8 min readβ’november 3, 2020

Peter Apps

The acceleration of the center of mass of a system is related to the net force exerted on the system, where a = F/m.

βΆ A force may be applied to this point to cause a linear acceleration without angular acceleration occurring.

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

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.

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,Β but it is imperative to understand the **relationships **each graph has to one another.Β

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.Β Β

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.Β

**Linearization π€ΈββοΈ**

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 = Β½ a, students often make the mistake of writing m = a which is double the acceleration.Β

βΆ Still feeling a little confused on Linearization? Donβt worry! Check out __this video__ from AP Physics 1 Online for more practice!

In Kinematics there are four major equations you must understand to begin calculations. They relate **acceleration, displacement, initial and final velocity, **and **time **together**.Β **

βΆ In order to solve for a variable without having all four other quantities known, we look at the βVariable Missingβ column to pick the equation that best suits our question.Β

EXAMPLE:A super car races by at a speed of 68 m/s and slows down as a rate of 4 m/s/s.Β How much runway is needed to stop the plane? - We are given the variables a, Vo, and Vf but we are missing Ξx and t
- Vf^2 =Vo^2 + 2aΞx
- STEP 1: Cross out Vf because the car will stop at a velocity of 0 m/s
- STEP 2: Plug in the known variables: 0 = (68 m/s^2) + 2 (-4 m/s/s^2) (Ξx)
Your final answer should be Ξx = 578 m |

Equation | Formula | Variable Missing |

Big Four #2 | Vf = Vo + gt | Ξy |

Big Four #3 | y = Vot + 1/2 gt2 | Vf |

Big Four #4 | Vf2 = Vo2 + 2gy | t |

βΆ In free fall equations, we now replace Ξx with Ξy and a with g giving us a modified list of The Big Four as seen in the table above.Β

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.Β What is the displacement of the ball? - We are given the variables g, t, and Vo, but we are missing Ξy and Vf
- Ξy = Vot + Β½ gt2
- STEP 1: Cross out Vo because an object that is dropped has an initial velocity of 0 m/s
- STEP 2: Plug in the known variables: Ξy = Β½ (10 m/s/s) (2.8 s2)
Your final answer should be Ξy = 39.2 m What is the ballβs final velocity? - We are given the variables g, t, and Vo, but we are missing Ξy and Vf
- Vf = Vo + gt
- STEP 1: Cross out Vo because an object that is dropped has an initial velocity of 0 m/s
- STEP 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 Ξy and t
- Vf2 = Vo2 + 2gΞy
- STEP 1: Cross out Vf because an object thrown up has a final velocity of 0 m/s
- STEP 2: Plug in known variables: 0 = (20 m/s2) + 2(10 m/s/s) (Ξy)
Your final answer should be Ξy = 20 m |

- Horizontal Velocity is constant
- Vertical Velocity is accelerated
- Gravity affects the vertical motion
- Air resistance is ignored

βΆ 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.Β

Formula | Type | Variable Missing |

y = Voyt + 1/2 gt2 | Vertical | Vfy |

Vfy = Voy + gt | Vertical | Ξy |

Vfy2 = Voy2 + 2gy | Vertical | t |

x = Vxt | Horizontal | N/A |

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 Ξy, Voy and Vx, but we are missing t, Ξx, and Vfy
- Ξy = Voyt + Β½ gt2
- STEP 1: Cross out Voy because there is no initial vertical velocity
- STEP 2: Plug in known variables: 1.5m = Β½ (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 Ξy, Voy and Vx, but we are missing Ξx and Vfy
- ΞxΒ = Vxt
- STEP 1: Plug in known variables: ΞxΒ = (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 Ξy, Voy and Vx, but we are missing VfyΒ
- Vfy = Voy + gt OR Vfy2 = Voy2 + 2gΞy
- STEP 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 |

βΆ Still feeling a little confused on Projectile Launches? Donβt worry! Check out __this live stream__ from Fiveable for more practice!

Angled launches require you to find the Vox and Voy, or vector components, based on the initial velocity Vo and the angle Σ¨. Now you can solve for the following: Vo, Vox, Voy, Σ¨, t, X, Ymax, and Vf.Β

- 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/Σ¨ |

sin(Vo) = Voy/Σ¨ |

Vox = Vocos(Σ¨) |

Voy = Vosin(Σ¨) |

t = 2(Voy)/g(This only applies if the starting and ending heights are the same) |

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 Σ¨ and Vo, but we are missing t, Vox, and Voy
- t = 2(Voy)/g AND Voy = Vosin(Σ¨)
- STEP 1: Solve for Voy: Voy = 20sin(30), Voy = 10 m/s
- STEP 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 Σ¨ and Vo, but we are missing Vox
- ΞxΒ = Vxt
- STEP 1: Solve for Vx: Vox = 20cos(30), Vox = 17.3 m/s
- STEP 2: Plug in known variables: ΞxΒ = (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 Σ¨, Vfy, and Vo, but we are missing Vox
- Vfy2 = Voy2 + 2gΞyΒ
- STEP 1: Plug in known variables: 0 = (10 m/s)2 + 2(10 m/s/s)Ξy
Your final answer should be 5 meters |

βΆ Still feeling a little confused on Angled Launches? Donβt worry! Check out __this video__ from Khan Academy for more practice!Β
Want more practice - Check out the Fiveable Live streams on this topic:

- π₯ Watch AP Physics 1 - 2D Motion & Freefall
- π₯ Watch AP Physics 1 - Horizontal Launch Problems
- π₯ Watch AP Physics 1 - Angle Launch Problems

Sign up now for instant access to 2 amazing downloads to help you get a 5

Browse Study Guides By Unit

π

AP Physics Essentials

π

Big Reviews: Finals & Exam Prep

βοΈ

Free Response Questions (FRQs)

π§

Multiple Choice Questions (MCQs)

π

Unit 10: Mechanical Waves & Sound

π

Unit 1: Kinematics

π

Unit 2: Dynamics

π

Unit 3: Circular Motion

β‘οΈ

Unit 4: Energy

β³οΈ

Unit 5: Momentum

πΈ

Unit 6: Simple Harmonic Motion

π‘

Unit 7: Torque & Rotational Motion

π‘

Unit 8: Electric Charges & Electric Force

π

Unit 9: DC Circuits

Practice your typing skills while reading Representations of Motion

Start GameTake this quiz for a progress check on what youβve learned this year and get a personalized study plan to grab that 5!

START QUIZStudying with Hours = the ultimate focus mode

Start a free study session