# Two-Dimensional Motion and Vectors

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Two-Dimensional Motion and Vectors
Chapter 3 pg

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 Make sure when adding vectors that You use the same unit Describing similar quantities

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.

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.

3.2 Vector Operations

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.

Pythagorean Theorem and Tangent Function
We can use the inverse of the tangent function to find the angle. θ= tan-1 (opp/adj) Another way to look at our triangle d2 =Δx2 + Δy2 d Δy θ Δx

Example An archaeologist climbs the great pyramid in Giza. The pyramid height is 136 m and width is 2.30 X 102m. What is the magnitude and direction of displacement of the archaeologist after she climbs from the bottom to the top?

Example Given: Unknown: Δy= 136m
width is 2.30 X 102m for whole pyramid So, Δx = 115m Unknown: d = ?? θ= ??

Example d2 =Δx2 + Δy2 d = √Δx2 + Δy2 Calculate: d = √ (115)2 +(136)2
d = 178m θ= tan-1 (opp/adj) θ= tan-1 (136/115) θ= 49.78°

Example While following the directions on a treasure map a pirate walks 45m north then turns and walks 7.5m east. What single straight line displacement could the pirate have taken to reach the treasure?

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

Resolving Vectors into Components
Component: the horizontal x and vertical yparts that add up to give the actual displacement 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 the angle is 35.0˚. Review the first solution with students, and then let them solve for the second component. 35°

Example Given: v= 95 km/h θ= 35.0° Unknown vx=??vy= ??
Rearrange the equations sin θ= opp/ hyp opp=(sin θ) (hyp) cosθ= adj/ hyp adj= (cosθ)(hyp)

Example vy=(sin θ)(v) vx= (cosθ)(v) vy= (sin35°)(95) vy= 54.49 km/h
vx = km/h

Example How fast must a truck travel to stay beneath an airplane that is moving 105 km/h at an angle of 25° to the ground?

3.3 Projectile Motion

Projectile Motion Projectiles: objects that are launched into the air
tennis balls, arrows, baseballs, javelin Gravity affects the motion Projectile motion: The curved path that an object follows when thrown, launched or otherwise projected near the surface of the earth 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.

Projectile Motion Path is parabolic if air resistance is ignored
Path is shortened under the effects of air resistance

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,idoes change (because of the acceleration due to gravity).

Projectile Motion Projectile motion is free fall with an initial horizontal speed. Vertical and horizontal motion are independent of each other. Vertically the acceleration is constant (-10 m/s2 ) We use the 4 acceleration equations Horizontally the velocity is constant We use the constant velocity equations The 4th and 5th summary points are essential for problem solving. Emphasize these points now, and return to them as students work through problems.

Projectile Motion 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).

Example The Royal Gorge Bridge in Colorado rises 321 m above the Arkansas river. Suppose you kick a rock horizontally off the bridge at 5 m/s. How long would it take to hit the ground and what would it’s final velocity be?

Example Given: d = 321m a = 10m/s2 vi= 5m/s t = ?? vf = ??
REMEMBER we need to figure out : Up and down aka free fall (use our 4 acceleration equations) Horizontal (use our constant velocity equation)

Classroom Practice Problem (Horizontal Launch)
People in movies often jump from buildings into pools. If a person jumps horizontally by running straight off a rooftop from a height of 30.0 m to a pool that is 5.0 m from the building, with what initial speed must the person jump? Answer: 2.0 m/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).

Projectiles Launched at an Angle
We will make a triangle and use our sin, cos, tan equations to find our answers Vy = V sin θ Vx = V cosθ tan = θ(y/x)

Classroom Practice Problem (Projectile Launched at an Angle)
A golfer practices driving balls off a cliff and into the water below. The edge of the cliff is 15 m above the water. If the golf ball is launched at 51 m/s at an angle of 15°, how far does the ball travel horizontally before hitting the water? Answer: 1.7 x 102m (170 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.