30.f(x)=4(3-x)2 31.f(x)=(x' +2x)(x+1) X

2

32. f(x)= x + 2x 33.y=x +cotx 34.y=½ +4secx 35. f(x)=5sec x + tan x X

Basic Integral Practice

I. J4xdx 2. f x 3dx 3. f 3x4dx 4. f (3x+4)dx

5. f Sdx 6. f -7dx

7. f 4pdx

8. Jdx 9. f x 2dx 10.f x3dx

1J.J xdx 12.f 2xdx

13.f 5x2dx

14.f 2/3 x 4dx 15.f 3x2+5dx 16.f (2x3 + 4x2 + 3x + l)dx

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

Poi1'.ior. (m~!i rs )

~ I I I3 I'1 ! ! ("-.l"J I' 1 •~''"'"' 10

·2

a.) Detennine the object's total distance traveled and displacement. b.) What is the object's velocity at the following times: I= I, t = 3, and t = 6. c.) Detennine the object's average velocity and average speed from I= 0 tot= 10. d.) What is the object's acceleration at I = 5?

..

2. An object's velocity during a IO second time interval is shown by the graph below: V1:~1!y{m'i )

/~ I I I """"~''' 1f 2f ~ ! 10! j ! l \"',

a.) Detennine the object's total distance traveled and displacement. b.) At I= 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 = 10.

3. An object's position during a given time interval is shown by the graph below: P n~,o,,fr,il~'lj

I/\: ./

I \J, ,.,."'~~,('- C'

ii C

a.) At which of the marked points is the object's velocity the greatest? The least? b.) ls the object's acceleration positive or negative between points A and B? c.) Suppose this curve can be modeled by the function x(t) = 13

- 9.512 + 23t - 9. Find the object's

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

Velocity and Acceleration Practice {Using Derivatives and Integrals)

I . An object is moving in one dimension according to the formula x(t) = 2t3+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=Oat time t=O.

3. An object moves vertically according to y(t)= l2-4t+2r'. 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. Ir'. 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=[IO-(t-5)2] . Find the expression for acceleration.

7. An object moving along the x axis has its velocity described by tl1e function v=2t' mis, where t is in seconds. Its initial position is x(O)=lm at time zero t(O)=Os. at t=I 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.

I j i

PRACTICE EXERCISES • SECTION I M ULTIPLE CHOICE

'j 1. A ball is thrown straight ' up in the ai r. When air resisrance is ignored, which of the

following is t rue at the ball's highest point? I. The veloc ity is 0.

II. The acceleration is 0. III. The acceleration is directed downward.I

(Al I only (B) II only (C) III on ly (D) I and II only (E) I and Ill only

Questions 2 and 3 Objects Rand 5 start at the same origin, and their velociry vs. time graphs are shown on the

same set of axes.

l t(sec)

I 2. After 2 s, the two objects are

(A) 25 m apart (B) at the same position (C) 5 m apart (D) 20 m apart (E) moving at constant speed


V (~)20

IO l t? I ::::S:,,,. k

-JO S

- 20

figure 6

3. After 4 s,

(A) R has traveled 3 5 m (B) R is instantaneously at rest (C) 5 has recurned to its original position (D) A and B o nly (E) A, B, and C

Q u estions 4 and 5

~ }1,~JL, J==-,

figure 7

-:-.\::. A car with an initial positive velocity slows to a stop with a constant acceleration .

4. Which graph best represents its position vs. time graph'

(A) A (ll) 8 (C) C (D J D (E) E

5. Which graph bes t represents the velocity vs. time graph?

(A) A (B) B (C) C (D) D (E) E

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 in a time closest to

(A) 2 s (B) 3 s (C) 4 s (D) 5 s (E) 7 s

Chapter I

,\],

_

figure 8

• PRACTICE EXERCISES

5mm, II FREE Rmo,,, •

l. A two stage rocket leaves its launch pad moving vertically with an average acceleration of 4 m/s2

. At 10 s after launch, the first stage of the rocket (now without fuel) is released. The second stage now has an acceleration of 6 m /s2

.

(a) How high is the rocket when the first stage separates' (b) How fast is the rocket moving upon first stage separation ? (c) What will be the maximum height artained by the first stage after separation? (d) What will be the d istance between the first and second stages 2 s after separation?7. An object moves with a velocity vs. time graph as shown. The position vs. rime

graph for the same time period would be

v(~) 10

(ci"_T _ ''" :~-; bI l~. (B)

xl-- - - -+------1

8 1(sec) _L__

(D) (E) I -, -t-

figu re 9

8. An object moves venically according to y(t) = 12 - 4t + 2f. At t = 3 s, its acceleration is -Jo+---+---+-~-~----+-

(A) 4 m/s2 (B) 50 m/s2 (C) 54 m/s2 (D) 36 m /s2 (E) 12 m/s2 figure 10

2. A vehicle moves in one dimension wi th the velocity vs . time graph shown. (a) Over what time intervals is the velocity increasing?9. A car currently moving at 10 mis accelerates nonuniformly according to a(t) = 3t2. (b) Over what time intervals is the velocity decreasing?After 2 s, its velocity is (c) Determine the displacement during the interval from t = 2 to t = 5 s. (d) On the axes below, sketch the a~celeration vs . time graph of the vehicle over the(A) 22 mis (B) 18 mis (C) 3 mis (D) 8 mis (E) 12 mis

entire time.

.l. 10. An object moves in one d imension such that x(t) is proportional to t 2 • This means

v2 will be proportional to

(A) t l. 2 (B) t

L 2 (C) f (D) t7 (E) t .


a c~)

-

2 4 6 8 sec)

figure 11

I

5111/2016 PhysicsLAB: Chase Problems #2

Worksheet~ .. Physics Chase Problems #2 ~Vantine LAB

Spursuer = "gap" + SJeader

v0 t + ½at2 number v0 t + ½at2

vt vt

Refer to the following information for the next three questions.

Two llalls are rolling in Uie same drreclion. Wl1e11 first ollsser,ed, tl1e slower ball is 1.5 meters in front of faster moving ball. The slower ball is travelin,J at 30 crnisec while the faster ball is rnovinfJ at 50 crn/sec.

Cl) Sketch (using general shapes) and then describe the position-time and veloci ty-time graphs for this

chase problem. Circle on your position-time graph where the two balls coll ide.

position (rn) ve locity (cm/sec)

time (sec) time (sec)

How long wi ll 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-moving ball?

Refer to the following information for the next eight questions.

Interpret this graph showing two cars traveling nortl1 on 1-95 in adjacent lanes. One car has just Joined the flow of traffic after fixing a flat tire.

http://www.physicslab.Cf~DocumeotPrintaspx?doctype=5&filename=Kinematics_ChaseProblems2xml 1/3

5111/2016 PhysicsLAB: Chase Proolems #2

veloc ity B(m/sec)

30 1----- ---..,,C.--- - ---A

trme~--------~---------(sec) 0 10

Althougb_ t~e_c_c1~_c1~_s_i_de-by-side at time t :' 0, w~~h car wi ll_i!]i ti_c1lly ta_l{e ! he lead] Support your answer.

What acceleration does car B experience?

At what time wi ll the cars be traveling at the same speed?

At what time w ill they once again be side-by-side?

\/V_119t speed wiHe_ach car be m_ov ing 9tthe exact n_iomentthatttiey pass each_CJ_th?r?

How far did each vehicle travel during the chase?

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

rest of the trip?

Sketch (using general shapes) and then describe the position-time graph of this problem.

posi ti on (rn)

time (sec)

http:/lwww.physi cslab.or!)'DocumentPrintaspx?doctype=5&fi lenam e=Kinematics_ChaseProbfem s2.xml 2/3

- - - ---

- - -

- - -

- - -

- - -

5111/2016 PhysicsLAB: SVA Relationships #4 5111/2016 PhysicslAB: SVA R~ationships #4

position( 11elocity accelerationWorksheet (ml Cm/sec) Cm/sec')A .. Physics

SVA Relationships #4 35..,~nfine LAB

s-t v-t a-t

instantaneous position

y-coordinate of point

displacement difference in two y-

coordinates area between graph and

x-axis

instantaneous velocity slope of tangent (to the graph) at specified time


change in velocity difference in two y-

coordinates area between graph and

x-axis

instantaneous acceleration

slope of graph at specified time


-7:'"7:31- I

-, II I V lJ

2 8 2 8 2 8

C)v

p

Refer to the following information for the next two questions.

velocity-lime, and acceleration-time graphs to determine the velocity, v. These graphs show an object lcsing speed while

Refer to the following information for the next two questions. position accele ..ation

Group #1: Use tile s,iven position-t,me. veloc1ty-t1me. and acceler'0t,on-t1me w"phs to detenrnne the Cml 11e loc ity

(m/sec')Cm/sec) missinq values for position, P. and ttie final velocity, v. These graphs show an object gaininq speed while

1 72501-~traveling in a negative direction.

v1· v-·,____ 4 position( 11elocity acceleration

Cml Cm/sec) Cm/sec•) ...L~ -35 1 -2 9 9

1 7 1 7

-13- ~ ,b"'Ll- -p;_:_ -V

I I V - - -

2 9 p

C)v

C) P PhysicsLAB Copyright© 1997-2016Refer to the following information for the next two questions. Catharine H. Colwell Al rights reserved.

Group #2: Use the given position-time, velocity-time, and acceleration-time graphs to determine the Application Prog ra mmer missing values for pos ition, P, and the final velocity, v. These graphs show an object gaining speed whi le Mark Acton

traveling in a positive direction.

hft+J:/fwww.physicslab.org1DocumentPrinlaspx?dcx:type=5&filename=Kinematics_SVA4.xml 1/2 http://www.physicslab.org'OocumentPrinLaspx?dcx:type=5&filename=Kinematics_SVA4.xml 2/2

5111/2016 Physicsl.AB: Freefall #1 5111/2016 PhysicsLAB: Freefal l #1

Worksheet Scenario #4: A rock is thrown straight up into the air and returns to its original release pos it ion after 3 ..t ... Physics seconds.Freefall #1""Vonline 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 v 0 O vr O a O s D t initia l velocity final veloc ity acceleration displacement time interval


O v 0 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 kinematics equation should you use?

0 a = v, - v, 1 1 lO s=- (v +v1 )t 0 v; =v; + 2as 0 s =v t +- at2 • • 2t

At what velocity does it strike the water? How fast was it released?

Which kinematics equation did you use to solve this problem?

0 a= v, -v, 1O s= -(v +v1 )t 0 v; =v; + 2as

2 •t 0 1 l s= v t+-~• 2

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

O a= v1 -v, 1 1 l0 s = 2(v, +v1 )t O v;=v;+2as 0 s=vt+-at• 2t

Based on the ORIGINAL givens, which equation would you now use to determine how mucl1 time it takes for the rock to reach the water?

0 a= v, - v, 10 s=

2(v, +v1 )t 0 v; =v; + 2as 0 s =v t +2-at1

• 2t

How high did it rise?

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


Scenario 11'2: The rock is now tossccl upwards at 6 miscc from a 20 meter bridge falls into tile river below.

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

Mark Acton

At what velocity does it strike the water?



Scenario #'3: The rock is now thrown downward at 6 m/sec from a 20 meter bridge falls into the river below.

At what velocity does it strike the water?



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5111/2016 PhysicslAB: Freefall #3 (Honors)

Worksheet.A .. Physics Freefall #3 (Honors) "' Vantine LAB


Wl1ile holding llis rifle at 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 misec 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 falling back down to earth?

If he does not move fast enougl1, at what velocity would the descending bullet strike his shoulder?


When rising to spike a ball in a volleyball 9arne. a player jumps verlically 1.5 meters off tt,e iloor.

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

ground.

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


Two students are tossing a set of keys from one to the other. Tl1e first student (wl10 initially l1as the 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: Freefal l #3 (Honors)

~ o motc:ts

1.8 motors

ground

The student on the ground tosses the keys upward, re leas ing them exac tl y as his hand reaches the top of his head, with just the right velocity so tl1at their apex coincides with the second student's outreached hand.


O v 0 D v 1 D a O s O t initial velocity final veloc ity acceleration displacement time interval

How fast were the keys tossed?

How much time did the keys spend in the air?


Suppose that the student on t11e iJalcony was distracted and failed to catc,; the keys and they fall back dow 11 to the ground.

With what velocity will they strike the grass at the feet of the f irst student?

True or False. The keys spent the same amount of time fa ll ing to the ground as they spent rising towards the balcony.

O True 0 False

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

O True 0 False

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5111/2016 Phys;csLAB: Freefal l #3 (Honors)

Refer to the following information for the next four questions.

Three siudents are standing side-by-side next io the railing on a fifth floor balcony. Simultaneously. the three students release their pennies.

One student proceeds to drop a penny to the ground below. rhe second student tosses his penny straight downwards at 15 m/sec, while Tl1e third student tosses his penny straight upwards at 15 misec.

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?

0 the penny that was dropped



Which penny or pennies strike(s) the ground witl1 the greatest final velocity?

D the penny that was dropped



4D 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 v 0 O v1 D a O s O t initial velocity final velocity acceleration displacement time inteival

With what velocity with the pencil hit the floor?

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

4D If the room temperature is 25 °C , how much total time passed between when he lost his g1ip on the

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

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5111/2016 Phys;csLAB: Freefall #3 (Honors)

4D Modeling this situation, how deep is a well if the 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.

PhysicsLAB Copyr~ht(!:) 1997-2016 Catharine H. Cotwell Al rights reserved. Application Programmer

Mark Acton

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5/11/2016 PhysicslAB: Position-,Time Graph "Story~ Combinations

Worksheet~ .. Physics Position-Time Graph "Story" Combinations"'Vantine LAB

Refer to the following information for the next four questions.

For each of the graphs shown below. choose the description thai 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 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 f inal rest.

0 whi le 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 orig inally at rest, uniformly 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

~ 0 a cart originally at rest, uniformly gains speed in a positive direction, maintains that speed, and then

uniformly loses its speed finall y 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 (111 iformly 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, unifom1ly 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

http://www.physicslab.or9'DocumentPrintaspx?doctype=5&filename=Kinematics_SVAcombinations.xml 1/3

5111/2016 PhysicsLAB: Position-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 whi le 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 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 veloc ity, then uniformly loses speed and comes to a final rest.

0 w hile 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 fi nally coming to a stop.

0 none of the above

Refer to the following information for the next question.

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

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

t-dtf,:/lwww.physicslab.or9'DocumentPrinlaspx?doctype= 5&filename=Kinematics_SVAcornbinations .xml 2/3

5111/2016 PhysicsLAB: Position-Time Graph "Story" Combinations

~ "walk" light turns "green," then walks the next block more slowly, and finally runs the final two blocks very ~ rapidly.

block 5 11-----l---+--+--+----+--+--+--

block 4 t-----+---+---+--+-->-----+----+--

block 3 1-------1---+--+---+--+-----+---+-

block 2 t-----+--+----+-----,>----+---+--+--

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

5 10 15 20 25 30 35 ti me(min)

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Mark Acton

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5111/2016 Pll'jsicsLAB: Derivatives: Instantaneous vs Average Vek1cities

Resource Lesson~ .. Physics Derivatives: Instantaneous vs Average Velocities...,Vonfine LAB

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 the length of t11e path traveled and displacement representing the length of the straight-line vector that begins at the starting point and ends at the finishing point, independent of the path taken.

BA- ------------

c~D

In both of these examples, the displacements are identical whereas the distance traveled between points C and Dis much greater than the distance traveled between points A and B since the direction switches back and forth causing the path len gth 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 througt1out the internal then the average speed during that interval would only occasionally equal its instantaneous speed at any one moment in the interval. Obvious ly, the notion of a constant speed would no longer be discussed under those conditions.

6 8 sL------ L _

64 3 4

2 2

-1 -2

-2

-3 -4 -4

-6-5

-6 -8

Both of these graph represent position vs time graphs for objects travel in g along straight-line paths. On the

5111/2016 Physicsl.AB: Dwivatives: Instantaneous vs Average Velocities

left, the first object travels at a constant speed beginning at x = 5 and ending at x = -6 taking approximately 5.5 seconds to complete the trip. On the right graph, a second object is traveling at a variable rate beginning at x = 1 traveling down to x = 0 then up to x =6 and finally all the way back to x = -8 taking approximately 5 seconds to complete 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 the 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 achieve 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 ass ign positive/negative directions in setting up the problem.

On the left graph displayed above the first object's average speed (11 units/ 5.5 seconds) equals t11e magnitude of its average velocity (11 units/ 5.5 seconds, down}. While on the right graph the second object's average velocity (9 units 15 seconds, down) is much smaller than Its average speed (15 units/ 5 seconds).

• On the other hand, an object's instantaneous speed wi ll always equal the magnitude of its instantaneous velocity.

PhysicsLAB Copyright a:> 1997-2016 Cath arine H. Colwell Al rights reserved. Application Programmer

Mark Acton

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5111/2016 Physicsl.AB: Average Vetcx:ity • A Calcllus Approach

Resource Lesson~ ... Physics Average Velocity - A Calculus Approach ""Vonline LAB

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

Suppose a particle is moving back and forth along the x-axis with a position function (the coordinate giving the location of the particle on the x-axis) given by x(t) = t2 - 2t - 3.

4-,l-r--·--.

Assume that tis in seconds, t < 0 is not possible, and the coordinates on the x-axis are in meters. We already know the fo llowing 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 part ic le is at the origin when x(t) =0 =(t -3)(1 + 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 partic le 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 I =2 and t =5?

@ What is the instantaneous 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) = 2t - 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 Physics LAB: Average Vetocity - A Calcllus Approach

1

-8

6-e-- ---·-·•··

·19-r--,-· -- ---,-,--

Some basic facts:

1. v(t) = 2t - 2 is t11e 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 detennine 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 pos ition of the particle after the first 5 seconds of motion?

@ What is the total distance traveled by t11e particle in the first 5 seconds?

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

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

Mark Acton

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5111/2016 PhysicsLAB: Acceteratecl MOOon: Practice with Data Analysis

Worksheet~ .. Physics Accelerated Motion: Practi ce with Data Analysis ...,Vantine LAB


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

I I position

sec I m 0 I 0

D.5 I D.25

1 I 1

1.5 I 2.25

2 I 4

2.5 I 6.25

3 I 9

3,5 I 12.25

4 I 16

4.5 I 20.25

f'> I 25

Cl) What was the ball's acceleration?


The following table represents a group's data for the motion of their cart clown the hallway.

I I position

sec I m

2 I 5

2.8 I 9.8

3.6 I 16.2

4.4 I 24.2

5.2 I 33.6 5 I 45

6.8 I 57.8

7.6 I 72.2

8.4 I 88.2

9.2 I 105.8

10 I 125

Cl) What acceleration did the cart experience?


Tl1e follow ing table represents a group's data for the motion of box sliding down a s lide.

5111/2016 Physics LAB: Accelerated MOOon: Practice with Data Analysis

t I position

sec I cm 0 I 0

1 I 0.8

2 I 3.2

3 I 7.2

4 I 13,8

5 I 23 6 I 32.8

7 I -1 6.2

6 I 60.2

9 I 75.9

10 I 92

c:) What acceleration did the box experience?

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

Mark Acton

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graphs and /or Cal cul Ot-ho-ns

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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.

Graph shape Written relat ionship Modification

requirecf fo linearize graph

Algebraic representation

.. --- .i l - . . •·· - · · ·· ·- •·

y

~

. -X

As x increases, y re mains the same. There is no re lationship between the variables.

.. .. -~ . .. .. · ·

None :,.,.- ..

y = b, or y · is constant

.a y V_

As X increases, y increases proportionally. Y is directly proportional to x.

None y '= mx+b

- .... X

. h

y As X increases, y decreases.

1Graph y vs - , or

X

( l ~ y=m-+b\x)

~ Y is inversely proportional to X .

-1 y V S X

. -X

_j

y

-~-....

X

..

Y is proportional to the square of x.

..

x2Graph y VS

'

2 y = mx +b

~ L

y

. ~

X

The square of y is proportional to x .

Graph y2 VS X 2 y = mx +b

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

due the first day of class. kgazq9j - shelbyed.k12.al.us · Morion is all around us, so it's easy...

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