Ch3_MalleyJ

Ch3_MalleyJe toc

**Vectors and Direction**
A study of motion will involve the introduction of a variety of quantities that are used to describe the physical world. All these quantities can by divided into two categories - [|vectors and scalars]. A vector quantity is a quantity that is fully described by both magnitude and direction. On the other hand, a scalar quantity is a quantity that is fully described by its magnitude. Examples of vector quantities that have been [|previously discussed] include [|displacement], [|velocity], [|acceleration], and [|force]. Each of these quantities are unique in that a full description of the quantity demands that both a magnitude and a direction are listed. For example, suppose your teacher tells you "A bag of gold is located outside the classroom. To find it, displace yourself 20 meters." This statement may provide yourself enough information to pique your interest; yet, there is not enough information included in the statement to find the bag of gold. The displacement required to find the bag of gold has not been fully described. On the other hand, suppose your teacher tells you "A bag of gold is located outside the classroom. To find it, displace yourself from the center of the classroom door 20 meters in a direction 30 degrees to the west of north." This statement now provides a complete description of the displacement vector - it lists both magnitude (20 meters) and direction (30 degrees to the west of north) relative to a reference or starting position (the center of the classroom door). 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. Such diagrams are commonly called as [|free-body diagrams]. An example of a scaled vector diagram is shown in the diagram at the right. The vector diagram depicts a displacement vector. Observe that there are several characteristics of this diagram that make it an appropriately drawn vector diagram. **Conventions for Describing Directions of Vectors** Vectors can be directed due East, due West, due South, and due North. But some vectors are directed //northeast// (at a 45 degree angle); and some vectors are even directed //northeast//, yet more north than east. Thus, there is a clear need for some form of a convention for identifying the direction of a vector that is __not__ due East, due West, due South, or due North. There are a variety of conventions for describing the direction of any vector. The two conventions that will be discussed and used in this unit are described below: > Two illustrations of the second convention (discussed above) for identifying the direction of a vector are shown below.
 * a scale is clearly listed
 * a vector arrow (with arrowhead) 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. In this case, the diagram shows the magnitude is 20 m and the direction is (30 degrees West of North).
 * 1) The direction of a vector is often expressed as an angle of rotation of the vector about its "[|tail]" from east, west, north, or south. For example, a vector can be said to have a direction of 40 degrees North of West (meaning a vector pointing West has been rotated 40 degrees towards the northerly direction) of 65 degrees East of South (meaning a vector pointing South has been rotated 65 degrees towards the easterly direction).
 * 2) The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "[|tail]" from due East. Using this convention, a vector with a direction of 30 degrees is a vector that has been rotated 30 degrees in a counterclockwise direction relative to due east. A vector with a direction of 160 degrees is a vector that has been rotated 160 degrees in a counterclockwise direction relative to due east. A vector with a direction of 270 degrees is a vector that has been rotated 270 degrees in a counterclockwise direction relative to due east. This is one of the most common conventions for the direction of a vector and will be utilized throughout this unit.

Observe in the first example that the vector is said to have a direction of 40 degrees. You can think of this direction as follows: suppose a vector pointing East had its [|tail] pinned down and then the vector was rotated an angle of 40 degrees in the counterclockwise direction. Observe in the second example that the vector is said to have a direction of 240 degrees. This means that the tail of the vector was pinned down and the vector was rotated an angle of 240 degrees in the counterclockwise direction beginning from due east. A rotation of 240 degrees is equivalent to rotating the vector through two quadrants (180 degrees) and then an additional 60 degrees //into the// third quadrant.

**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.



In conclusion, vectors can be represented by use of a scaled vector diagram. On such a diagram, a vector arrow is drawn to represent the vector. The arrow has an obvious tail and arrowhead. The magnitude of a vector is represented by the length of the arrow. A scale is indicated (such as, 1 cm = 5 miles) and the arrow is drawn the proper length according to the chosen scale. The arrow points in the precise direction. Directions are described by the use of some convention. The most common convention is that the direction of a vector is the counterclockwise angle of rotation which that vector makes with respect to due East.

**Vector Addition**
A variety of mathematical operations can be performed with and upon vectors. One such operation is the addition of vectors. 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. That is the [|net force] was the result (or [|resultant]) of adding up all the force vectors. During that unit, the rules for summing vectors (such as force vectors) were kept relatively simple. Observe the following summations of two force vectors:

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. In this unit, the task of summing vectors will be extended to more complicated cases in which the vectors are directed in directions other than purely vertical and horizontal directions. For example, a vector directed up and to the right will be added to a vector directed up and to the left. The //vector sum// will be determined for the more complicated cases shown in the diagrams below. **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. The Pythagorean theorem is a mathematical equation that relates the length of the sides of a right triangle to the length of the hypotenuse of a right triangle. 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. This problem asks to determine the result of adding two displacement vectors that are at right angles to each other. The result (or resultant) of walking 11 km north and 11 km east is a vector directed northeast as shown in the diagram to the right. Since the northward displacement and the eastward displacement are at right angles to each other, the Pythagorean theorem can be used to determine the resultant (i.e., the hypotenuse of the right triangle). The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15.6 km. [|Later], the method of determining the direction of the vector will be discussed.

**Using Trigonometry to Determine a Vector's Direction** The direction of a //resultant// vector can often be determined by use of trigonometric functions. Most students recall the meaning of the useful mnemonic SOH CAH TOA from their course in trigonometry. SOH CAH TOA is a mnemonic that helps one remember the meaning of the three common trigonometric functions - sine, cosine, and tangent functions. These three functions relate an acute angle in a right triangle to the ratio of the lengths of two of the sides of the right triangle. The **sine function** relates the measure of an acute angle to the ratio of the length of the side opposite the angle to the length of the hypotenuse. The **cosine function** relates the measure of an acute angle to the ratio of the length of the side adjacent the angle to the length of the hypotenuse. The **tangent function** relates the measure of an angle to the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. The three equations below summarize these three functions in equation form. These three trigonometric functions can be applied to the [|hiker problem] in order to determine the direction of the hiker's overall displacement. The process begins by the selection of one of the two angles (other than the right angle) of the triangle. Once the angle is selected, any of the three functions can be used to find the measure of the angle. Write the function and proceed with the proper algebraic steps to solve for the measure of the angle. The work is shown below. Once the measure of the angle is determined, the direction of the vector can be found. In this case the vector makes an angle of 45 degrees with due East. Thus, the direction of this vector is written as 45 degrees. (Recall from [|earlier in this lesson] that the direction of a vector is the counterclockwise angle of rotation that the vector makes with due East.) The measure of an angle as determined through use of SOH CAH TOA is __not__ always the direction of the vector. The following vector addition diagram is an example of such a situation. Observe that the angle within the triangle is determined to be 26.6 degrees using SOH CAH TOA. This angle is the southward angle of rotation that the vector R makes with respect to West. Yet the direction of the vector as expressed with the CCW (counterclockwise from East) convention is 206.6 degrees.

**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. A common Physics lab involves a //vector walk//. Either using centimeter-sized displacements upon a map or meter-sized displacements in a large open area, a student makes several consecutive displacements beginning from a designated starting position. Suppose that you were given a map of your local area and a set of 18 directions to follow. Starting at //home base//, these 18 displacement vectors could be //added together// in consecutive fashion to determine the result of adding the set of 18 directions. Perhaps the first vector is measured 5 cm, East. Where this measurement ended, the next measurement would begin. The process would be repeated for all 18 directions. Each time one measurement ended, the next measurement would begin. In essence, you would be using the head-to-tail method of vector addition.

The head-to-tail method involves [|drawing a vector to scale] on a sheet of paper beginning at a designated starting position. Where the head of this first vector ends, the tail of the second vector begins (thus, //head-to-tail// method). The process is repeated for all vectors that are being added. Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. Once the resultant is drawn, its length can be measured and converted to //real// units using the given scale. The [|direction] of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East.

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 **Topic Sentence**: Vectors and vector diagrams are useful for determining the location of an object after displacement is accounted for.

**Vector Resolution** As mentioned [|earlier in this lesson], any vector directed at an angle to the horizontal (or the vertical) can be thought of as having two parts (or components). That is, any vector directed in two dimensions can be thought of as having two components. For example, if a chain pulls upward at an angle on the collar of a dog, then there is a tension force directed in two dimensions. This tension force has two components: an upward component and a rightward component. As another example, consider an airplane that is displaced northwest from O'Hare International Airport (in Chicago) to a destination in Canada. The displacement vector of the plane is in two dimensions (northwest). Thus, this displacement vector has two components: a northward component and a westward component. In this unit, we learn two basic methods for determining the magnitudes of the components of a vector directed in two dimensions. 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 **Parallelogram Method of Vector Resolution** The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. Briefly put, the method involves drawing the vector to scale in the indicated direction, sketching a parallelogram around the vector such that the vector is the diagonal of the parallelogram, and determining the magnitude of the components (the sides of the parallelogram) using the scale. 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. A step-by-step procedure for using the parallelogram method of vector resolution is: The step-by-step procedure above is illustrated in the diagram below to show how a velocity vector with a magnitude of 50 m/s and a direction of 60 degrees above the horizontal may be resolved into two components. The diagram shows that the vector is first [|drawn to scale] in the indicated direction; a parallelogram is sketched about the vector; the components are labeled on the diagram; and the result of measuring the length of the vector components and converting to m/s using the scale. (NOTE: because different computer monitors have different resolutions, the actual length of the vector on your monitor may not be 5 cm.) **Trigonometric Method of Vector Resolution** The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. [|Earlier in lesson 1], the use of trigonometric functions to determine the direction of a vector was described. Now in this part of lesson 1, trigonometric functions will be used to determine the components of a single vector. [|Recall from the earlier discussion] that 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. As such, trigonometric functions 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. The method of employing trigonometric functions to determine the components of a vector are as follows: The above method is illustrated below for determining the components of the force acting upon Fido. As the 60-Newton tension force acts upward and rightward on Fido at an angle of 40 degrees, the components of this force can be determined using trigonometric functions.
 * the parallelogram method
 * [|the trigonometric method]
 * 1) Select a scale and accurately draw the vector to scale in the indicated direction.
 * 2) 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) 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 (up, down, left, right).
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward velocity component might be labeled vx; etc.
 * 5) 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.
 * 1) 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) Draw a rectangle about the vector such that the vector is the diagonal of the rectangle. 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.
 * 3) 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) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward force velocity component might be labeled vx; etc.
 * 5) 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) Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

In conclusion, a vector directed in two dimensions has two components - that is, an influence in two separate directions. The amount of influence in a given direction can be determined using methods of vector resolution. Two methods of vector resolution have been described here - [|a graphical method] (parallelogram method) and a [|trigonometric method]. **Topic Sentence**: (The above sentence actually gives a beautifully summary, but I'll re-summarize). Vectors are important to show the influences of direction and speed/distance; and it can be determined many ways, but we'll use two: a graphical method, and a trigonometric method.

**Relative Velocity and Riverboat Problems**
On occasion objects move within a medium that is moving with respect to an observer. For example, an airplane usually encounters a wind - air that is moving with respect to an observer on the ground below. As another example, a motorboat in a river is moving amidst a river current - water that is moving with respect to an observer on dry land. In such instances as this, the magnitude of the velocity of the moving object (whether it be a plane or a motorboat) with respect to the observer on land will not be the same as the speedometer reading of the vehicle. That is to say, the speedometer on the motorboat might read 20 mi/hr; yet the motorboat might be moving relative to the observer on shore at a speed of 25 mi/hr. Motion is relative to the observer. The observer on land, often named (or misnamed) the "stationary observer" would measure the speed to be different than that of the person on the boat. The observed speed of the boat must always be described relative to who the observer is. To illustrate this principle, consider a plane flying amidst a **tailwind**. A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity. If the plane is traveling at a velocity of 100 km/hr with respect to the air, and if the wind velocity is 25 km/hr, then what is the velocity of the plane relative to an observer on the ground below? The resultant velocity of the plane (that is, the result of the wind velocity contributing to the velocity due to the plane's motor) is the vector sum of the velocity of the plane and the velocity of the wind. This resultant velocity is quite easily determined if the wind approaches the plane directly from behind. As shown in the diagram below, the plane travels with a resulting velocity of 125 km/hr relative to the ground. If the plane encounters a headwind, the resulting velocity will be less than 100 km/hr. Since a headwind is a wind that approaches the plane from the front, such a wind would decrease the plane's resulting velocity. Suppose a plane traveling with a velocity of 100 km/hr with respect to the air meets a headwind with a velocity of 25 km/hr. In this case, the resultant velocity would be 75 km/hr; this is the velocity of the plane relative to an observer on the ground. This is shown in the diagram below. Now consider a plane traveling with a velocity of 100 km/hr, South that encounters a **side wind** of 25 km/hr, West. Now what would the resulting velocity of the plane be? This question can be answered in the same manner as the previous questions. The resulting velocity of the plane is the vector sum of the two individual velocities. To determine the resultant velocity, the plane velocity (relative to the air) must be added to the wind velocity. This is the same procedure that was used above for the headwind and the tailwind situations; only now, the resultant is not as easily computed. Since the two vectors to be added - the southward plane velocity and the westward wind velocity - are at right angles to each other, the [|Pythagorean theorem] can be used. This is illustrated in the diagram below. In this situation of a side wind, the southward vector can be added to the westward vector using the [|usual methods of vector addition]. The magnitude of the resultant velocity is determined using Pythagorean theorem. The algebraic steps are as follows: (100 km/hr)2 + (25 km/hr)2 = R2 10 000 km2/hr2 + 625 km2/hr2 = R2  10 625 km2/hr2 = R2  SQRT(10 625 km2/hr2) = R  **103.1 km/hr = R**

The direction of the resulting velocity can be determined using a [|trigonometric function]. Since the plane velocity and the wind velocity form a right triangle when added together in head-to-tail fashion, the angle between the resultant vector and the southward vector can be determined using the sine, cosine, or tangent functions. The tangent function can be used; this is shown below:

tan (theta) = (opposite/adjacent) tan (theta) = (25/100) theta = invtan (25/100) **theta = 14.0 degrees** If the resultant velocity of the plane makes a 14.0 degree angle with the southward direction (theta in the above diagram), then the direction of the resultant is 256 degrees. Like any vector, the resultant's [|direction] is measured as a counterclockwise angle of rotation from due East.

**Analysis of a Riverboat's Motion** The affect of the wind upon the plane is similar to the affect of the river current upon the motorboat. If a motorboat were to head straight across a river (that is, if the boat were to point its bow straight towards the other side), it would not reach the shore directly across from its starting point. The river current influences the motion of the boat and carries it downstream. The motorboat may be moving with a velocity of 4 m/s directly across the river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the downstream direction. While the speedometer of the boat may read 4 m/s, its speed with respect to an observer on the shore will be greater than 4 m/s. 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. Since the boat heads straight across the river and since the current is always directed straight downstream, the two vectors are at right angles to each other. Thus, the [|Pythagorean theorem] can be used to determine the resultant velocity. Suppose that the river was moving with a velocity of 3 m/s, North and the motorboat was moving with a velocity of 4 m/s, East. What would be the resultant velocity of the motorboat (i.e., the velocity relative to an observer on the shore)? The magnitude of the resultant can be found as follows: (4.0 m/s)2 + (3.0 m/s)2 = R2 16 m2/s2 + 9 m2/s2 = R2  25 m2/s2 = R2  SQRT (25 m2/s2) = R  **5.0 m/s = R**

The [|direction] of the resultant is the counterclockwise angle of rotation that the resultant vector makes with due East. This angle can be determined using a trigonometric function as shown below. tan (theta) = (opposite/adjacent) tan (theta) = (3/4) theta = invtan (3/4) **theta = 36.9 degrees**

Given a boat velocity of 4 m/s, East and a river velocity of 3 m/s, North, the resultant velocity of the boat will be 5 m/s at 36.9 degrees.

Motorboat problems such as these are typically accompanied by three separate questions: The first of these three questions was answered above; the resultant velocity of the boat can be determined using the Pythagorean theorem (magnitude) and a trigonometric function (direction). The second and third of these questions can be answered using the [|average speed equation] (and a lot of logic). **ave. speed = distance/time** ** Topic Sentence ** : Word problems using vectors can be solved in the same way as ordinary problems that we've already discussed.
 * 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?

**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. That is to say, if you pull upon an object in an upward and rightward direction, then you are exerting an influence upon the object in two separate directions - an upward direction and a rightward direction. These two parts of the two-dimensional vector are referred to as [|components]. A **component**  describes the affect of a single vector in a given direction. Any force vector that is exerted at an angle to the horizontal can be considered as having two parts or components. The vector sum of these two components is always equal to the force at the given angle. This is depicted in the diagram below. 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. Consider the pull upon Fido as an example. If the horizontal pull upon Fido increases, then Fido would be accelerated at a greater rate to the right; yet this greater horizontal pull would not exert any vertical influence upon Fido. Pulling horizontally with more force does not lift Fido vertically off the ground. A change in the horizontal component does not affect the vertical component. This is what is meant by the phrase "perpendicular components of vectors are independent of each other." 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. The resulting motion of a plane flying in the presence of a crosswind is the combination (or sum) of two simultaneous velocity vectors that are perpendicular to each other. Suppose that a plane is attempting to fly northward from Chicago to the Canada border by simply directing the plane due northward. If the plane encounters a crosswind directed towards the west, then the resulting velocity of the plane would be northwest. The northwest velocity vector consists of two components - a north component resulting from the plane's motor (the //plane velocity//) and a westward component resulting from the crosswind (the //wind velocity//). These two components are independent of each other. An alteration in one of the components will not affect the other component. For instance, if the wind velocity increased, then the plane would still be covering ground in the northerly direction at the same rate. It is true that the alteration of the wind velocity would cause the plane to travel more westward; however, the plane still flies northward at the same speed. Perpendicular components of motion do not affect each other. Now consider an air balloon descending through the air toward the ground in the presence of a wind that blows eastward. Suppose that the downward velocity of the balloon is 3 m/s and that the wind is blowing east with a velocity of 4 m/s. The resulting velocity of the air balloon would be the combination (i.e., the vector sum) of these two simultaneous and independent velocity vectors. The air balloon would be moving downward and eastward.If the wind velocity increased, the air balloon would begin moving faster in the eastward direction, but its downward velocity would not be altered. If the balloon were located 60 meters above the ground and was moving downward at 3 m/s, then it would take a time of 20 seconds to travel this vertical distance. **d = v • t** So **t = d / v** (60 m) / (3 m/s) **20 seconds** During the 20 seconds taken by the air balloon to fall the 60 meters to the ground, the wind would be carrying the balloon in the eastward direction. With a wind speed of 4 m/s, the distance traveled eastward in 20 seconds would be 80 meters. If the wind speed increased from the value of 4 m/s to a value of 6 m/s, then it would still take 20 seconds for the balloon to fall the 60 meters of downward distance. A motion in the downward direction is affected only by downward components of motion. An alteration in a horizontal component of motion (such as the eastward wind velocity) will have no affect upon vertical motion. Perpendicular components of motion are independent of each other. A variation of the eastward wind speed from a value of 4 m/s to a value of 6 m/s would only cause the balloon to be blown eastward a distance of 120 meters instead of the original 80 meters. In the [|most recent section of Lesson 1], the topic of relative velocity and riverboat motion was discussed. A boat on a river often heads straight across the river, perpendicular to its banks. Yet because of the flow of water (i.e., the current) moving parallel to the river banks, the boat does not land on the bank directly across from the starting location. The resulting motion of the boat is the combination (i.e., the vector sum) of these two simultaneous and independent velocity vectors - the boat velocity plus the river velocity. In the diagram at the right, the boat is depicted as moving eastward across the river while the river flows southward. The boat starts at Point A and heads itself towards Point B. But because of the flow of the river southward, the boat reaches the opposite bank of the river at Point C. The time required for the boat to cross the river from one side to the other side is dependent upon the boat velocity and the width of the river. Only an eastward component of motion could affect the time to move eastward across a river.Suppose that the boat velocity is 4 m/s and the river velocity is 3 m/s. The magnitude of the resultant velocity could be determined to be 5 m/s using the Pythagorean Theorem. The time required for the boat to cross a 60-meter wide river would be dependent upon the boat velocity of 4 m/s. It would require 15 seconds to cross the 60-meter wide river. **d = v • t** So **t = d / v** (60 m) / (4 m/s) **15 seconds** The southward river velocity will not affect the time required for the boat to travel in the eastward direction. If the current increased such that the river velocity became 5 m/s, then it would still require 15 seconds to cross the river. Perpendicular components of motion are independent of each other. An increase in the river velocity would simply cause the boat to travel further in the southward direction during these 15 seconds of motion. An alteration in a southward component of motion only affects the southward motion. All vectors can be thought of as having perpendicular components. In fact, any motion that is at an angle to the horizontal or the vertical can be thought of as having two perpendicular motions occurring simultaneously. These perpendicular components of motion occur independently of each other. 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. **Topic Sentence**: We can solve word problems by graphing with vectors and by using equations. Also, knowing d = v*t makes a huge difference when trying to find the distance !

Lab: Vector Mapping
The percent error of our calculations shows that our graph and analytical analysis of our data was fairly accurate. What's interesting, however, it that the graph shows a more accurate percent error than the analytical does, which is usually not true. This can most likely be contributed to the mistakes might have been made while measuring out distances, due to the fact that a large unknown was how tightly the measuring tape was held when it was originally measured out. These measuring inaccuracies may have been essentially counter-acted in the measuring inaccuracies with the graph, as it was very easy to accidentally shift the ruler. Overall, our measurement appeared to be very precise and accurate.
 * Legs Given to Us By Other Group (Data):**
 * Graph**:
 * Analytical Vector Addition**:
 * Percent Error Calculation**:
 * Conclusion**:

Lesson 2 Notes
This is simple to remember and an effective way to read a passage. P = //Preview// to identify the main parts of the reading Q = Develop //questions// to which you want to find answers. R = //Read// the material, twice if possible. S = //State// the central idea or theme. T = //Test// yourself by answering the questions.
 * Method 3: P-Q-R-S-T**

**What is a projectile?**

 * Questions and Answers
 * How are kinematics and laws of motion related?
 * These can be easily connected in problems where an object is moving in //two dimensions//, like a projectile.
 * Why is the only force acting upon a projectile gravity?
 * If there were any other forces acting upon an object, then it would not be a projectile.
 * Why do people often assume that gravity isn't the only force acting upon an object?
 * They misunderstand Newtonian physics and think that if an object is moving upwards, there must be a force acting on it making it go that way.
 * Does gravity affect how far an object is moving horizontally?
 * No! Gravity only affects vertical motion.
 * Main Ideas and Central Themes
 * A projectile is an object upon which the only force acting is gravity.
 * Newton's laws suggest that forces are only required to cause an acceleration (not a motion).
 * Newton's first law of physics: an object in motion will remain in motion unless something else acts upon it
 * Horizontal forces are not required to keep a projectile moving horizontally.

**Characteristics of a Projectile's Trajectory**

 * Questions and Answers
 * What are the two components of the projectile's motion?
 * They are: horizontal and vertical motion, and they must be discussed separately.
 * Why do projectiles travel with a constant horizontal velocity and a downward vertical acceleration?
 * Gravity is the only force acting upon a projectile object, and therefore only the vertical acceleration is going to change. Because no force is acting on the object horizontally, it remains constant.
 * What difference does gravity make?
 * Gravity influences the acceleration and is eventually the force the makes objects stop moving.
 * What do projectiles travel with a parabolic trajectory?
 * This is because the downward force of gravity accelerates them downward from their otherwise straight-line, gravity-free trajectory.
 * Why does the horizontal velocity remain constant?
 * There are no forces acting upon it.
 * Main Ideas and Central Themes
 * Projectiles travel with a constant horizontal velocity and a downward vertical acceleration.
 * Parabolic trajectory - the downward force of gravity acts upon the object to cause the same vertical motion (downward acceleration) even if the object is shot upward
 * Parabolic trajectory - the downward force of gravity acts upon the object to cause the same vertical motion (downward acceleration) even if the object is shot upward

**Describing Projectiles with Numbers**
>>>>
 * **Horizontal and Vertical //Velocity//**
 * Questions and Answers
 * How can you show the components of a projectile trajectory without drawing everything out?
 * You can make a table. In this table, you should show //time//, //horizontal velocity//, and //vertical velocity//. This numerical information helps to illustrate the same points, just without an illustration.
 * Why does gravity only affect the vertical vector?
 * Gravity is a vertical force, therefore, it only changes the vertical acceleration.
 * How would the horizontal and vertical velocities change with time with a horizontally launched projectile?
 * The horizontal velocity //always always always// remains constant during the course of the trajectory, and the vertical velocity changes by 9.81 m/s/s. //They have two different velocities, and are essentially two different paths just being combined//.
 * **Horizontal and Vertical //Displacement//**
 * Questions and Answers
 * What's the equation for vertical displacement for a horizontally launched projectile?
 * y = 0.5 • g • t2
 * What's the equation for horizontal displacement for a horizontally launched projectile?
 * x = vix • t
 * What's the equation for vertical displacement for an angle-launched projectile?
 * y = viy • t + 0.5 • g • t2

**Horizontally Launched Projectile Problems**

 * Questions and Answers
 * What properties make up problem type 1?
 * A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.
 * [[image:number_one]]
 * What properties make up problem type 2?
 * A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it's at its peak.
 * [[image:NUMBER_2]]
 * What equations can be used most often for these types of problems?
 * [[image:equations width="362" height="164"]]
 * What equations can be used for the horizontal components of motion?
 * [[image:horizontal_equations]]
 * What equations can be used for the vertical components of projectile motion?
 * [[image:vert_comps]]
 * Main Ideas and Central Themes
 * Summary of problem solving:
 * Carefully read the problem and list the known and unknown information in terms of the symbols of the kinematic equations. For convenience sake, make a table with horizontal information on one side and vertical information on the other side
 * Identify the unknown quantity that the problem requests you to solve for
 * Select either a horizontal or vertical equation to solve for the time of flight of the projectile
 * With the time determined, use one of the other equations to solve for the unknown. (Usually, if a horizontal equation is used to solve for time, then a vertical equation can be used to solve for the final unknown quantity.)

Lab: Target Practice

 * Objectives**:
 * Measure the initial velocity of a ball.
 * Apply concepts from two-dimensional kinematics to predict the impact point of a ball in projectile motion.
 * Take into account trial-to-trial variations in the velocity measurement when calculating the impact point.


 * Pre-Lab Questions**:
 * If you were to drop a ball, releasing it from rest, what information would be needed to predict how much time it would take for the ball to hit the floor?
 * The only information you would need to be given would be the distance from the dropping point to the floor. You can assume the the velocity is -9.81 m/s/s (which is that of gravity) and that the initial velocity is 0 m/s because it's starting off from rest. You can also assume that you're only dealing with the vertical distance because the object is being dropped, not thrown.
 * If the ball in question 1 is traveling at a known horizontal velocity when it starts to fall, explain how you would calculate how far it will travel before it hits the ground.
 * You can calculate the horizontal distance by using the equation d=vt. The velocity of the horizontal vector remains constant because there is no force acting upon it, so whatever you're given can be used. Additionally, you can use the time that you found from the previous question in here, because it will hit the ground at the same time. This is enough information to find the distance.
 * A single Photogate can be used to accurately measure the time interval for an object to break the beam of a Photogate. If you wanted to know the velocity of the object, what additional information would you need?
 * Assuming you're only worrying about finding the velocity of the horizontal vector, you would also need to know the horizontal distance that the object started away from the Photogate.
 * What data will you need to collect for this lab before you begin?
 * You will need to find the distance that the shooting machine is away from the ground. More information that you know is: horizontal velocity of 0 m/s, horizontal acceleration of 0 m/s, vertical acceleration of -9.81 m/s/s, and the angle from which it's shooting is 0 degrees.
 * How will you analyze your results in terms of precision and/or accuracy?
 * You can account for precision by doing multiple trials and seeing if the ball hits the ground at around the same spot every time. You can account for accuracy by doing a different test (like the ball in the cup) and predicting where the ball will land based solely on your previously collected data. While doing this, you can measure where the ball actually hits (if not in the cup), and find the percent error to account for accuracy.

Page One: Page Two:
 * Laboratory Calculations**:

media type="file" key="Ball in a Cup - Mobile.m4v" width="300" height="300" Analysis of this: This test run did not work, as the ball did not go in the cup. As can be seen, the ball hits the //side// of the cup. This is because the height of the cup is not accounted for, which means that the cup is going to be too far - the ball is landing //just// before it gets to the rim of the cup, which is why it's not going in.
 * Test Run Before Accounting for the Height of the Cup**:

media type="file" key="working ball in a cup.m4v" width="300" height="300"
 * Test Run WHEN IT ACTUALLY WORKED!!!!!**

This experiment was designed to test how accurately a projectile launch could be measured. What my group discovered is that it could be tested extremely accurately with proper equipment and measuring, as our percent error was less than half of a percent. By using your knowns and finding the velocity, this was entirely possible. Some sources of error may be contributed to inaccuracies when measuring the distance between the table and the projectile launcher, and measuring the range; this would probably be due to a lack of extraordinarily precise measuring tools.
 * Conclusion**:

Lab: Shoot Your Grade!
UNITS USED: meters, seconds, meters per second, and meters per second squared We used the initial velocity and the measurement of how far the ball had traveled in its projectory to find time that we could use to find the y distance so we could hang up our rings. The boxes labeled 1-5 are calculations for where we should hang the rings, and the final calculation (6) is where we should place our cup.
 * Partners**: Maddy Weinfeld, Dani Rubenstein, and John Chiavelli
 * Objective**: To launch a ball from the launcher at a given angle and speed setting so that the ball passes consecutively through five rings and lands in a cup on the floor.
 * Purpose and Rationale**: The purpose of this lab is to launch a ball at a 30 degree angle with a certain speed and get it to go through 5 measured rings consecutively, and finally land in a cup on the floor. This lab is based on the principles of projectile motion. To understand we has to do, we had to understand the concept of projectile trajectory to know the path that the ball would follow. Additionally, we had to understand the concept of the x and y vectors to be able to calculate theoretical positions for the rings and cup. Originally, we gathered data from a previous lab (Ball in a Cup) to find the speed.
 * Materials and Methods**: The main materials that we used were 5 rings (tape rings), a small yellow ball, a projectile launcher (directed at a 30 degree angle), string (cut to hang up the rings), tape (to keep the string attached to the ceiling), and a plastic cup. Firstly, we calculated the initial velocity, as shown below. Next, we used this information to calculate how far apart each hoop should be, as well as how high each ring should be held and where the cup should be placed. After that, we hung up all of the rings and adjusted based on error for the ball to go through the hoops. When we had finished, we measured where the rings were and proceeded to do more calculations to calculate the error. *Pictures of the procedure are posted after calculations (I did everything in order)*
 * Our Given Angle**: 30%
 * Calculations**:


 * Organized Physics Calculations**:
 * || Vertical Position (m) *from launcher* || Horizontal Position (m) *from launcher* || Time (s) ||
 * Ring One || .2331 || .557 || .133 ||
 * Ring Two || .2998 || 1.114 || .265 ||
 * Ring Three || .191 || 1.671 || .398 ||
 * Ring Four || -.089 || 2.228 || .530 ||
 * Ring Five || -.543 || 2.785 || .663 ||
 * Cup || -1.075 || 3.27 || .778 ||

We set our launcher to 30 degrees, our given angle. We measured where to put the rings We hung up the rings We tested the rings that we hung up We proceeded to adjust the rings and hang more. We did a test run to see if it would go through... Some of them worked!!! media type="file" key="6 test run for procedure.mov" width="300" height="300" We repeated this until we got the ball through four rings.
 * Procedure**:

(only worked once on the first day) || 2.80 m || -.546 m || media type="file" key="Shoot for Your Grade - 4 Hoops - Medium.m4v" width="300" height="300"
 * Recorded Data for When it Actually Worked**: (vertical measurements are from the launcher to the hoop [not the ground], horizontal measurements are a straight path from the launcher to the hoop)
 * || Horizontal Placement || Vertical Placement ||
 * First Ring || .557 m || .235 m ||
 * Second Ring || 1.13 m || .299 m ||
 * Third Ring || 1.69 m || .195 m ||
 * Fourth Ring || 2.25 m || -.092 m ||
 * Fifth Ring
 * Cup || never got it || -1.08 m ||
 * Video of us Getting it through Four Hoops**:
 * Percent Error Calculations**:

In conclusion, it is possible to trace the path of a projectile shot at a 30 degree angle. Although there are some percent errors (as seen above), none of them are more than 5%. There were some errors due to minor shifting of the launcher, and the fact that the horizontal distances were hard to measure to a precise .557m with the measuring utensils that were provided to our group. This also made it more difficult to get the left and right measurements 100 percent accurately for the launcher. For the most part, the closer it was to the launcher, the more accurate the measurements were. If I were to do this lab again, I would find a more precise way of measuring, most likely with a meter stick, so that I would have a flat measurement and wouldn't have to worry about incorrectly pulling on measuring tape. Additionally, I would make it so that the string couldn't move as easily that was holding up the hoops because it was difficult to keep them 100 percent steady when the ball was shot. I would also test it multiple times before moving the rings, because often the launcher had some minor inconsistencies. This would be relevant to a real life situation to someone that was, for example, working in the military. If they had to shoot a bomb from the top of their base to a nearby village, they would want to have exact data about where it was going to land, and where it would travel over. Additionally, knowing how high up it would be by tracing its path would be important if there was something like an airplane in the air that they could hit if they weren't careful. It's important to understand the physics concept of projectiles for other reasons as well, like if you worked in the circus and had an act where you were going to be launched through hoops and end by grabbing on to a bar, you'll want to know that your angle and velocity are going to guarantee you a working act.
 * Conclusion**:

Gourd-o-Rama Contest

 * Cart #6!! Partner: Sarah Malley**
 * Picture of Final Project**:


 * Calculations**:


 * Results**:
 * __Vehicle Mass__ with Pumpkin: 1.41 kg
 * __Time__: 1.38 s
 * __Distance Traveled__: 1.5 m
 * __Acceleration Value__: -1.57 m/s^2
 * __Velocity Value__: 2.17 m/s

If I had to remake this cart, I would definitely make a smaller base and get better wheels. It was obvious that the weight was incorrectly balanced, as to put it nicely, this cart won for hang time with a nice flip after coming off of the ramp. The cart that worked extremely well in our class only had three wheels, one center front and two on the back sides. This seems like a good design because when it goes down the ramp, the front wheel picks up most of the momentum and keeps its moving even when it reaches the flat hallway. I would try to design something like this, and I would also try to find something to make the cart go straight (like lining up the axles correctly).
 * Conclusion**: