Ch3_bonnettd

toc 1CHAPTER 3

= ONLINE NOTES =

** __Lesson 1: A&B__ **
10.12.11
 * __Vectors- Fundamentals and Operations (A & B)__ **

__**Vectors and Direction**__ Quantities used to describe physics include distance, displacement, speed, velocity, acceleration, force, mass, momentum, energy, work, power, etc. All these quantities can by divided into two categories - **vectors and scalars **.
 * Vector quantity** = is a quantity that is fully described by both magnitude and direction.
 * Examples of vector quantities include displacement, velocity , acceleration , and force.
 * Vector quantities are not fully described unless both magnitude and direction are listed.
 * Vector quantities are often represented by scaled vector diagrams.
 * Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction.
 * Vector diagrams were introduced and used in earlier units to depict the forces acting upon an object.
 * Characteristics:
 * a scale is clearly listed
 * a vector arrow is drawn in a specified direction. The vector arrow has a //head// and a //tail//.
 * the magnitude and direction of the vector is clearly labeled
 * Scalar quantity** = is a quantity that is fully described by its magnitude.

Two illustrations of the second convention for identifying the direction of a vector are shown below.

**Representing the Magnitude of a Vector** The magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow. The arrow is drawn a precise length in accordance with a chosen scale. For example, the diagram at the right shows a vector with a magnitude of 20 miles. Since the scale used for constructing the diagram is __1 cm = 5 miles__, the vector arrow is drawn with a length of 4 cm. That is, 4 cm x (5 miles/1 cm) = 20 miles. Using the same scale (__1 cm = 5 miles__), a displacement vector that is 15 miles will be represented by a vector arrow that is 3 cm in length. Similarly, a 25-mile displacement vector is represented by a 5-cm long vector arrow. And finally, an 18-mile displacement vector is represented by a 3.6-cm long arrow. See the examples shown below.



__**Vector Addition:**__ Two vectors can be added together to determine the result (or resultant). Recall in our discussion of Newton's laws of motion, that the //net force// experienced by an object was determined by computing the vector sum of all the individual forces acting upon that object. Example:

These rules for summing vectors were applied to free-body diagrams in order to determine the net force (i.e., the vector sum of all the individual forces). Sample applications are shown in the diagram below.  There are a variety of methods for determining the magnitude and direction of the result of adding two or more vectors. The two methods that will be discussed in this lesson and used throughout the entire unit are: **The Pythagorean Theorem** The Pythagorean theorem is a useful method for determining the result of adding two (and only two) vectors __that make a right angle__ to each other. The method is not applicable for adding more than two vectors or for adding vectors that are __not__ at 90-degrees to each other. To see how the method works, consider the following problem: >> Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement.
 * the Pythagorean theorem and trigonometric methods
 * the head-to-tail method using a scaled vector diagram

**Using Trigonometry to Determine a Vector's Direction** The direction of a //resultant// vector can often be determined by use of trigonometric functions. Sine, cosine, and tangent relate an acute angle in a right triangle to the ratio of the lengths of two of the sides of the right triangle. Equations:


 * Give us degrees, but not always direction*

 **Use of Scaled Vector Diagrams to Determine a Resultant** The magnitude and direction of the sum of two or more vectors can also be determined by use of an accurately drawn scaled vector diagram. Using a scaled diagram, the **head-to-tail method** is employed to determine the vector sum or resultant.easurement ended, the next measurement would begin.

A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below.
 * 1) Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a diagram that is as large as possible, yet fits on the sheet of paper.
 * 2) Pick a starting location and draw the first vector //to scale// in the indicated direction. Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).
 * 3) Starting from where the head of the first vector ends, draw the second vector //to scale// in the indicated direction. Label the magnitude and direction of this vector on the diagram.
 * 4) Repeat steps 2 and 3 for all vectors that are to be added
 * 5) Draw the resultant from the tail of the first vector to the head of the last vector. Label this vector as **Resultant** or simply **R**.
 * 6) Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale (4.4 cm x 20 m/1 cm = 88 m).
 * 7) Measure the direction of the resultant using the counterclockwise convention discussed earlier in this lesson.

An example of the use of the head-to-tail method is illustrated below. The problem involves the addition of three vectors: 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg.SCALE: 1 cm = 5 m The head-to-tail method is employed as described above and the resultant is determined (drawn in red). Its magnitude and direction is labeled on the diagram. SCALE: 1 cm = 5 m

Interestingly enough, the order in which three vectors are added has no affect upon either the magnitude or the direction of the resultant. The resultant will still have the same magnitude and direction. For example, consider the addition of the same three vectors in a different order.
 * 15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg.****SCALE: 1 cm = 5 m**

When added together in this different order, these same three vectors still produce a resultant with the same magnitude and direction as before (20. m, 312 degrees). The order in which vectors are added using the head-to-tail method is insignificant. SCALE: 1 cm = 5 m



** __Lesson 1: C & D:__ **
**__Resultants:__** The **resultant** is the vector sum of two or more vectors. It is //the result// of adding two or more vectors together. If displacement vectors A, B, and C are added together, the result will be vector R. As shown in the diagram, vector R can be determined by the use of an accurately drawn, scaled, vector addition diagram.

A + B + C = R The above discussion pertains to the result of adding displacement vectors. When displacement vectors are added, the result is a //resultant displacement//.

**__Vector Components:__**

A vector is a quantity that has both magnitude and direction. When there was a free-body diagram depicting the forces acting upon an object, each individual force was directed in //one dimension// - either up or down or left or right. When an object had an acceleration and we described its direction, it was directed in //one dimension// - either up or down or left or right. Now in this unit, we begin to see examples of vectors that are directed in //two dimensions//- upward and rightward, northward and westward, eastward and southward, etc.



In situations in which vectors are directed at angles to the customary coordinate axes, a useful mathematical trick will be employed to //transform// the vector into two parts with each part being directed along the coordinate axes. For example, a vector that is directed northwest can be thought of as having two parts - a northward part and a westward part.

Each part of a two-dimensional vector is known as a ** component **.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">If Fido's dog chain is stretched upward and rightward and pulled tight by his master, then the tension force in the chain has two components - an upward component and a rightward component. To Fido, the influence of the chain on his body is equivalent to the influence of two chains on his body - one pulling upward and the other pulling rightward. If the single chain were replaced by two chains. with each chain having the magnitude and direction of the components, then Fido would not know the difference.



__** Lesson 1: E **__
__**Vector Resolution**__

-Any vector directed at an angle to the horizontal (or the vertical) can be thought of as having two parts (or components). The process of determining the magnitude of a vector is known as **vector resolution**. The two methods of vector resolution that we will examine are
 * the parallelogram method
 * the trigonometric method

__Parallelogram Method of Vector Addition:__

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. If one desires to determine the components as directed along the traditional x- and y-coordinate axes, then the parallelogram is a rectangle with sides that stretch vertically and horizontally.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Steps:


 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Select a scale and accurately draw the vector to scale in the indicated direction.
 * 2) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Sketch a parallelogram around the vector: beginning at the tail of the vector, sketch vertical and horizontal lines; then sketch horizontal and vertical lines at the head of the vector; the sketched lines will meet to form a rectangle (a special case of a parallelogram).
 * 3) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Draw the components of the vector. The components are the //sides// of the parallelogram. The tail of the components start at the tail of the vector and stretches along the axes to the nearest corner of the parallelogram. Be sure to place arrowheads on these components to indicate their direction
 * 4) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Meaningfully label the components of the vectors with symbols to indicate which component represents which side
 * 5) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Measure the length of the sides of the parallelogram and ** use the scale to determine the magnitude ** of the components in //real// units. Label the magnitude on the diagram.

__Trigonometric Method of Vector Resolution__: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. Trigonometric functions relate the ratio of the lengths of the sides of a right triangle to the measure of an acute angle within the right triangle. They can be used to determine the length of the sides of a right triangle if an angle measure and the length of one side are known.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Steps:


 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Construct a //rough// sketch (no scale needed) of the vector in the indicated direction. Label its magnitude and the angle that it makes with the horizontal.
 * 2) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Draw a rectangle about the vector such that the vector is the diagonal of the rectangle. Beginning at the <span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">tail of <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> the vector, sketch vertical and horizontal lines. Then sketch horizontal and vertical lines at the head of the vector. The sketched lines will meet to form a rectangle.
 * 3) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> Draw the components of the vector. The components are the //sides// of the rectangle. The tail of each component begins at the tail of the vector and stretches along the axes to the nearest corner of the rectangle. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right)
 * 4) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Meaningfully label the components of the vectors with symbols to indicate which component represents which side
 * 5) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">To determine the length of the side opposite the indicated angle, use the sine function. Substitute the magnitude of the vector for the length of the hypotenuse. Use some algebra to solve the equation for the length of the side opposite the indicated angle
 * 6) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">


 * <span style="color: #008080; font-family: 'Trebuchet MS',Helvetica,sans-serif;">Lesson 1: F **

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; line-height: 0px; overflow: hidden;">__**Component Method of Vector Addition:**__

When the two vectors are added head-to-tail as shown below, the resultant is the hypotenuse of a right triangle. The sides of the right triangle have lengths of 11 km and 11 km. The resultant can be determined using the Pythagorean theorem; it has a magnitude of 15.6 km. The solution is shown below the diagram.



= = __ Addition of three or more right angle vectors: __ <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">A student drives his car 6.0 km, North before making a right hand turn and driving 6.0 km to the East. Finally, the student makes a left hand turn and travels another 2.0 km to the north. What is the magnitude of the overall displacement of the student?

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The head-to-tail vector addition diagram is shown below.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The order in which the three vectors are added must be changed in order to make a right triangle. If the three vectors are added in the order 6.0 km, N + 2.0 km, N + 6.0 km, E, then the diagram will look like this:



<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">After rearranging the order in which the three vectors are added, the resultant vector is now the hypotenuse of a right triangle. The lengths of the perpendicular sides of the right triangle are 8.0 m, North (6.0 km + 2.0 km) and 6.0 km, East. The magnitude of the resultant vector (R) can be determined using the Pythagorean theorem. <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px; text-align: center;">R2 (8.0 km)2+ (6.0 km)2 R264.0 km2+ 36.0 km2  R2 <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">100.0 km2  RSQRT (100.0 km2) ** R = 10.0 km ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Adding vectors **A + B + C** gives the same resultant as adding vectors **B + A + C** or even **C + B + A**. As long as all three vectors are included with their specified magnitude and direction, the resultant will be the same.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The resultant of the addition of three or more right angle vectors can be easily determined using the Pythagorean theorem. Doing so involves the adding of the vectors in a different order.

__**SOH-CAH-TOA and the direction of Vectors:**__ The convention is known as the counter-clockwise from east convention, often abbreviated as the **CCW** convention. Using this convention, the direction of a vector is often expressed as a counter-clockwise angle of rotation of the vector about its //tail// from due East.

Ex: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> 6.0 km, N + 6.0 km, E + 2.0 km, N. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">In this problem, we know the length of the side opposite theta (Θ) - 6.0 km - and the length of the side adjacent the angle theta (Θ) - 8.0 km. The tangent function will be used to calculate the angle measure of theta (Θ). <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">Tangent(Θ) = Opposite/Adjacent Tangent(Θ) = 6.0/8.0 Tangent(Θ) = 0.75 Θ = tan-1(0.75) Θ = 36.869 …° ** Θ =37° **

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The **result** is 37° east of north. Since the angle that the resultant makes with east is the complement of the angle that it makes with north, we could express the direction as 53° CCW.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">__Addition of non-perpendicular vectors:__ <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">It is possible to force two (or more) non-perpendicular vectors to be transformed into other vectors that do form a right triangle. The trick involves the ** concept of a vector component ** and the ** process of vector resolution. ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Ex: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Max plays middle linebacker for South's football team. During one play in last Friday night's game against New Greer Academy, he made the following movements after the ball was snapped on third down. First, he back-pedaled in the southern direction for 2.6 meters. He then shuffled to his left (west) for a distance of 2.2 meters. Finally, he made a half-turn and ran downfield a distance of 4.8 meters in a direction of 240° counter-clockwise from east (30° W of S) before finally knocking the wind out of New Greer's wide receiver. Determine the magnitude and direction of Max's overall displacement.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">As is the usual case, the solution begins with a diagram of the vectors being added.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The resultant is the vector sum of these three vectors; a head-to-tail vector addition diagram reveals that the resultant is directed southwest. Of the three vectors being added, vector C is clearly the //nasty// vector. Its direction is neither due south nor due west. The solution involves resolving this vector into its components.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Vector C makes a 30° angle with the southern direction. By sketching a right triangle with horizontal and vertical legs and C as the hypotenuse, it becomes possible to determine the components of vector C. This is shown in the diagram below. The side adjacent this 30° angle in the triangle is the vertical side; the vertical side represents the vertical (southward) component of C - C <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px;">y <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">. So to determine C <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px;">y <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">, the cosine function is used. The side opposite the 30° angle is the horizontal side; the horizontal side represents the horizontal (westward) component of C – C <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px;">. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> The cosine function is used to determine the southward component since the southward component is adjacent to the 30° angle. The sine function is used to determine the westward component since the westward component is the side opposite to the 30° angle.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Now our vector addition problem has been transformed from the addition of two nice vectors and one nasty vector into the addition of four nice vectors.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">With all vectors oriented along are customary north-south and east-west axes, they can be added head-to-tail in any order to produce a right triangle whose the hypotenuse is the resultant.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The triangle's perpendicular sides have lengths of 4.6 meters and 6.756 meters. The length of the horizontal side (4.6 m) was determined by adding the values of B (2.2 m) and C <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px;">x <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> (2.4 m). The length of the vertical side (6.756… m) was determined by adding the values of A (2.6 m) and C <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px;">y <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> (4.156… m). The resultant's magnitude (R) can now be determined using the Pythagorean theorem.

<span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px; text-align: center;">R2= (6.756… m)2+ (4.6 m)2

<span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px; text-align: center;"> R2 = 45.655… m2+ 21.16 m2

<span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px; text-align: center;"> R2= 66.815… m2

<span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;"> R= SQRT(66.815… m2) R = 8.174 … m ** R = ~8.2 m **

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">

tangent(Θ) = (6.756… m)/(4.6 m) = 1.46889… Θ = tan-1 (1.46889…) = 55.7536… ° ** Θ = ~56° **

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">This 56° SW angle is the angle between the resultant vector. The (CCW) can be determined by adding 180° to the 56°= 236

<span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">**Example:** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Cameron Per (his friends call him Cam) and Baxter Nature are on a hike. Starting from home base, they make the following movements. <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**A: 2.65 km, 140° CCW B: 4.77 km, 252° CCW C: 3.18 km, 332° CCW** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">This is the angle measure diagrams.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Values:

2.030… km, West || (2.65 km)•sin(40°) = 1.703… km, North ||
 * Vector ||  East-West Component  ||  North-South Component  ||
 * = **A**
 * 2.65 km**
 * 140° CCW** ||= (2.65 km)•cos(40°)
 * = **B**
 * 4.77 km**
 * 252° CCW** ||= (4.77 km)•sin(18°)

1.474… km, West || (4.77 km)•cos(18°) = 4.536… km, South ||
 * = **C**
 * 3.18 km**
 * 332° CCW** ||= (3.18 km)•cos(28°)

2.808… km, East || (3.18 km)•sin(28°) = 1.493… km, South ||
 * = **Sum of**
 * A + B + C** ||= 0.696 km, West ||= 4.326 km, South ||

<span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px; text-align: center;">R2= (0.696 km)2+ (4.326 km)2

<span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px; text-align: center;"> R2= 0.484 km2 + 18.714 km2

<span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px; text-align: center;"> R2= 19.199 km2

<span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;"> R= SQRT(19.199 km2) ** R = ~4.38 km ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Tangent(Θ) = opposite/adjacent <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Tangent(Θ) = (4.326 km)/(0.696 km) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Tangent(Θ) = 6.216 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Θ = tan <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 13.3333px;">-1 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">(6.216) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** Θ = 80.9° ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">It would be worded as 80.9° SW. Since west is 180° counterclockwise from east, the direction could also be expressed in the counterclockwise (CCW) from east convention as 260.9°.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">**FINAL RESULTANT:** 4.38 km with a direction of 260.9° (CCW).

**Lesson 1: G & H**
Relative Velocity and Riverboat Problems

On occasion objects move within a medium that is moving with respect to an observer.







The affect of the wind upon a plane is similar to the affect of a river current upon a motorboat:

The resultant velocity of the motorboat can be determined in the same manner as was done for the plane. The resultant velocity of the boat is the vector sum of the boat velocity and the river velocity. Motorboat problems such as these are typically accompanied by three separate questions: 1) What is the resultant velocity (both magnitude and direction) of the boat 2) If the width of the river is //X// meters wide, then how much time does it take the boat to travel shore to shore? 3) What distance downstream does the boat reach the opposite shore? The difficulty of the problem is conceptual in nature; the difficulty lies in deciding which numbers to use in the equations. That decision emerges from one's conceptual understanding (or unfortunately, one's misunderstanding) of the complex motion that is occurring. The motion of the riverboat can be divided into two simultaneous parts - a motion in the direction straight across the river and a motion in the downstream direction. These two parts (or components) of the motion occur simultaneously for the same time duration.  Independence of Perpendicular Components of Motion  A force vector that is directed upward and rightward has two parts - an upward part and a rightward part. the vector sum of these two components is always equal to the force at a given angle.
 * the resultant velocity of the boat can be determined using the Pythagorean theorem (magnitude) and a trigonometric function (direction).
 * using the average speed equation for both the x-component and y-component.
 * **ave. speed = distance/time**
 * using the average speed equation for both the x-component and y-component.
 * **ave. speed = distance/time**

Any vector - whether it is a force vector, displacement vector, velocity vector, etc. - directed at an angle can be thought of as being composed of two perpendicular components. These two components can be represented as legs of a right triangle formed by projecting the vector onto the x- and y-axis.

The two perpendicular parts or components of a vector are independent of each other. A change in the horizontal component does not affect the vertical component. A change in one component does not affect the other component. Changing a component will affect the motion in that specific direction. While the change in one of the components will alter the magnitude of the resulting force, it does not alter the magnitude of the other component.

Any component of motion occurring strictly in the horizontal direction will have no affect upon the motion in the vertical direction. Any alteration in one set of these components will have no affect on the other set.

** Lesson 2: A&B **
<span style="font-family: Verdana,Geneva,sans-serif;">__Questions:__ <span style="font-family: Verdana,Geneva,sans-serif;">What is a projectile? <span style="font-family: Verdana,Geneva,sans-serif;">What types of projectiles are there? <span style="font-family: Verdana,Geneva,sans-serif;">What is a horizontal projectile? <span style="font-family: Verdana,Geneva,sans-serif;">What other kinds of projectiles are there? <span style="font-family: Verdana,Geneva,sans-serif;">What types of problems do you do with projectiles?
 * <span style="font-family: Verdana,Geneva,sans-serif;">a projectile is an object upon which the only acting force is gravity.
 * <span style="font-family: Verdana,Geneva,sans-serif;">anything thrown upwards, dropped from rest, or thrown at a horizontal angle (provided there is no air resistance).
 * <span style="font-family: Verdana,Geneva,sans-serif;">a projectile thrown at a horizontal angle.
 * <span style="font-family: Verdana,Geneva,sans-serif;">things thrown upwards.
 * <span style="font-family: Verdana,Geneva,sans-serif;">things thrown at angles.
 * <span style="font-family: Verdana,Geneva,sans-serif;">dropping, throwing, projecting things and seeing how long it takes to reach the ground, or things like that.

<span style="font-family: Verdana,Geneva,sans-serif;">__central idea__: a projectile is an object upon which the only acting force is gravity. If there was no gravity, objects would continue motion at constant velocity because forces are only required for acceleration.

** Lesson 2: C **
Questions: <span style="display: block; font-family: Verdana,Geneva,sans-serif; font-size: 90%; text-align: left;">List two components of projectile motion. <span style="display: block; font-family: Verdana,Geneva,sans-serif; font-size: 90%; text-align: left;">What is the acceleration of gravity? <span style="display: block; font-family: Verdana,Geneva,sans-serif; font-size: 90%; text-align: left;">Must there be a horizontal force for horizontal acceleration? <span style="display: block; font-family: Verdana,Geneva,sans-serif; font-size: 90%; text-align: left;">How would an object travel in constant motion at constant speed? <span style="display: block; font-family: Verdana,Geneva,sans-serif; font-size: 90%; text-align: left;">How would a projectile cause a projectile to move? <span style="display: block; font-family: 'Times New Roman',Times,serif; font-size: 14.4px; text-align: left;">__central idea:__ Gravity only affects the vertical displacement of a projectile, not the horizontal.
 * <span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">Horizontal and vertical motion
 * <span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">Free falling objects fall at an acceleration of 9.8 m/s
 * <span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">Yes because gravity acts perpendicular to the horizontal motion although perpendicular components of motion are independent, so object moves with constant horizontal velocity at downward acceleration.
 * <span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">If there was no unbalanced force, or absence of gravity; object would travel in parabolic trajectory because the downward force of gravity accelerates them downward from an otherwise straight-line, gravity-free trajectory.
 * <span style="font-family: Verdana,Geneva,sans-serif; font-size: 90%;">Gravity causes the object to move downward vertically. Gravity doesn't affect horizontal displacement.

= LABS & ACTIVITIES: =

**Vector Activity:**
__Data:__
 * __Graphically:__**

__Scale:__ 1cm = 100cm __Resultant:__ 15.6 = R

Results: 1562.72 @ 1.16 degrees
 * __Analytically:__**


 * __% Error:__**

__Conclusion:__ With a 0.11% error, our length proved that our calculations were not very far off. However, our angle error was 5.98%, which could have been due to not being precise when measuring.

**Ball in Cup Activity:**
__**Finding Velocity/Where the ball will hit the ground:**__ Velocity = 7.1 m/s Where the ball will hit the ground = 3.26 m After we found this distance out, we had to adjust our calculations using the height of the cup. __**Figuring out where to place the cup: (new calculations)**__

Final placement of where to place the cup = 3.09 m

__**Percent error between the two calculations:**__ __Conclusion:__ In the beginning of the lab, we forgot to take the height of the cup into account so as a result our calculations were completely wrong. After we realized this, we then re-calculated and got the right results. Another error that we could have made could have been the angle of the launcher. It was suppose to be at 20 degrees the whole time, but it could have easily been moved.

Shoot Your Grade Lab:
__Partners:__ Andrew Chung, Sarah Malley, and Caroline Braunstein __Objective:__ With a given angle and speed, launch a ball from the launcher and successfully get it to travel through five rings hanging from the ceiling and then land in a cup on the floor. __Hypothesis:__ The ball will launch successfully through all five hoops based on the heights that we calculated. __Materials:__ launcher, ball, carbon paper __Given angle:__ 20<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">°

__Video of Final Results:__ (4/5 rings) media type="file" key="Movie on 2011-11-09 at 09.13.mov" width="300" height="300"

__Calculations:__ Finding initial velocity: Calculating where we should place the cup: Finding out where to place the rings: __Percent Error:__ __Results:__ __Conclusion:__ After completing this lab, we have learned that it is possible to trace the path of a projectile. Some of the high percent errors could have been caused by many things. The main thing is the measurements because it was hard to get a precise number. There was also a large amount of measurements that we had to take into account, so it became a little confusing. Overall, we did a goo job of organizing the data because most of our percent errors were very low. We only finished the lab completing four out of the five rings due to lack of time. We did figure out where it should be placed, just ran out of time. To fix some of our mistakes, I believe that we should have been more cautious of time. We also should have used more tape on the ceiling. This would have prevented the rings from moving over the weekend.

Gordarama Contest:
__**Partner:**__ Dani Rubenstein __**Front view:**__ __**Side view:**__ __**Calculations:**__

__**Results:**__ The total mass of our project was 1.31 kg. After dropping it down the ramp, we found that our project traveled a total distance of 4.5 meters in 3.2 seconds. Using this information, we calculated the acceleration and got -0.88 m/s2. We also found that the velocity of our project was 2.81 m/s.

I think that if Dani and I had to change something about our project, it would be to have wheels that were a little bigger. This would help us because the box containing our pumpkin was not that small, so if we were to make the wheels a little larger then it may have been able to travel a farther distance without going off to the side and crashing so quickly.
 * __Summary:__**