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Motion In 2 Dimensions Homework 101

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Motion in two dimensions

Motion in two dimensions

Sections 3.5 - 3.7

Extending things from 1 dimension

In 1 dimension, we wrote down some general equations relating velocity to displacement, and relating acceleration to the change in velocity. We also wrote down the four equations that apply in the special case where the acceleration is constant. We're going to do the same thing in 2 dimensions, and the equations will look similar; this shouldn't be surprising because, as we will see, a two (or three) dimensional problem can always be broken down into two (or three) 1-dimensional problems.

When we're dealing with more than 1 dimension (and we'll focus on 2D, but we could use these same equations for 3D), the position is represented by the vector r. The velocity will still be represented by v and the acceleration by a. In general, the average velocity will be given by:

The instantaneous velocity is given by a similar formula, with the condition that a very small time interval is used to measure the displacement.

A similar formula gives the average acceleration:

Again, the instantaneous acceleration is found by measuring the change in velocity over a small time interval.

The constant acceleration equations When the acceleration is constant, we can write out four equations relating the displacement, initial velocity, velocity, acceleration, and time for each dimension. Like the 1D equations, these apply under the following conditions:
  1. the acceleration is constant
  2. the motion is measured from t = 0
  3. the equations are vector equations, but the variables are not normally written in bold letters. The fact that they are vectors comes in, however, with positive and negative signs.

If we focus on two dimensions, we get four equations for the x direction and four more for the y direction. The four x equations involve only the x-components, while the four y equations involve only the y-components.

One thing to notice is that the time, t, is the only thing that doesn't involve an x or a y. This is because everything else is a vector (or a component of a vector, if you'd rather look at it that way), but time is a scalar. Time is the one thing that can be used directly in both the x and y equations; everything else (displacement, velocity, and acceleration) has to be split into components.

This is important!

Something that probably can't be emphasized enough is that even though an object may travel in a two-dimensional path (often following a parabola, in the standard case of an object moving under the influence of gravity alone), the motion can always be reduced to two independent one-dimensional motions. The x motion takes place as if the y motion isn't happening, and the y motion takes place independent of whatever is happening in the x direction.

One good example of this is the case of two objects (e.g. baseballs) which are released at the same time. One is dropped so it falls straight down; the other is thrown horizontally. As long as they start at the same height, both objects will hit the ground at the same time, no matter how fast the second one is thrown.

What we're ignoring

We will generally neglect the effect of air resistance in most of the problems we do. In some cases that's just fine. In other cases it's not so fine. A feather and a brick dropped at the same time from the same height will not reach the ground at the same time, for example. This has nothing to do with the weight of the feather compared to the brick. It's simply air resistance; if we took away all the air and dropped the feather and brick, they would hit the ground at exactly the same time.

So, remember that we're often analyzing ideal cases, especially this early in the semester. In reality, things might be a little different because of factors we're neglecting at the moment.

An example

Probably the simplest way to see how to apply these constant acceleration equations is to work through a sample problem. Let's say you're on top of a cliff, which drops vertically 150 m to a flat valley below. You throw a ball off the cliff, launching it at 8.40 m/s at an angle of 20° above the horizontal.
(a) How long does it take to reach the ground?
(b) How far is it from the base of the cliff to the point of impact?

It's a good idea to be as systematic as possible when it comes to analyzing the situation. Here's an example of how to organize the information you're given. First, draw a diagram.

Then set up a table to keep track of everything you know. It's important to pick an origin, and to choose a coordinate system showing positive directions. In this case, the origin was chosen to be the base of the cliff, with +x being right and +y being up. You don't have to choose the origin or the positive directions this way. Pick others if you'd like, and stick with them (an origin at the top of the cliff, and/or positive y-direction down would be two possible changes).

Now that everything's neatly organized, think about what can be used to calculate what. You know plenty of y-information, so we can use that to find the time it takes to reach the ground. One way to do this (definitely not the only way) is to do it in two steps, first calculating the final velocity using the equation:

This gives vy 2 = 2.873 2 + 2 (-9.8) (-150) = 2948.3 m 2 / s 2. Taking the square root gives: vy = +/- 54.30 m/s.

Remember that the square root can be positive or negative. In this case it's negative, because the y-component of the velocity will be directed down when the ball hits the ground.

Now we can use another equation to solve for time:

So, -54.30 = 2.873 - 9.8 t, which gives t = 5.834 seconds. Rounding off, the ball was in the air for 5.83 s.

We can use this time for part (b), to get the distance traveled in the x-direction during the course of its flight. The best equation to use is:

duffy/PY105/5g.GIF" />
So, from the base of the cliff to the point of impact is 46.0 m.

A point about symmetry

At some point in its flight, the ball in the example above returned to the level of the top of the cliff (you threw it from the top of the cliff, it went up, and on its way down it passed through a point the same height off the ground as the top of the cliff). What was the ball's velocity when it passed this height? Its speed will be the same as the initial speed, 8.40 m/s, and its angle will be the same as the launch angle, only measured below the horizontal.

This is not just true of the initial height. At every height the ball passes through on the way up, there is a mirror-image point (at the same height, with the same speed, and the same angle, just down rather than up) on the downward part of the path.

Other articles

Ch 4 – Motion in 2-D

Ch 4 – Motion in 2-D. Welcome back. Homework Reminders: One problem per page Sketches, sketches, sketches Blurbs Start with formula from equation sheet. Presentation on theme: "Ch 4 – Motion in 2-D. Welcome back. Homework Reminders: One problem per page Sketches, sketches, sketches Blurbs Start with formula from equation sheet. "— Presentation transcript:

2 Welcome back. Homework Reminders: One problem per page Sketches, sketches, sketches Blurbs Start with formula from equation sheet.

3 To solve 2-d problems: Draw a sketch, and add vectors by placing them tip-to-tail (or parallelogram) and drawing the Resultant. Add vectors more precisely by: breaking them into x- and y- components; adding x- and y-components separately to get components of Resultant converting Resultant components back to polar notation Using vector-based kinematic equations (like r=v i t+(1/2)at 2 ) with vectors written in unit-vector (i, j) notation.

4 Example 1 – Projectile Motion A long jumper leaves the ground with a velocity of 11.0m/s at 20.0° above the horizontal. a.What is the maximum height he reaches? b.How far horizontally does he jump? Solve using components. c.Then, once you know time, solve again using i, j notation.

5 Example 1 – Solutions a.How far does she jump? Get components of initial velocity: Consider vertical and horizontal situations independently! Lets start with vertical situation: how long will she be in the air?

7 Example 1 – Solutions c.If we calculate (using y direction) that the time for the leap is t=0.767s, lets solve for horizontal distance.

8 Example 2 – More theoretical A theoretical problem: A particle starts from the origin at time t=0, with initial velocity v x =-10m/s, and v y =+5m/s. The particle is accelerating at 3 m/s 2 in the x direction. a.Find v components as a function of time. b.Find v as a function of time. c.Find v inst at time t=5.00s. d.Find position coordinates x and y as a function of time. e.Find r as a function of time. f.Find displacement vector(in i,j notation) at time t=5.00s. a.v x = 5-3t; v y = 5 b.v=((5-3t)i+5j)m/s c.v inst =11.2m/s d.x=v i t+1/2at 2 = (- 10)t+1/2(3)t 2 ; y=5t e.r=((- 10)i+5j)t+1/2(3i)t 2 f.r=-12.5i+25j

9 Projectile Motion Weve already discussed projectile motion briefly. The general idea is this: Objects thrown in the air are considered projectiles; objects thrown at some angle to the vertical follow a curved parabolic path called a trajectory. We can easily solve all sorts of projectile problems if we make two assumptions: free-fall acceleration is constant throughout the trajectory, and air resistance is negligible. The reason these problems are easy to solve is because we can consider the horizontal and vertical motions independently. In fact, were forced to this by the nature of the problem: horizontally, there is no force that causes the velocity of the object to change, while vertically, the force of gravity causes the object to accelerate toward the earth.

13 Example 3 A stone is thrown from the top of a building, with an initial velocity of 20m/s at 30° above the horizontal. If the building is 45.0 m tall. a. how long is the stone in flight? b. What is the speed of the stone just before it hits the ground? c. Where does the stone strike the ground?

14 Uniform Circular Motion Weve defined acceleration a to be a change in velocity over a period of time: Because velocity is a vector, it should be clear that there are two ways that we can change velocity: a.changing speed (speeding up or slowing down) b.changing direction (even if the speed remains constant) Thus, an object traveling in a circle, even at constant speed, is accelerating.

15 Direction of centripetal a We can determine a formula for calculating centripetal (center-seeking) acceleration as follows.

17 Centripetal acceleration An object moving in a circle of radius r with constant speed v has an acceleration directed toward the middle of the circle, with a magnitude

18 Example 4 – Hammer time At the beginning of this hammer throw, a 5 kg mass is swung in a horizontal circle of 2.0 m radius, at 1.5 revolutions per second. What is the acceleration of the mass (magnitude and direction)?

19 Example 4 – Hammer time At the beginning of this hammer throw, a 5 kg mass is swung in a horizontal circle of 2.0 m radius, at 1.5 revolutions per second. What is the acceleration of the mass (magnitude and direction)?

20 Example 4 – Nice acting, Keanu 1.There is a gap in the freeway. Convert the gaps width to meters. 2.How fast is the bus traveling when it hits the gap? What is its velocity in m/s? 3.Keanu hopes that there is some "incline" that will assist them. Assume that the opposite side of the gap is 1 meter lower than the takeoff point. Also, the stunt drivers that launch this bus clearly have the assistance of a "takeoff ramp" from which the bus launches at an angle. Assume that the ramp is angled at 3.00° above the horizontal. Prove whether or not the bus will make it to the opposite side. (Consider bus as a particle.)

21 Tangential & radial a Radial acceleration a r is due to the change in direction of the velocity vector: a r =v 2 /r Tangential acceleration a t is due to the change in speed of the particle: a t = dv/dt.

22 More unit vectors? Its sometimes convenient to be able to write the acceleration of a particle moving in a circular path in terms of two new unit vectors: is a unit vector tangent to the circular path, with positive in ccw direction r is a unit vector along the radius vector, and directed radially outward. So, if the bowling ball above had a radial acceleration of 4m/s 2, and a tangential acceleration of 2m/s 2, what would its net acceleration be at that point?

23 Example 5 – A little trickier A bowling ball pendulum is tied to the end of a string 3.00 m in length, and allowed to swing in a vertical circle under the influence of gravity. When the hanging ball makes an angle of 15.0° with vertical, it has a speed of 2.00 m/s. a.Find the radial acceleration at this instant b.When the ball is at an angle relative to vertical it has tangential acceleration of g sin. Find the net acceleration of the ball at = 15.0°,(using same data as above) and express it in,r form. Then express that net a in polar notation. (

24 Relative Motion Observers with different viewpoints (frames of reference) may measure different displacements, velocities, and accelerations for any given particle, especially if the two observers are moving relative to each other.

27 Relative Velocity in 2-D What happens when we try to aim the boat directly across the river?

28 Relative Motion Analysis For relative motion problems, its useful to label the vectors very clearly, in such a way that we can describe an objects velocity relative to a certain reference frame. If the boats velocity (relative to the water) is 3m/s @ 90°, and the rivers velocity is 4m/s @ 0° (relative to the shore), what is the boats velocity relative to the shore?

29 Example 6 A boat can travel with a velocity of 20.0 km/hr in water. a.If the boat needs to travel straight across a river with a current flowing at 12.0 km/hr, at what upstream angle should the boat head? b.If the river is 6 km across, how long will it take the boat to get there?

30 Example 7 A plane wants to fly at 300 km/h, 135°, but a 50 km/h wind blows from the north. What velocity should the plane fly at (magnitude & direction) to achieve its desired velocity?

31 Feeback on First Lab Some points to keep in mind when doing labs: 1.Name goes at upper right corner of every sheet 2.Blurbs (brief written descriptions) should accompany all calculations, especially where source of values used may not be obvious. 3.When using graphs for calculations, identify on the graph which data points youre using for your calculations. 4.Follow the outline given in the lab writeup description (online) 5.Space out your writing as much as possible, to leave room for corrections & blurbs (you) & comments (me) 6.In summary, state error percentages as evidence that the lab was (or was not) successful.

32 Projectile Motion The Monkey & the Hunter (Demo) Question: will bullet hit: a) above target; b) on target, or c) below target?

33 Projectile Motion Explanation: Solving for the targets position as a function of time, and the bullets position as a function of time, will allow you to determine the y- coordinates of each object when their x- coordinates are equal.

PPT - Motion in Two Dimensions PowerPoint Presentation

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Presentation Transcript

Motionin Two Dimensions

Homework

Projectile Motion
  • What is the path of a projectile as it moves through the air?
  • Straight up and down?

    • Yes, both are possible.
  • What forces act on projectiles?
    • Only gravity, which acts only in the negative y-direction.
  • Air resistance is ignored in projectile motion.

    Choosing Coordinates & Strategy
    • For projectile motion:
    • Choose the y-axis for vertical motion where gravity is a factor.
  • Choose the x-axis for horizontal motion. Since there are no forces acting in this direction (of course we will neglect friction due to air resistance), the speed will be constant (a = 0).

  • Analyze motion along the y-axis separate from the x-axis.

  • If you solve for time in one direction, you automatically solve for time in the other direction.

  • Motion in two dimensions

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    Solution to Example 3

    Solution to example 3. motion in two or three

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    Solution to Example 3. Motion in Two or Three Dimensions Identify: Take the origin of coordinates at the room and let the + y direction be upward. The rock moves in projectile motion, with a x = 0 and a y =  g. g v t gt v v i i h h i i y       sin 0 sin   m s m s m h g v h g v g g v v h y i i i i i i i i f 6. 13 ) / 80. 9 ( 2 ) 0. 33 ( sin / 0. 30 2 sin sin 2 1 sin sin 2 2 2 2 2 2                    yi v xi v i 

    (b) Calculate the magnitude of the velocity of the rock just before it strikes the ground. In our chosen coordinates the ground is located at -15 m. along x: along y: Solution to Example 3. Motion in Two or Three Dimensions (continued) s m v v i i x / 2. 25 cos    g i i y gt v v    sin   s m s m s m v s m v m s m v y g v v m y y g a y y a v v yg o yg g i i yg g i y i g y yi yg / 6. 34 ) / 7. 23 ( ) / 2. 25 ( / 7. 23 ) 0. 15 )( / 80. 9 ( 2 0. 33 sin ) 0. 30 ( ) ( 2 sin 15 ; 0 ; 2 2 2 2 2 2 2 2 2 2                       

    END OF LECTURE 4

    PHY101 Physics Solved MCQs Motion in two and Three Dimensions 1

    Virtual University MCQs BANK - MCQs Collection from Online Quizzes

    PHY101 Physics Solved MCQs Motion in two and Three Dimensions

    1. Velocity is defined as:
    A. rate of change of position with time
    B. position divided by time
    C. rate of change of acceleration with time
    D. a speeding up or slowing down
    E. change of position

    2. Acceleration is defined as:
    A. rate of change of position with time
    B. speed divided by time
    C. rate of change of velocity with time
    D. a speeding up or slowing down
    E. change of velocity

    3. Which of the following is a scalar quantity?
    A. Speed
    B. Velocity
    C. Displacement
    D. Acceleration
    E. None of these

    4. Which of the following is a vector quantity?
    A. Mass
    B. Density
    C. Speed
    D. Temperature
    E. None of these

    5. Which of the following is NOT an example of accelerated motion?
    A. Vertical component of projectile motion
    B. Circular motion at constant speed
    C. A swinging pendulum
    D. Earth’s motion about sun
    E. Horizontal component of projectile motion

    6. A particle goes from x = −2m, y =3m, z =1mto x =3m, y = −1m, z = 4m. Its displacement is:
    A. (1m)ˆi + (2m)ˆj + (5m) ˆk
    B. (5m)ˆi −(4m)ˆj + (3m) ˆk
    C. −(5m)ˆi + (4m)ˆj −(3m) ˆk
    D. −(1m)ˆi −(2m)ˆj −(5m) ˆk
    E. −(5m)ˆi −(2m)ˆj + (3m) ˆk

    7. A jet plane in straight horizontal flight passes over your head. When it is directly above you, the sound seems to come from a point behind the plane in a direction 30◦
    from the vertical. The speed of the plane is:
    A. the same as the speed of sound
    B. half the speed of sound
    C. three-fifths the speed of sound
    D. 0.866 times the speed of sound
    E. twice the speed of sound

    8. A plane traveling north at 200m/s turns and then travels south at 200m/s. The change in its velocity is:
    A. zero
    B. 200m/s north
    C. 200m/s south
    D. 400m/s north
    E. 400m/s south

    9. Two bodies are falling with negligible air resistance, side by side, above a horizontal plane. If one of the bodies is given an additional horizontal acceleration during its descent, it:
    A. strikes the plane at the same time as the other body.
    B. strikes the plane earlier than the other body
    C. has the vertical component of its velocity altered
    D. has the vertical component of its acceleration altered
    E. follows a straight line path along the resultant acceleration vector

    10. The velocity of a projectile equals its initial velocity added to:
    A. a constant horizontal velocity
    B. a constant vertical velocity
    C. a constantly increasing horizontal velocity
    D. a constantly increasing downward velocity
    E. a constant velocity directed at the target

    PPT - Physics 103: Lecture 5 Vectors Motion in Two Dimensions PowerPoint presentation

    Physics 103: Lecture 5 Vectors Motion in Two Dimensions - PowerPoint PPT Presentation

    Transcript and Presenter's Notes


    Title: Physics 103: Lecture 5 Vectors Motion in Two Dimensions


    1
    Physics 103 Lecture 5 Vectors - Motion in Two
    Dimensions
    • Todays lecture will be on
    • Vectors
    • Two dimensions
    • projectile motion

    2
    1D Kinematics Equations for Constant Acceleration
    _
    • ?x vt (linear)
    • ?x v0t 1/2 at2 (parabolic)
    • ?v at (linear)
    • v2 v02 2a ?x
    • (independent of time)

    3
    Two Dimensions
    • Position can be anywhere in the plane
    • Select an origin
    • Draw two mutually perpendicular lines meeting at
      the origin
    • Select /- directions for horizontal (x) and
      vertical (y) axes
    • Any position in the plane is given by two signed
      numbers
    • A vector points to this position
    • The square of its length is, R2 Rx 2 Ry 2
    • The angle of that vector is. tan-1(Ry / Rx)

    R
    Ry
    Rx R cos. Ry R sin ?
    q
    Rx
    4
    Vector Algebra
    • Analytical method
    • Add the components separately to get the
      components of sum vector
    • Rx R1x R2x
    • Ry R1y R2y
    • Scalar multiplication of vector
    • Can change magnitude and sign
    • Multiply all components by scalar
    • Components of sR are sRx and sRy
    • Negation of vector (multiplying by -1)
    • Reverse signs of both components
    • Vector points in
      opposite direction

    y
    x
    5
    Kinematics in Two Dimensions
    • x x0 v0xt 1/2 axt2
    • vx v0x axt
    • vx2 v0x2 2ax ?x
    • y y0 v0yt 1/2 ayt2
    • vy v0y ayt
    • vy2 v0y2 2ay ?y

    x and y motions are independent! They share a
    common time t
    6
    Kinematics for Projectile Motion ax 0 ay
    -g -9.8
    • y y0 v0yt - 1/2 gt2
    • vy v0y - gt
    • vy2 v0y2 - 2g ?y
    • x x0 vxt
    • vx v0x

    x and y motions are independent! They share a
    common time t
    7
    Question?
    Without air resistance, an object dropped from a
    plane flying at constant speed in a straight line
    will 1. Quickly lag behind the plane. 2.
    Remain vertically under the plane. 3. Move
    ahead of the plane
    There is no acceleration along horizontal -
    object continues to travel at constant speed
    (same as that of the plane) along horizontal. Due
    to gravitational acceleration the objects speed
    downwards increases.
    8
    Projectile Motion
    Look at maximum height reached Time taken for
    getting there
    9
    Projectile Motion Maximum Range
    10
    Projectile Motion
    11
    Shooting the Monkey.
    x v0 t y -1/2 g t2. g 9.8
    x x0 y -1/2 g t2
    No monkeys were harmed during the making of this
    slide
    12
    Shooting the Monkey.
    y y0 - 1/2 g t2
    • At an angle, still aim at the monkey!

    y vy0 t - 1/2 g t2 g 9.8
    13
    Shooting the Enemy Paratrooper.
    • If a soldier wants to shoot down an enemy
      paratrooper descending at uniform speed, s, where
      should he aim?
    • Above the enemy
    • At the enemy
    • Below the enemy
    • Answer depends on the enemys position and
      vertical speed.

    y y0 - st
    y vy0 t 1/2 g t2
    Miss the enemy
    14
    Review 2D Motion
    • x x0 v0xt 1/2 axt2
    • vx v0x axt
    • vx2 v0x2 2ax ?x
    • y y0 v0yt 1/2 ayt2
    • vy v0y ayt
    • vy2 v0y2 2ay ?y
    • Solving 2D motion problems
    • Make a pictures and write out the equations
    • Mark everything you know and what you want to
      know
    • Simplify, eliminate 0 terms
    • Think, is there an obvious answer to the problem?
    • Shoot the Monkey The dart would hit the monkey
      without gravity so if you add the same gravity
      term to both the dart will still hit the target
    • Solve remembering time is in both sets of
      equations

    15
    Summary
    Projectile Motion
    • y y0 v0yt - 1/2 gt2
    • vy v0y - gt
    • vy2 v0y2 - 2g ?y
    • x x0 v0t
    • v v0x

    16
    Reference Frames Relative Motion
    VCBVCAVAB
    Velocity of B relative to ground ( C )
    VCB Velocity of A relative to ground ( C )
    VCA Velocity of B relative to A VAB
    17
    Relative Motion
    • If an airplane flies in a jet stream, depending
      on the relative orientation of the airplane and
      the jet stream, the plane can go faster or slower
      than it normally would in the absence of the jet
      stream
    • If a person rows a boat across a rapidly flowing
      river and tries to head directly for the shore,
      the boat moves diagonally relative to the shore
    • Velocity is a vector - add velocities like
      vectors
    • Sum the components
    • Vx V1x V2x
    • V1x V1 cosq
    • Vy V1y V2y
    • V1y V1 sinq

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    Physics 3305

    Class Information
    • Instructor: Professor Jodi Cooley
    • When/Where: Tuesdays and Thursdays, 9:30 am - 10:50 am, 155 Fondren Science Building
    • Teaching Assistants: Mr. Hang Qiu and Mr. Matthew Rispoli
    • Office Hours:
      • Cooley: Tuesdays from 2-3 pm and Wednesdays from 2-3 pm in FS 151
      • Qiu: Thursdays from 4-5 pm in FS 041
      • Rispoli: Friday 1 - 3:00 pm in FS 026
      • Other times available by appointment
    • Help Sessions:
      • Wednesdays 7 - 8 pm in Dallas 101 (Qiu)
      • Thursdays 7 - 8 pm in Fondren Science 158 (Rispoli)
    • Course Syllabus
    • Regrade Policy
    • Homework Policy
    • Homework Rubric
    • WileyPLUS Information:
      • Getting Started
      • For help with WileyPLUS software contact tech support. (http://wileyplus.custhelp.com/)
    • Addition information on Significant Figures
      • Handout on Significant Figures
    Class Schedule News
    • Final exam is on Friday, December 9 at 8 am in FOSC 155.
    • Quiz 9 covering chapter 13 will be in class on Tuesday, November 29.
    • Extra help sessions will be held in FOSC 026 on Monday, Nov. 14 at 7 pm and Tueaday, Nov. 15 at 6 pm.
    • Midterm exam 3 covering chapters 10 - 12 will be in-class on Thursday, November 17.
    • Midterm exam 2 covering chapters 5 - 9 will be in-class on Thursday, October 27. Study hard!
    • Quiz 6 covering chater 8 will be in-class on Thursday, October 20.
    • Quiz 5 covering chapter 7 will be in-class on Thursday, October 6.
    • Quiz 4 covering chapters 5 - 6 will be in-class on Thursday, September 29.
    • Review sessions for Exam 1 will be Tuesday, Sept. 20 from 6 - 7 pm in FS 153 and Wednesday, Sept. 21 from 7 - 8 pm in Dallas Hall 101. Bring your questions.
    • Exam 1 covering chapters 1 - 4 will be in-class on Thursday, Sept. 22
    • Quiz 3 covering chapters 2 and 3 will be in-class on Thursday, Sept. 15
    • Quiz 2 covering chapters 2 and 3 will be in-class on Thursday, Sept 8.
    • Don't foget to bring your calculator for in-class quiz 1 covering chapter 1 on Thursday Sept 1.
    • First day of class is Tuesday, August 23!

    Copyright © 2013 Jodi Cooley.