KINEMATICS

=SEPTEMBER 10 (Notes by Mrs. Price)= __**Error Analysis**__ - uncertainty in every measured value - expressed in 2 ways: eg. 24.08 +/- .005 (half of the smallest division of the scale on the instrument) or eg. 24.08 +/- 5% ( a % of the whole value)
 * TOPIC - Math Review (Cont.)**
 * Text: pg 755**

__**% Error**__ - used to compare the experimental value with the most commonly accepted value

%Error = [measured value - accepted value] / accepted value x 100%

__**% Difference**__ - Used to determine if a column of values within an experiment yeild a constant value - helps to determine the consistancy of the experimental proceedure

% Difference = (highest value - lowest value) / average value x 100%

Use the 5% Rule for both % Difference and Error. i.e. % Difference - if the value calculated is less than or equal to 5% then the value can be considered **THE SAME AS** the accepted value

% Error - if the value calculated is less than or equal to 5% then you can state that you have determined a constant value through your experiment.

//SEPTEMBER 11 **(Notes by Balthazar)**//

__**1.1 Speed and Velocity in 1 & 2 D's**__ __**Position and Velocity Graphs**__ - On a position time graph (d-t graph), a straight line with a constant slope, represents an object at a constant velocity. - Finding the slope of this line gets you the velocity of this object. The formula is as so. If you want to get really fancy about it, you can think about the delta(triangle) d as d2 – d1, and delta t as t2-t1. Though it's infinitely easier just to choose your first point as 0 and eliminate the need for subtraction. But it's a free country! So choose wisely. You might only be given two specific points on that line, so now you will know what to do if caught in that situation.
 * Topic: D-t and v-t graphs, displacement and acceleration**
 * Text pg 11-16, 20-21, today's notes**

- (Note: Because the object has a constant velocity, the average velocity is the same as the instantaneous velocity) - Now, let's say our object is accelerating. On a d-t graph, you would see it as a curve moving upwards, as it's traveling a faster distance in a shorter time interval. - To measure instantaneous velocity, you would first draw a line tangential to a point on this curve. That is, a line that intersects with one AND I MEAN PRECISELY ONE, point on the curve. Find the slope of that line and you have the velocity at the value for t at that point. Graph time again!

-Notation for the instantaneous velocity is slightly different:

__**Displacement and Velocity in 2D**__ -Problems involving displacement, velocity and time can take place on a horizontal or vertical plane, or at an angle to the horizontal plane -Our vectors need direction, so the conventional use here is cardinal directions. Know your norths, souths, easts, and wests! If the direction is between those, use an angle to be specific. Eg: A car travelling 40 m/s [N 14**°** E] -To round things off, here's a slightly new equation called total displacement. In a problem involving more than one displacement, the total displacement is the sum total of all of those displacements. Vectors are added tip to tail as always. Moving on now.. wait you want a formula!? Fine, I guess I can write it out...
 * What's with the lim thing you ask? I'm pretty sure that's calculus notation for a limit to this function. In this case, I think that means that delta t is going to infinitely approach 0, because anything less would make no sense in this physics context. There has to be a change in time for anything to well... move. That part is important

__**1.2 Acceleration in 1 and 2 Dimensions**__

__**Graphing Motion With Constant Acceleration**__ -What do those three graphs share in common? They're all describing an object at constant acceleration. -On that v-t graph, you can find displacement by calculating the area underneath the line. With the acceleration-time (a-t) graph, you can find the change in velocity by finding that area -Also on a v-t graph, finding the slope gets the acceleration for that period. This can be expressed very nicely mathematically as:

__**AND A FEW OTHER THINGS**__ -The slope of a secant between two points on an acceleration curve in a d-t graph, is equal to the average velocity of that time interval.

-Was that chart causing you grief in trying to remember what a positive acceleration does to a negative velocity? Here's it handily filled up to ease up that confusion. Try and remember it like sign rules in math. + and + = +, - and - = +, positive meaning that you're speeding up. + and - = -, - and + = -, which means the object is slowing down. If that velocity be positive, you're moving in a positive direction, and if it be negative, you're moving in the negative direction.

(You DON'T know how much work went into getting those pictures here)
 * Speeding up || Slowing down || a || v || v-t graph || d-t graph ||
 * Yes, in positive direction ||  || + || + || [[image:http://i147.photobucket.com/albums/r292/Underqwertyuiop/Physics9.jpg width="174" height="103"]] || [[image:http://i147.photobucket.com/albums/r292/Underqwertyuiop/Physics10.jpg width="139" height="111"]] ||
 * || Yes, in negative direction || + || - || [[image:http://i147.photobucket.com/albums/r292/Underqwertyuiop/Physics11.jpg width="164" height="109"]] || [[image:http://i147.photobucket.com/albums/r292/Underqwertyuiop/Physics12.jpg width="162" height="101"]] ||
 * || Yes, in positive direction || - || + || [[image:http://i147.photobucket.com/albums/r292/Underqwertyuiop/Physics13.jpg width="159" height="122"]] || [[image:http://i147.photobucket.com/albums/r292/Underqwertyuiop/Physics14.jpg width="154" height="120"]] ||
 * Yes, in negative direction ||  || - || - || [[image:http://i147.photobucket.com/albums/r292/Underqwertyuiop/Physics15.jpg width="164" height="103"]] || [[image:http://i147.photobucket.com/albums/r292/Underqwertyuiop/Physics16.jpg width="161" height="102"]] ||

//SEPTEMBER 12 **(Notes by your friendly neighbourhood Spider-man)**//

//**Topic: Continued 2D acceleration**// //**Text pgs 24-27**// __**1.2 Accelerating in 1 & 2 D's (Cont.)**__ __**Solving Constant Acceleration Problems**__ -When it comes to acceleration, there are two defining equations. One for acceleration, and another for displacement. Through rearranging and substituting them into each other, you end up with three other fancy and very useful equations. The first two are: -With some good old (some might call it fun!) math, you can derive these formulae: **Examples** (and they're even part of the homework, two birds with one stone!)

A badminton shuttle, or "birdie," is struck, giving it a horizontal velocity of 73 m/s [W] Air resistance causes a constant acceleration of 18 m/s2[E]. Determine it's velocity after 1.6 s

vi= 73m/s [W]a = -18 m/s2 [E]t = 1.6 svf = ? a = vf-vi/t vf = at + vi= (-18m/s2) (1.6s) + 73 m/svf = 44 m/s [W]

An electron travelling at 7.72 x 107 m/s [E] enters a force field that reduces its velocity to 2.46 x 107 m/s [E]. The acceleration is constant. The displacement during the acceleration is 0.478 m [E]. Determinea) the electron's accelerationb) the time interval over which the acceleration occurs vi = 7.72 x 107 m/s [E]vf = 2.46 x 107 m/s [E]d = 0.478 ma = ?t = ? a) vf2 – vi2 = 2ada = vf2 – vi2/2d= (2.46 x 107)2 – (7.72 x 107)2/ 2(0.478)= -5.35 x 1015/0.956a = 5.60 x 1015 m/s2 [W] b) a = vf – vi/tt = vf – vi/a= (2.46 x 107) – (7.72 x 107)/ -5.60 x 1015= -5.26 x 107/-5.60 x 1015t = 0.9 x 10-8 s

SEPTEMBER 14 (Notes by Gehrig) **TOPIC - Solving One- and Two-Body-Problems** **Text Book Pages: Unknown**

**__Solving a One-Body Problem__** *NOTE: Diagrams are often very helpful, but not required.
 * 1) Define a reference point and direction
 * 2) State all given information
 * 3) Make any necessary conversions to appropriate units
 * 4) Identify unknown variable
 * 5) Choose appropriate equation(s)
 * 6) Substitute in given values
 * 7) Solve for unknown variable
 * 8) Include a therefore (:.) statement to show and justify selected answer

**__Solving Two-Body Problems__** *NOTE: Diagrams are often very helpful, but not required.
 * 1) Treat each body as independent one-body problem
 * 2) Define a reference point and direction for each One-Body Problem
 * 3) State all given information for each body problem
 * 4) Make any necessary conversion to appropriate unites
 * 5) Identify unknown variable
 * 6) Choose appropriate equations(s)
 * 7) Substitute in given values
 * 8) Find relation between a variable in each One-Body Problem
 * 9) Link the two equations
 * 10) Solve for unknown variable
 * 11) Include a therefore (:.) statement to show and justify selected answer

Friday the 14... Apparently (Notes By: Gehrig) __**Calculations Involving Free Fall, a very Brief note**__**Text Book Reference: Pg. 35-37**Define up or down as positive (very important to do so)Then use the kinematics equations Balthazar posted (I believe), this is possible since, in free fall acceleration is constant.Make sure you are aware of what the variables are now referring to in these sorts of problems.

SEPTEMBER 17 (Notes by Gehrig)


 * TOPIC - d-t graphs for given Accelerations and Velocities**


 * Text: pg UNKNOWN**

Without knowing too many specifics the basic shape (magnitude, direction etc.) should be able to be determined, see below.

If more information is known about a given problem, then the graph becomes more precise as the axes (scales) and points along the graphs are filled in with values.



SEPTEMBER 18 (Notes by Gehrig)
 * TOPIC - Thought Experiments, or Co-Operative Group Problem Solving**


 * Text: pg Handout**

__**FIVE STEPS**__

__STEPS A-C Set-UP__

A. Picture --- Draw a clear diagram to show what’s happening --- Attach important information using simple phrases --- Make any important measurements --- Identify any unknown information if possible --- Indicate the co-ordinate system and sign convention (ref point and ref direction)

B. The Question --- Create a specific physics question that will give the answer to the problem --- Indicate which quantities will allow you to answer the question

C. The Plan --- Summarize how it works; mention the key physics ideas involved --- Outline the key steps in solving the problem --- List any useful “textbook” equations (i.e. found in the bold/box in the text) and any other relationships (equations) that will be used

__STEPS D-E Set-UP__

D. The Work --- Create the specific equations you will use- write them down in a simple statement explaining what you are doing. You should only use symbols that appear in Part A. --- Perform the algebraic manipulations first, whenever practical --- Verify the units of the final derived expression --- NO NUMBER MATH IN THIS STEP!!!.

E. The Results --- Substitute the numbers into the manipulated equations and calculate a result --- State the final answer in response to the question you created --- Write brief statements explaining why the answer seems reasonable in size and units and direction.

__September 19 (notes by Jonathan)__

__Challenge:__ Set the washers on a string so that when they hit the ground, they strike in equal time intervals --must have a washer at both ends --bottom washer will be just above the ground when the string is released --the string cannot have its length changed
 * The Washer Activity (for those who missed it or forget it)**

Picture
 * Method**



Question At what distances between the washers will give an equal time interval between each drop?

Plan Find total time using the displacement equation Divide total time by 4 to get individual drop times Find the velocities of the other washers Find displacement of other washers

Work--write out simplified equations and conversions to be used --(not writing this down, this is only a guideline)

Calculations--do the math to get a final answer, make sure LS=RS

**Vectors part 1 (pg 756-760)**
__Vectors__ –are represented by an arrow or line segment  –arrow length= magnitude –direction of vector is the same as arrow –tail= initial point –head= end point –magnitude is always positive

ex. 32m



–directions are marked in square brackets: [ and ] –compass directions are [E], [W], [N], and [S]  –other examples are [down], [forward], [11.5° below horizontal], [24° N of W]

–computers and calculators use angles to measure directions



<span style="font-family: Arial,sans-serif;">__Multiplying a Vector by a Scalar__ –<span style="font-family: Arial,sans-serif;">multiply both magnitudes –<span style="font-family: Arial,sans-serif;">same direction* <span style="font-family: Arial,sans-serif;"> *if the scalar is negative, reverse direction

<span style="font-family: Arial,sans-serif;">__Components of Vectors__ –<span style="font-family: Arial,sans-serif;">it is the projection of a vector along the coordinate system –<span style="font-family: Arial,sans-serif;">any vector can be defined by x,y,z coordinates or by rectangular components <span style="font-family: Arial,sans-serif; text-decoration: none;"> –all axes are perpendicular to each other <span style="font-family: Arial,sans-serif; text-decoration: none;"> –can be called orthogonal components (Greek: ortho=right, gonia=angle) <span style="font-family: Arial,sans-serif; text-decoration: none;"> <span style="font-family: Arial,sans-serif; text-decoration: none;">Fy= F sin Φ or F cos β Fx= F cos Φ or F sin β <span style="font-family: Arial,sans-serif; text-decoration: none;">–the two angles always equal 90° <span style="font-family: Arial,sans-serif; text-decoration: none;"> Φ= Greek capital letter phi <span style="font-family: Arial,sans-serif; text-decoration: none;"> β= Greek lowercase letter beta

–<span style="font-family: Arial,sans-serif;">it is important to choose an appropriate coordinate system related to the problem

<span style="font-family: Arial,sans-serif;">__Vector Addition__ –<span style="font-family: Arial,sans-serif;">to add vectors, first join them head to tail. Shift them if necessary. –<span style="font-family: Arial,sans-serif;">the result is usually called the resultant, or the vector sum <span style="font-family: Arial,sans-serif; text-decoration: none;"> –be sure to distinguish the resultant vector

<span style="font-family: Arial,sans-serif; text-decoration: none;">ex. A= 24 km [32° N of E] <span style="font-family: Arial,sans-serif; text-decoration: none;"> B= 18 km [E] = 59 km [21° S of E] <span style="font-family: Arial,sans-serif; text-decoration: none;"> C= 38 km [25° E of S]

R= A + B + C

<span style="font-family: Arial,sans-serif; text-decoration: none;">–Accuracy of vector addition is greatly improved by applying trigonometry –<span style="font-family: Arial,sans-serif;">if perpendicular: use Pythagorean theorem for magnitudes <span style="font-family: Arial,sans-serif; text-decoration: none;"> use primary trig ratios for angles –<span style="font-family: Arial,sans-serif;">if not, use sine law and cosine law to find angles and magnitudes

<span style="font-family: Arial,sans-serif; text-decoration: none;">Pythagorean theorem: a^2 + b^2 = c^2

<span style="font-family: Arial,sans-serif; text-decoration: none;">Primary trig ratios: **sinθ= opp/hyp** //and// **cosθ= adj/hyp** //and// **tanθ= opp/adj**

<span style="font-family: Arial,sans-serif; text-decoration: none;">Sine law: <span style="font-family: Arial,sans-serif; text-decoration: none;">sinA/a = sinB/b = sinC/c <span style="font-family: Arial,sans-serif; text-decoration: none;">or <span style="font-family: Arial,sans-serif; text-decoration: none;"> a/sinA = b/sinB = c/sinC

<span style="font-family: Arial,sans-serif; text-decoration: none;">Cosine law: a^2= b^2 + c^2–2bc cosA and cosA= (a2–b2–c2) /-2bc

//Why won't multiple spaces or superscripts or subscripts show up?//

September 20 (notes by Jonathan)

**Vectors (part 2)**
<span style="font-family: Arial,sans-serif;">__Vector Addition–Components__ –<span style="font-family: Arial,sans-serif;">every vector can be divided into its x and y components <span style="font-family: Arial,sans-serif; text-decoration: none;">**Steps:** <span style="font-family: Arial,sans-serif; text-decoration: none;">1.define and x-y coordinate system and indicate reference direction <span style="font-family: Arial,sans-serif; text-decoration: none;">2.determine the x and y components of the vectors <span style="font-family: Arial,sans-serif; text-decoration: none;">–remember: <span style="font-family: Arial,sans-serif; text-decoration: none;"> sin<span style="font-family: Arial,sans-serif;">θ= opp/hyp --> sin <span style="font-family: Arial,sans-serif;">θ= y/v --> **v sinθ= y** <span style="font-family: Arial,sans-serif; text-decoration: none;">cosθ= adj/hyp --> cos <span style="font-family: Arial,sans-serif;">θ <span style="font-family: Arial,sans-serif; text-decoration: none;">= x/v --> **<span style="font-family: Arial,sans-serif; text-decoration: none;">v cosθ= x **

<span style="font-family: Arial,sans-serif; text-decoration: none;">3.add all the x components and y components <span style="font-family: Arial,sans-serif; text-decoration: none;">4.find the final magnitude and direction using trigonometry <span style="font-family: Arial,sans-serif; text-decoration: none;">ex. determine displacement <span style="font-family: Arial,sans-serif; text-decoration: none;">A= 10.5m [E] <span style="font-family: Arial,sans-serif; text-decoration: none;">B= 14.0m [21.5° E of S] <span style="font-family: Arial,sans-serif; text-decoration: none;">C= 25.6m [18.9° S of W] –<span style="font-family: Arial,sans-serif;">sketching the resultant will result in the dispplacement somewhere SW  <span style="font-family: Arial,sans-serif; text-decoration: none;"> –for convenience, use SW as positive

<span style="font-family: Arial,sans-serif; text-decoration: none;">yes, I tried to make it as rainbow-y as possible ☺

<span style="font-family: Arial,sans-serif;">Ax= -10.5m <span style="font-family: Arial,sans-serif; text-decoration: none;">Bx= -14.0m sin21.5°= -5.13m <span style="font-family: Arial,sans-serif; text-decoration: none;">Cx= 25.6m cos18.9°= 24.2m A<span style="font-family: Arial,sans-serif;">y= 0m B<span style="font-family: Arial,sans-serif;">y= 14.0m cos21.5°= 13.0m C<span style="font-family: Arial,sans-serif;">y= 25.6m sin18.9°= 8.29m

<span style="font-family: Arial,sans-serif; text-decoration: none;">Rx = Ax + Bx + Cx <span style="font-family: Arial,sans-serif; text-decoration: none;"> = -10.5m –5.13m +24.5m <span style="font-family: Arial,sans-serif; text-decoration: none;">Rx = 8.6m <span style="font-family: Arial,sans-serif;">Ry = Ay + By + Cy = 0m +13.0m +8.29m R<span style="font-family: Arial,sans-serif;">y = 21.3m

<span style="font-family: Arial,sans-serif;">

<span style="font-family: Arial,sans-serif; text-decoration: none;">above: the x and y components of the resultant

<span style="font-family: Arial,sans-serif; text-decoration: none;">R= √¯¯(Rx2 + Ry2) <span style="font-family: Arial,sans-serif; text-decoration: none;">= √¯¯(8.62 m+ 21.32 m) <span style="font-family: Arial,sans-serif; text-decoration: none;">R= 23m

<span style="font-family: Arial,sans-serif; text-decoration: none;">tanθ= Rx/Ry <span style="font-family: Arial,sans-serif; text-decoration: none;"> θ= tan-1 (8.6/21.3) <span style="font-family: Arial,sans-serif; text-decoration: none;">θ= 22°

<span style="font-family: Arial,sans-serif; text-decoration: none;">Therefore the displacement is 23m [22° W of S]

September 24 2012 (Notes by Maria)

<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Topic: <span style="background-color: white; font-family: 'Times New Roman',serif; font-size: 12pt;">Vectors Continued <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">RAT's and non-RAT's

<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Practice Problems: <span style="background-color: white; font-family: 'Times New Roman',serif; font-size: 12pt;">pg 66 # 33, 34, 35


 * 1) <span style="font-family: Arial,sans-serif; font-size: 9pt;">1. <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Fun Reminder, (Not that anyone would forget or anything) we have our first test tomorrow! The test will contain 11 multiple choice questions, some graph analysis questions but will mostly focus on the 5 kinematics equations (Similar to the ones that were completed for homework on September 24 2012)
 * 2) **__<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Adding Vectors __**

- <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Vectors added “Tip-to-Tail” <span style="font-family: Arial,sans-serif; font-size: 9pt;"> 3. **__<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Vector Components __**

- <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">2 Vectors

- <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Perpendicular to each other

- <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Added to make the resultant vector

- <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Always form right triangle. <span style="font-family: Arial,sans-serif; font-size: 9pt;"> 4. **__<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Drawing Vector Diagrams __**

- <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Begin by drawing a set of polar axes (Defining North, South, East, and West)

- <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Draw Vectors.

- <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">The size of the vector should be relative to its magnitude

- <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Label all known values

- <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">When doing river questions begin by drawing river banks and indicate known displacement values using dotted lines perpendicular to the river banks. <span style="font-family: Arial,sans-serif; font-size: 9pt;"> 5. **__<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Sample Questions From Class __** <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> 1. A train moves straight North at 10 m/s. A person is walking across the train (East) at 4 m/s. Draw a Vector Diagram to find the resultant vector.

<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Use trigonometric ratios or the Pythagorean theorem to solve for the persons’ resultant velocity (Resultant Vector) <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">2. A river flows south at 10 m/s. A boat sets a compass course (E 45° N) with a speedometer reading of 18 m/s. Draw a vector diagram making sure that the vectors are **__relative in size.__** (10 m/s should be smaller than 18 m/s)



Unit: Solving Vector Problems with Non-RATS September 26 2012 (Notes By Maria) In Class: - Solved Sample vector problems with non-RATS. - <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">A river flows south at 10 m/s. A boat sets a compass course (E 45° N) with a speedometer reading of 18 m/s. Draw a vector diagram making sure that the vectors are **__relative in size.__** (10 m/s should be smaller than 18 m/s)



<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">There are two main ways to solve this type of problem. <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> 1. Cosine Law <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> 2. Sine Law. <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">First we can use angle properties to determine the angle between the velocity of the boat relative to the water and the velocity of water relative to the land (45°) <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Using that information we can then use the Cosine Law to determine the Resultant Vector.



<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">C= 13 m/s

<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">Next we can use the Sine Law and Properties of Angles to solve the rest of the problem.



<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">θ=32°

<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">β= 45°-32° <span style="font-family: 'Times New Roman',serif; font-size: 12pt;">β= 13°

1. River flow [W] at 8 km/hr. A boat leaves the north bank with a compass course [S 36.9o W] and speedometer marking at 25 km/hr. It takes the boat ¼ hour to cross the river.
a) Vector Diagram

b) Width of the River

c) Distance Downstream

d) Actual Velocity of the boat

B VW <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">⊥ = B VW cos 36.9o = 25 cos 36.9o = 20 km/hr

D <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">⊥ = V <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">⊥ x t = 20 (0.25) = 5 km

B VW <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">∥ = B VW sin 36.9o = 15 km/hr

B VL <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">∥ = W VL + B VW = 8 +15 = 23 km/hr

D <span style="background-color: white; font-family: 'Cambria Math',serif; font-size: 13pt;">∥ = V <span style="background-color: white; font-family: 'Cambria Math',serif; font-size: 13pt;">∥ x t = 23 x 0.25 = 5.75 km

2. A boat is headed [N 30 o E] with a speedometer reading of 20 m/s. The river’s current is 4.0 m/s [E] and it is 250 m wide.

a. Draw a Vector Diagram to represent this situation. b. Determine the components of the boat’s velocity parallel and perpendicular to the riverbank contributed by the boat’s motor. c. Determine the components of the boat’s overall velocity parallel and perpendicular to the riverbank. d. Determine the overall velocity of the boat. e. How long does it take to cross the river? f. How far downstream does the boat end up?




 * b. ** B VW <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">∥ = B VW sin 30 o

= 10 m/s

B V W <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">⊥ = B VW cos 30 o

= 17.3 m/s

= B VW <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">⊥ = 17.3 m/s
 * c. ** B VL <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">⊥

B VL <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">∥ = B VW <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">∥ + W VL

­­­= 10 + 4

= 14 m/s


 * d. ** | B VL | = (B V2L <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">⊥ + B VL <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">∥ )

= (17.32 + 142 )

= 22.3 m/s

Tan θ = B VL <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">⊥ / B VL <span style="font-family: 'Cambria Math',serif; font-size: 13pt;">∥

θ = 51 o

B VL = 22.3 m/s [ E 51 o N]

// (end) //


 * DAY 16** - Vector Problem Solving (continued): Sept 28 / 12 10:21 PM

Weekend Worksheet! Make vector diagrams. Use significant digits. Use proper format.

Be prepared for wordy variables due to the simplicity of the word formatter... :(

1. A small boat is crossing a river which flows at 10.0 km/h [E]. The driver of the boat keeps its heading directly (at 90.0<span style="font-family: 'Times New Roman',serif; font-size: 16px;">°) away from the river bank at a speed of 24 km/h relative to the water. The river is 0.5 meters wide.

(Figure 1: //made using ms paint, which somehow in its broken state is still more manageable than this writing program)// a. speed of boat relative to land? = velocity of boat relative to land b. actual heading of boat? = direction of boat (θ in diagram) c. time to cross river? d. distance downstream? (total horizontal displacement)

No reference on North or South directions given, therefore the boat may be coming from either direction.

Givens: Speed (Velocity) of boat relative to water: 24 km/h Speed (Velocity) of water relative to land: 10.0 km/h [E] Width of River (Total vertical displacement of boat): 0.50 km

__ab)__ Required: Velocity of boat relative to land Direction of boat

Approach: Pythagorean theorem (a^2 = b^2 + c^2) Trigonometric ratio (tanθ = opposite side / adjacent side)

Solution: Velocity of boat<span style="font-family: Arial,sans-serif;"> relative to land= √¯(velocity of boat relative to water^2 + velocity of water relative to land^2) <span style="font-family: Arial,sans-serif;"> = √¯(24^2 + 10.0^2) <span style="font-family: Arial,sans-serif;"> **= 26 km/h**

Direction of boat = tan^(-1) x (velocity of boat relative to water / velocity of water relative to land) = tan^(-1) x (10.0/24) <span style="font-family: 'Times New Roman',serif; font-size: 16px;">Paraphrase: <span style="font-family: 'Times New Roman',serif; font-size: 16px;">Therefore, the description of the boat (the speed & actual heading) is either 26 km/h [N 23° E] or 26 km/h [S 23° E].
 * =** 22.6<span style="font-family: 'Times New Roman',serif; font-size: 16px;">°
 * = 23<span style="font-family: 'Times New Roman',serif; font-size: 16px;">° **<span style="font-family: 'Times New Roman',serif; font-size: 16px;">(2 significant digits)

__c)__ Required: Time

Approach: velocity = distance / time

Solution: time = total vertical displacement of boat / velocity of boat relative to water = 0.50 / 24
 * = 0.020 h**
 * = 1.2 min**

Paraphrase: Therefore, it takes the boat 1.2 min to cross the river.

__d)__ Required: Total horizontal displacement of boat

Approach: velocity = distance / time

Solution: Total horizontal displacement = velocity of water relative to boat x time = 10.0 x 0.020 ((make sure units are the same: km))
 * = 0.20 km**

Paraphrase: Therefore, the distance the boat travelled downstream is 0.20 km.

--- Addendum 10:24 PM: Notes made by Eddiiiiiiiii ---

Next week, I believe we discuss Projectile Motion, which takes on many concepts from two lessons back on kinematics. Here are some videos to get ahead.

media type="youtube" key="RcSadoSQhdA?feature=player_detailpage" height="360" width="640"

media type="youtube" key="cEOxZWGp-8E?feature=player_detailpage" height="360" width="640"

_ =**OCTOBER 1, 2012** (Notes by Zaldy)= Topic: //Relative Velocity/Frames of Reference/Projectile Motion Intro// Textual Reference: //Nelson Physics 12, pages 41 to 57// _

__**INTRODUCTION TO PROJECTILE MOTION**__

 * Projectile motion involves the motion of an object near the earth's surface.
 * In a vertical direction, projectiles experience uniform acceleration due to gravity //a (g)// of 10 metres per second squared.
 * In a horizontal direction, it experiences constant speed.
 * The concept of air resistance is ignored.
 * In general, it is a motion with a constant horizontal velocity combined with a constant vertical acceleration caused by gravity.

Projectile: an object moving through the air along a trajectory without a propulsion system Horizontal Range: the horizontal displacement of a projectile
 * Key terms:**
 * Initial velocity is divided into a vertical (iy) and horizontal (ix) component.
 * Final velocity has the same horizontal component (ignoring air resistance), but has a different vertical component due gravity's acceleration.

//HORIZONTAL MOTION// is calculated by only one equation, vx = d divided by t, however //VERTICAL MOTION// can be calculated with the five kinematic equations, including displacement, acceleration, and speed.

//**General Rules in Solving a Projectile Motion Problem:**//
 * 1) Indicate the reference point and direction of the situation.
 * 2) Divide the given initial velocity into vertical and horizontal components.
 * 3) Divide all information given into horizontal (parallel to ground) or vertical (perpendicular to)
 * 4) Assign positive and negative values to all variables determined.
 * 5) Use kinematics equations to find the missing information.

The change in velocity between position 1 and position 2 is //v = v2 - v1.// //Projectiles accelerate because its direction of its instantaneous vertical component is continually changing.//

=October 2, 2012 (Notes by Zaldy)= Topic: //Projectile Motion (continued)// Textual Reference: //Nelson Physics 12, pages 46 to 48//




 * Generally:**
 * 1) As projectiles rise, their velocity decreases.
 * 2) As projectiles fall, their velocity increases.
 * 3) At maximum height, the velocity equals the horizontal component (v = vh).
 * 4) Horizontal/parallel* velocity remains constant (due to constant speed in horizontal direction).
 * 5) Vertical/perpendicular* velocity decreases as projectile rises and increases as it falls (due to acceleration done by gravity)
 * parallel and perpendicular to ground

When stating reference direction, it is important to look at what operations your values have. If positive is up, then vertical velocity is positive while acceleration due to gravity is negative. The opposite is also true. If down is positive, then vertical velocity is negative, while acceleration due to gravity is positive.

DRAWING A GENERAL DIAGRAM
FOR INITIAL PROJECTILE MOTION PROBLEMS



October 3, 2012 -- notes by Mikaila
=__Range__=