# Preview Section 1 Introduction to Vectors Section 2 Vector Operations

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Preview Section 1 Introduction to Vectors Section 2 Vector Operations
Section 3 Projectile Motion Section 4 Relative Motion

Introduction to Vectors
Scalar - a quantity that has magnitude but no direction Examples: volume, mass, temperature, speed Vector - a quantity that has both magnitude and direction Examples: acceleration, velocity, displacement, force Emphasize that direction means north, south, east, west, up, or down. It does not mean increasing or decreasing. Even though the temperature may be going “up”, it is really increasing and has no direction. To further emphasize the distinction, point out that it is meaningless to talk about the direction of temperature at a particular point in time, while measurements such as velocity have direction at each moment.

Vector Properties Vectors are generally drawn as arrows.
Length represents the magnitude Arrow shows the direction Resultant - the sum of two or more vectors

Finding the Resultant Graphically
Method Draw each vector in the proper direction. Establish a scale (i.e. 1 cm = 2 m) and draw the vector the appropriate length. Draw the resultant from the tip of the first vector to the tail of the last vector. Measure the resultant. The resultant for the addition of a + b is shown to the left as c. Ask students if a and b have the same magnitude. How can they tell?

Vector Addition Vectors can be moved parallel to themselves without changing the resultant. the red arrow represents the resultant of the two vectors Stress that the order in which they are drawn is not important because the resultant will be the same.

The resultant (d) is the same in each case Subtraction is simply the addition of the opposite vector.

Properties of Vectors Click below to watch the Visual Concept.

Sample Resultant Calculation
A toy car moves with a velocity of .80 m/s across a moving walkway that travels at 1.5 m/s. Find the resultant speed of the car. Use this to demonstrate the graphical method of adding vectors. Use a ruler to measure the two components and determine the scale. Then determine the size and direction of the resultant using the ruler and protractor. This would make a good practice problem for Section 2, when students learn how to add vectors using the Pythagorean theorem and trigonometry.

Now what do you think? How are measurements such as mass and volume different from measurements such as velocity and acceleration? How can you add two velocities that are in different directions? At this point, students should be able to answer both questions. Mass and volume are scalars, while velocity and acceleration are vectors. Vectors can be added together with the graphical method. Tell students that in the next section, they will learn another method for adding vectors.

Vector Operations Use a traditional x-y coordinate system as shown below on the right. The Pythagorean theorem and tangent function can be used to add vectors. More accurate and less time-consuming than the graphical method Direction means north, south, east, west, up, or down. It does not mean increasing or decreasing. So even though the temperature may be going “up,” it is really just increasing and has no direction.

Pythagorean Theorem and Tangent Function
Remind students that the Pythagorean theorem can only be used with right triangles.

Resolving Vectors Into Components
Review these trigonometry definitions with students to prepare for the next slide (resolving vectors into components).

Resolving Vectors into Components
Opposite of vector addition Vectors are resolved into x and y components For the vector shown at right, find the vector components vx (velocity in the x direction) and vy (velocity in the y direction). Assume that that the angle is 20.0˚. Answers: vx = 89 km/h vy = 32 km/h Review the first solution with students, and then let them solve for the second component.

Four steps Resolve each vector into x and y components Add the x components (xtotal = x1 + x2) Add the y components (ytotal = y1 + y2) Combine the x and y totals as perpendicular vectors Explain the four steps using the diagram. Show students that d1 can be resolved into x1 and y1 . Similarly for d2. Then, the resultant of d1 and d2 (dashed line labeled d) is the same as the resultant of the 4 components.

Click below to watch the Visual Concept. Visual Concept

Classroom Practice A camper walks 4.5 km at 45° north of east and then walks 4.5 km due south. Find the camper’s total displacement. Answer 3.4 km at 22° S of E For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful. Student should note that the 4.5 km south is subtracted from the y component of the first vector that is directed north.

Now what do you think? Compare the two methods of adding vectors.
What is one advantage of adding vectors with trigonometry? Are there some situations in which the graphical method is advantageous? Explain to students that they will primarily use the trigonometric method of vector addition for this course.

Projectile Motion Projectiles: objects that are launched into the air
tennis balls, arrows, baseballs, wrestlers Gravity affects the motion Path is parabolic if air resistance is ignored Path is shortened under the effects of air resistance Discuss the wide variety of projectiles. Tell students that the effect of air resistance is significant in many cases, but we will consider ideal examples with gravity being the only force. The effects of air were not very significant in the coin demonstration (see the Notes on the previous slide), but would be much more significant if the objects were traveling faster or had more surface area. Use the PHET web site to allow students to study projectile motion qualitatively. Go to simulations, choose “motion,” and choose then choose “projectile motion.” In this simulation, you can raise or lower the canon. Start with horizontal launches and note that the time in the air is only dependent on the height, and not on the speed of launch. You can change objects, and you can even launch a car. You also have the option of adding air resistance in varying amounts, as well as changing the launch angle. Have students determine which launch angles produce the same horizontal distance or range (complimentary angles) and find out which launch angle gives the greatest range (45°). Ask them to investigate the effect of air resistance on these results.

Components of Projectile Motion
As the runner launches herself (vi), she is moving in the x and y directions. Remind students that vi is the initial velocity, so it never changes. Students will learn in later slides that vx,i also does not change (there is no acceleration in the horizontal direction) but vy,i does change (because of the acceleration due to gravity).

Analysis of Projectile Motion
Horizontal motion No horizontal acceleration Horizontal velocity (vx) is constant. How would the horizontal distance traveled change during successive time intervals of 0.1 s each? Horizontal motion of a projectile launched at an angle: Since the initial velocity is constant, the change in x for each successive time interval (such as 0.1 s) will always be the same. Point out that the ball moves the same distance sideways between successive time intervals. Many students mistakenly believe that the ball is falling straight down eventually. In fact, it keeps moving sideways at a steady rate in the absence of air resistance. With air resistance, it can eventually reach a point where it is falling nearly straight down.

Analysis of Projectile Motion
Vertical motion is simple free fall. Acceleration (ag) is a constant m/s2 . Vertical velocity changes. How would the vertical distance traveled change during successive time intervals of 0.1 seconds each? Vertical motion of a projectile launched at an angle: Students should note that the vertical distance increases during each successive time interval. The equations above are simply equations (2), (5), and (4) from the previous section. You might want to write the “old” equations on the board prior to showing them these “new” equations.

Projectile Motion Click below to watch the Visual Concept.

Projectile Motion - Special Case
Initial velocity is horizontal only (vi,y = 0). Point out that these equations are the same as those on the previous slides with vi,y = 0 or a launch angle  = 0. These equations could be used for the coin as it fell off the table (see the Notes on the first slide of this section) or for an object dropped from an airplane flying at a level altitude. The previous equations (last two slides) are more general and apply to any projectile.

Projectile Motion – Launched at an angle
When projectiles are launched at an angle we can use the following equations and use acceleration of gravity. ΔX= Vi (cosθ)Δt Vx = Vi (cosθ) = constant Point out that these equations are the same as those on the previous slides with vi,y = 0 or a launch angle  = 0. These equations could be used for the coin as it fell off the table (see the Notes on the first slide of this section) or for an object dropped from an airplane flying at a level altitude. The previous equations (last two slides) are more general and apply to any projectile.

Projectile Motion Summary
Projectile motion is free fall with an initial horizontal speed. Vertical and horizontal motion are independent of each other. Horizontally the velocity is constant. Vertically the acceleration is constant (-9.81 m/s2 ). Components are used to solve for vertical and horizontal quantities. Time is the same for both vertical and horizontal motion. Velocity at the peak is purely horizontal (vy = 0). The 4th and 5th summary points are essential for problem solving. Emphasize these points now, and return to them as students work through problems.

Classroom Practice Problem (Horizontal Launch)
A pelican flying along a horizontal path drops a fish from a height of 5.4m while traveling 5.0m/s. how far does the fish travel horizontally before it hits the water below? Answer: 5.0m As the students look at the equations, they will not find a single equation that allows them to solve this problem. First, as is often the case, they must solve for time using the height of the building (y) and the acceleration of gravity (ag). Then, they can use this time with the horizontal distance (x) to find the horizontal speed (vx).

Classroom Practice Problem (Horizontal Launch)
Give both the horizontal and vertical components of the velocity of the fish from the previous problem before the fish enters the water. Answer: 5.0m/s, 9.8m/s As the students look at the equations, they will not find a single equation that allows them to solve this problem. First, as is often the case, they must solve for time using the height of the building (y) and the acceleration of gravity (ag). Then, they can use this time with the horizontal distance (x) to find the horizontal speed (vx).

Classroom Practice Problem (Horizontal Launch)
A movie director is shooting a scene that involves dropping a stunt dummy out of an airplane and into a swimming pool. The plane is 10.0 m above the ground, traveling at a velocity of 22.5 m/s in the positive x direction . The director wants to know where the plane’s path the dummy should be dropped so that it will land in the pool? Answer: 32.2m As the students look at the equations, they will not find a single equation that allows them to solve this problem. First, as is often the case, they must solve for time using the height of the building (y) and the acceleration of gravity (ag). Then, they can use this time with the horizontal distance (x) to find the horizontal speed (vx).

Classroom Practice Problem (Horizontal Launch)
Find the instantaneous velocity of the stunt dummy in the previous problem as it hits the water? Answer: 26.5 m/s at ° below the horizontal As the students look at the equations, they will not find a single equation that allows them to solve this problem. First, as is often the case, they must solve for time using the height of the building (y) and the acceleration of gravity (ag). Then, they can use this time with the horizontal distance (x) to find the horizontal speed (vx).

Classroom Practice Problem (Horizontal Launch)
A cat chases a mouse across a 1.0m high table. The mouse steps out of the way and the cat slides off the table at a speed of 5.0m/s. Where does the cat strike the floor? Answer: 2.2m from the table As the students look at the equations, they will not find a single equation that allows them to solve this problem. First, as is often the case, they must solve for time using the height of the building (y) and the acceleration of gravity (ag). Then, they can use this time with the horizontal distance (x) to find the horizontal speed (vx).

Classroom Practice Problem (Projectile Launched at an Angle)
In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building 4.0m away. After a running start, he leaps at an angle of 15° with respect to the flat roof while traveling at a speed of 5.0m/s. Will he make it to the other roof, which is 2.5 m shorter than the building he jumps from? Answer: Yes, -2.3 m One option is to first solve for t in the vertical motion equations. This requires the use of the quadratic equation. Then, t can be used to find the horizontal distance in the horizontal motion equations. The problem can also be divided into two parts and solved without a quadratic equation. First, find the time required to reach the peak where vy is zero. Then, find the height reached and add it onto the 15 m. Finally, find the time required to fall from this height, and use the total time to find the horizontal distance.

Classroom Practice Problem (Projectile Launched at an Angle)
A golfer can hit a golf ball a horizontal distance of over 300m on a good day. What maximum height would a 301.5m drive reach if it were launched at an angle of 25.0° to the ground? (At the top of its flight, the ball’s vertical velocity will be zero) Answer: 70.3m One option is to first solve for t in the vertical motion equations. This requires the use of the quadratic equation. Then, t can be used to find the horizontal distance in the horizontal motion equations. The problem can also be divided into two parts and solved without a quadratic equation. First, find the time required to reach the peak where vy is zero. Then, find the height reached and add it onto the 15 m. Finally, find the time required to fall from this height, and use the total time to find the horizontal distance.

Classroom Practice Problem (Projectile Launched at an Angle)
Salmon often jump waterfalls to reach their breeding grounds. Starting 2.00m from a waterfall 0.55m in height, at what minimum speed must a salmon jumping at an angle of 32.0° leave the water to continue upstream? Answer: 6.2 m/s One option is to first solve for t in the vertical motion equations. This requires the use of the quadratic equation. Then, t can be used to find the horizontal distance in the horizontal motion equations. The problem can also be divided into two parts and solved without a quadratic equation. First, find the time required to reach the peak where vy is zero. Then, find the height reached and add it onto the 15 m. Finally, find the time required to fall from this height, and use the total time to find the horizontal distance.

Classroom Practice Problem (Projectile Launched at an Angle)
A quarterback throws the football to a receiver who is 31.5m down the field. If the football is thrown at an initial angle of 40.0 ° to the ground, at what initial speed must the quarterback throw the ball? What is the ball’s highest point during its flight? Answer: 17.7 m/s, 6.60m One option is to first solve for t in the vertical motion equations. This requires the use of the quadratic equation. Then, t can be used to find the horizontal distance in the horizontal motion equations. The problem can also be divided into two parts and solved without a quadratic equation. First, find the time required to reach the peak where vy is zero. Then, find the height reached and add it onto the 15 m. Finally, find the time required to fall from this height, and use the total time to find the horizontal distance.

Relative Motion Click below to watch the Visual Concept.

Frames of Reference A falling object is shown from two different frames of reference: the pilot (top row) an observer on the ground (bottom row) You might want to discuss the old movies of WW I and II pilots after they released the bombs. They always turned to the right or left or sharply upward. This was particularly important if they were flying low. Students should be able to see why in the lower sequence of pictures.

Relative Velocity vac = vab + vbc
vac means the velocity of object “a” with respect to frame of reference “c” Note: vac = -vca When solving relative velocity problems, follow this technique for writing subscripts.

Sample Problem A boat is traveling downstream. The speed of the boat with respect to Earth (vbe) is 20 km/h. The speed of the river with respect to Earth (vre) is 5 km/h. What is the speed of the boat with respect to the river? Solution: vbr = vbe+ ver = vbe + (-vre) = 20 km/h + (-5 km/h) vbr = 15 km/h This can more easily be solved using a diagram or common sense, but it is useful to show the students how to manipulate the equation to switch velocities from one frame of reference to another.

Classroom Practice Problem
A passenger at the rear of a train traveling at 15m/s relative to the earth throws a baseball with a peed of 15m/s in the direction opposite the motion of the train. What is the velocity of the baseball relative to Earth as it leave the thrower’s hand? Answer: 0 m/s You might want to work through Problem F in the text before having students work on this problem. For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful. For this problem, vpg = vpa + vag Since both of these values are provided, it is a simple vector addition with two right-angle vectors. It might be a good idea to use the component method to add these, since it is unusual to have a right angle between the two vectors. It will also reinforce the notion of breaking vectors down into components before recombining them to get a resultant.

Classroom Practice Problem
A spy runs from the front to the back of an aircraft carrier at a speed of 3.5m/s. If the aircraft carrier is moving forward at 18,0 m/s, how fast does the spy appear to be running when viewed by an observer on a nearby stationary submarine? Answer: 14.5 m/s in the direction that the aircraft carrier is moving You might want to work through Problem F in the text before having students work on this problem. For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful. For this problem, vpg = vpa + vag Since both of these values are provided, it is a simple vector addition with two right-angle vectors. It might be a good idea to use the component method to add these, since it is unusual to have a right angle between the two vectors. It will also reinforce the notion of breaking vectors down into components before recombining them to get a resultant.

Classroom Practice Problem
A ferry is crossing a river. If the ferry is headed due north with a speed of 2.5m/s relative to the water and the river’s velocity is 3.0m/s to the east, what will the boat’s velocity be relative to the Earth? (Remember to include the direction in describing the velocity Answer: 3.9 m/s at 40.0° north of east You might want to work through Problem F in the text before having students work on this problem. For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful. For this problem, vpg = vpa + vag Since both of these values are provided, it is a simple vector addition with two right-angle vectors. It might be a good idea to use the component method to add these, since it is unusual to have a right angle between the two vectors. It will also reinforce the notion of breaking vectors down into components before recombining them to get a resultant.

Classroom Practice Problem
A passenger at the rear of a train traveling at 15m/s relative to the earth throws a baseball with a peed of 15m/s in the direction opposite the motion of the train. What is the velocity of the baseball relative to Earth as it leave the thrower’s hand? Answer: km/h at 40.1° north of east You might want to work through Problem F in the text before having students work on this problem. For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful. For this problem, vpg = vpa + vag Since both of these values are provided, it is a simple vector addition with two right-angle vectors. It might be a good idea to use the component method to add these, since it is unusual to have a right angle between the two vectors. It will also reinforce the notion of breaking vectors down into components before recombining them to get a resultant.

Classroom Practice Problem
A pet store supply truck moves at 25.0m/s north along a highway. Inside, a dog moves at 1.75 m/s at an angle 35.0° east of north. What is the velocity of the dog relative to the road? Answer: 26.4 m/s at 2.17° east of north You might want to work through Problem F in the text before having students work on this problem. For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful. For this problem, vpg = vpa + vag Since both of these values are provided, it is a simple vector addition with two right-angle vectors. It might be a good idea to use the component method to add these, since it is unusual to have a right angle between the two vectors. It will also reinforce the notion of breaking vectors down into components before recombining them to get a resultant.

Now what do you think? Suppose you are traveling at a constant 80 km/h when a car passes you. This car is traveling at a constant 90 km/h. How fast is it going, relative to your frame of reference? How fast is it moving, relative to Earth as a frame of reference? Does velocity always depend on the frame of reference? Does acceleration depend on the frame of reference? The second car is traveling at 10 km/h relative to your frame of reference, but 90 km/h relative to Earth. Students should now realize that velocity is always relative to a reference frame. The final question provides a way for you to extend the lesson content. Students may say yes because velocity does depend on a frame of reference, and acceleration is related to velocity. Try to start a class discussion about the two cars. Suppose the 90 km/h car speeds up to 110 km/h in 4 seconds. What acceleration would each observer calculate? It is 5 km/h/s for both observers because the CHANGE IN VELOCITY is the same for both. So, acceleration is NOT relative to the frame of reference.

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