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Unit 1

# 1.2 Representations of Motion Peter Apps

### AP Physics 1🎡

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## 4.A Enduring Understanding ⛓

The acceleration of the center of mass of a system is related to the net force exerted on the system, where a = F/m.
Key Vocabulary: Center of mass - a point on an object or system that is the mean position of the matter.
⟶ A force may be applied to this point to cause a linear acceleration without angular acceleration occurring.

### 4.A.1 Essential Knowledge ➡️

The linear motion of a system can be described by the displacement, velocity, and acceleration of its center of mass.

### 4.A.2 Essential Knowledge ⏰

The acceleration is equal to the rate of change of velocity with time, and velocity is equal to the rate of change of position with time.

## Graphical Representations of Motion 📈

As we covered in 1.1 Position, Velocity and Acceleration there are ways to represent all three quantities graphically. We also reviewed how to interpret these graphs,  but it is imperative to understand the relationships each graph has to one another. Image Courtesy of geogebra

From the image above, we can use position to find velocity for any given period of time by looking at the slope of the Position vs. Time Graph. Working from velocity to position we can look at the area underneath the curve to find displacement, but it is not possible to determine how far from the detector the object is located from a Velocity vs. Time Graph.
When working from velocity to acceleration we look at the slope of the Velocity vs. Time Graph. Similarly, when working from acceleration to velocity we look at the area under the curve to find velocity.

Linearization 🤸‍♀️

Many graphs in physics will not be perfectly straight lines, but we can turn the curve into a straight sloped line! We accomplish this by squaring the x-axis value. Image Courtesy of x-engineer.org

We start by graphing data points to confirm a nonlinear function (a curve), then we square the x-axis value and regraph! Keep in mind that once a graph is linearized finding the acceleration value is now m = ½ a, students often make the mistake of writing m = a which is double the acceleration.
⟶ Still feeling a little confused on Linearization? Don’t worry! Check out this video from AP Physics 1 Online for more practice!

## Mathematical Representations of Motion 📍

In Kinematics there are four major equations you must understand to begin calculations. They relate acceleration, displacement, initial and final velocity, and time together Variable Interpretation: Δx is horizontal displacement in meters, Vf is final velocity in meters/second, Vo is initial velocity in meters/second, t is time in seconds, and a is acceleration in m/s/s.
⟶ In order to solve for a variable without having all four other quantities known, we look at the ‘Variable Missing’ column to pick the equation that best suits our question.
 EXAMPLE:A super car races by at a speed of 68 m/s and slows down as a rate of 4 m/s/s. How much runway is needed to stop the plane?We are given the variables a, Vo, and Vf but we are missing Δx and tVf^2 =Vo^2 + 2aΔxSTEP 1: Cross out Vf because the car will stop at a velocity of 0 m/sSTEP 2: Plug in the known variables: 0 = (68 m/s^2) + 2 (-4 m/s/s^2) (Δx)Your final answer should be Δx = 578 m

## Free Fall ⚽️

Key Vocabulary: Free Fall - an object only under the influence of gravity
Equation: velocity = force of gravity x time
Key Vocabulary: Acceleration due to Gravity - 9.8 m/s/s (it is acceptable to round up to 10 m/s/s on the AP Physics 1 exam)
 Equation Formula Variable Missing Big Four #2 Vf = Vo + gt Δy Big Four #3 y = Vot + 1/2 gt2 Vf Big Four #4 Vf2 = Vo2 + 2gy t
Variable Interpretation: Δy is vertical displacement in meters, Vf is final velocity in meters/second, Vo is initial velocity in meters/second, t is time in seconds, and g is acceleration due to gravity in m/s/s.
⟶ In free fall equations, we now replace Δx with Δy and a with g giving us a modified list of The Big Four as seen in the table above.
Object Dropped (trip down)
• Vo (initial velocity) = 0 m/s
• g works in the direction of motion
Object Tossed (trip up)
• Vf (final velocity) = 0 m/s
• At maximum height of its trip, an object has a velocity of 0 m/s
• g works against the direction of motion
 EXAMPLE #1:A ball is dropped from the top of a building. It falls 2.8s. What is the displacement of the ball?We are given the variables g, t, and Vo, but we are missing Δy and VfΔy = Vot + ½ gt2STEP 1: Cross out Vo because an object that is dropped has an initial velocity of 0 m/sSTEP 2: Plug in the known variables: Δy = ½ (10 m/s/s) (2.8 s2)Your final answer should be Δy = 39.2 mWhat is the ball’s final velocity?We are given the variables g, t, and Vo, but we are missing Δy and VfVf = Vo + gtSTEP 1: Cross out Vo because an object that is dropped has an initial velocity of 0 m/sSTEP 2: Plug in the known variables: Vf = (10 m/s/s) (2.8 s)Your final answer should be Vf = 28 m/s
 EXAMPLE #2:A soccer ball is thrown straight up with an initial velocity of 20 m/s.What height did the soccer ball reach?We are given the variables g, Vo, and Vf, but we are missing Δy and tVf2 = Vo2 + 2gΔySTEP 1: Cross out Vf because an object thrown up has a final velocity of 0 m/sSTEP 2: Plug in known variables: 0 = (20 m/s2) + 2(10 m/s/s) (Δy)Your final answer should be Δy = 20 m

## Projectile Launches ☄️

Key Vocabulary: Projectile Launches - something is fired, thrown, shot, or hurled near Earth’s surface
• Horizontal Velocity is constant
• Vertical Velocity is accelerated
• Gravity affects the vertical motion
• Air resistance is ignored
⟶ Range describes the horizontal component, and height describes the vertical component
When dealing with horizontal projectile launches, launches that have an entirely horizontal initial velocity (ex. a ball rolling off a table), we break it up into two sets of equations: vertical and horizontal.
 Formula Type Variable Missing y = Voyt + 1/2 gt2 Vertical Vfy Vfy = Voy + gt Vertical Δy Vfy2 = Voy2 + 2gy Vertical t x = Vxt Horizontal N/A
Variable Interpretation: Δy is vertical displacement in meters, Δx is horizontal displacement in meters, Vfy is vertical final velocity in meters/second, Voy is vertical initial velocity in meters/second, Vx is horizontal velocity in m/s, t is time in seconds, and g is acceleration due to gravity in m/s/s.
As shown in the chart above, there are three equations we use for the vertical component of launches and one for the horizontal. You cannot put an x-component into a formula without a y-component!
 EXAMPLE:A tennis ball is rolling on a ledge with a velocity of 5 m/s. If the tennis ball rolls off the table, which is 1.5m high: (a) How long would it take to hit the ground?We are given variables Δy, Voy and Vx, but we are missing t, Δx, and VfyΔy = Voyt + ½ gt2STEP 1: Cross out Voy because there is no initial vertical velocitySTEP 2: Plug in known variables: 1.5m = ½ (10 m/s/s) t2 Your final answer should be 0.55 seconds(b) What will be the distance it travels before reaching the ground?We are given variables Δy, Voy and Vx, but we are missing Δx and VfyΔx  = VxtSTEP 1: Plug in known variables: Δx  = (5 m/s) (0.55 s) Your final answer should be 2.75 meters(c) What is the magnitude of the velocity right before the tennis ball reaches the ground?We are given variables Δy, Voy and Vx, but we are missing Vfy Vfy = Voy + gt OR Vfy2 = Voy2 + 2gΔySTEP 1: Plug in known variables: Vfy = (10 m/s/s) (0.55 s) OR Vfy2 = 2 (10 m/s/s) (1.5 m) Your final answer should be 5.5 m/s
⟶ Still feeling a little confused on Projectile Launches? Don’t worry! Check out this live stream from Fiveable for more practice!

## Angled Launches 🏹

Key Vocabulary: Angled Launches - launches at an angle that include both a horizontal and vertical component of initial velocity
Key Vocabulary: Vector Components - the horizontal and vertical parts of a vector
Angled launches require you to find the Vox and Voy, or vector components, based on the initial velocity Vo and the angle Ө. Now you can solve for the following: Vo, Vox, Voy, Ө, t, X, Ymax, and Vf.
• If an object is shooting upward, Voy is (+)
• If an object is shooting downward, Voy is (-)
• Vertical velocity is 0 m/s at the top
• Flight is symmetric if the projectile starts and ends at the same height
 Angled Launch Formulas cos(Vo) = Vox/Ө sin(Vo) = Voy/Ө Vox = Vocos(Ө) Voy = Vosin(Ө) t = 2(Voy)/g(This only applies if the starting and ending heights are the same)
Variable Interpretation: Voy is vertical initial velocity in meters/second, Vox is initial horizontal velocity in m/s, t is time in seconds, t is time in seconds, and g is acceleration due to gravity in m/s/s.
 EXAMPLE:A cannonball is shot at a 30 degree angle above the horizontal at 20 m/s.(a) How much time will the cannonball travel for in the air?We are given variables Ө and Vo, but we are missing t, Vox, and Voyt = 2(Voy)/g AND Voy = Vosin(Ө)STEP 1: Solve for Voy: Voy = 20sin(30), Voy = 10 m/sSTEP 2: Plug in known variables: t = 2(10 m/s)/(10 m/s/s)Your final answer should be 2 seconds(b) How far will the cannonball travel?We are given variables Ө and Vo, but we are missing VoxΔx  = VxtSTEP 1: Solve for Vx: Vox = 20cos(30), Vox = 17.3 m/sSTEP 2: Plug in known variables: Δx  = (17.3 m/s)(2 s)Your final answer should be 34.6 meters(c) What is the maximum height the cannonball will reach?We are given variables Ө, Vfy, and Vo, but we are missing VoxVfy2 = Voy2 + 2gΔy STEP 1: Plug in known variables: 0 = (10 m/s)2 + 2(10 m/s/s)ΔyYour final answer should be 5 meters
⟶ Still feeling a little confused on Angled Launches? Don’t worry! Check out this video from Khan Academy for more practice!  Want more practice - Check out the Fiveable Live streams on this topic:

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