Bell Ringer: Define Displacement. Define Velocity. Define Speed. Define Acceleration. What is constant acceleration? The slope of the line on a distance vs time graph represents what? The slope of the line on a velocity vs time graph represents what?

NOTES 2.2: One-Dimensional Motion Constant Acceleration Physics Honors I

OBJECTIVES: Interpret position-time graphs for motion with constant acceleration. Analyze one-dimensional motion when the acceleration is constant. Apply graphical and mathematical relationships to solve problems related to constant acceleration. Define acceleration due to gravity. Solve problems involving objects in free-fall.

Active Physics Reference: Negative Acceleration– Chapter 1, Section 5, Page 78-86 Constant Speed and Acceleration: Measuring Motion - Chapter 2, Section 2, Page 148-151 Physics: Principles and Problems (RED BOOK in classroom) Section 3.2 – Motion with Constant Acceleration – Page 65 Section 3.3 – Free Fall Acceleration – Page 72

Further Learning: Khan Academy – Constant Acceleration https://www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/v/average-velocity-for-constant-acceleration Physics Classroom http://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations

Constant Acceleration: https://www.youtube.com/watch?v=VrDFqmhTTVQ

Constant Acceleration: One common situation that often arises in one dimensional motion is the case of constant acceleration. As the name implies, constant acceleration is when acceleration is constant – meaning that an object moves, it increases (or decreases) its velocity at constant rate.

Automobile with constant Acceleration:

Constant Acceleration: Scenario: You are running the 40 yard dash. You are trying to run as fast as you can so that you make the team. From the time you start running, you give it all that you have. Let’s assume that it takes you 10 yards to reach your maximum speed. During those 10 yards, you are pushing as hard as you can to gain speed and accelerate at a constant rate.

Kinematic Equations: v f = v i +at v f 2 = v i 2 +2a ( x f − x i ) Kinematic equations are five equations that relate velocity, distance, time, and acceleration which are all dependent on the acceleration being constant. Important: Please take the time to remember these equations by heart because they will often play a part in further studies. v f = v i +at v f 2 = v i 2 +2a ( x f − x i ) x f − x i = v i t+ 1 2 a t 2 x f − x i = v f t− 1 2 a t 2 x f − x i = 1 2 v i + v f t

Example 1: An airplane accelerates down a run-way at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before take-off. Step 1: ????? Step 1: Determine what you have. We have: 𝑎=3.20 m/s2 – Constant Acceleration t = 32.8 s. v i =0 m/s 𝑑 i =0 m

Example 1: x f − x i = v i t+ 1 2 a t 2 An airplane accelerates down a run-way at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before take-off. Step 2: ????? Step 2: Choose an Equation. We need an equation that has acceleration, time, initial velocity, and displacement. What do we not need in an equation? Final velocity. The equation is not asking about final velocity so let’s pick an equation of the 5 constant acceleration equations without it. x f − x i = v i t+ 1 2 a t 2

Example 1: An airplane accelerates down a run-way at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before take-off. Step 3: ????? Step 3: Should be to isolate the variable you are searching for, in the case, we are looking for the change in position, x f − x i , so we don’t have to do any further equation manipulation. So leave the equation as it is.

Example 1: An airplane accelerates down a run-way at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before take-off. Step 4: ????? Step 4: Plug and Chug. You know what you are searching for, you have picked the right equation, you have isolated the variable you need, so just plug in and solve. As always, be careful putting it into the calculator. x f −(0 m s )= 0 m s t+ 1 2 (3.20 m/s2 )(32.8 𝑠) 2 x f = 1721 m

Check Point 1: A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration). An engineer is designing a runway for an airport. Several planes will use the runway and the engineer must design it so that it is long enough for the largest planes to become airborne before the runway ends. If the largest plane accelerates at 3.30 m/s2 and has a takeoff speed of 88.0 m/s, then what is the minimum allowed length for the runway? 11.2 m/s2 1.62 x 105 m/s2 (3 sig figs) 1173 m = 1170 m (3 sig figs)

Free-Fall Acceleration / Gravity: 11.2 m/s2 1.62 x 105 m/s2 (3 sig figs) 1173 m = 1170 m (3 sig figs)

Free-Fall Acceleration / Gravity: Free-fall acceleration (g) – the acceleration of an object towards the Earth. It is independent of the object’s characteristics, such as mass, density, or shape; instead it is the same for all objects so that: g=9.8 m s 2 =32 ft s 2 Up = negative Down = positive 11.2 m/s2 1.62 x 105 m/s2 (3 sig figs) 1173 m = 1170 m (3 sig figs)

Free-Fall Acceleration / Gravity: For Free-fall, we just substitute a g (for gravitational acceleration) instead of a (meaning constant acceleration). There is no difference, they both mean constant acceleration, only we always know the constant acceleration on the y-axis:  v f = v i +gt v f 2 = v i 2 +2𝑔 ( y f − y i ) y f − y i = v i t+ 1 2 𝑔 t 2 y f − y i = v f t− 1 2 𝑔 t 2 y f − y i = 1 2 v i + v f t  

Free-Fall Acceleration / Gravity: For Free-fall:   The directions of motion are now along a vertical y-axis instead of the x-axis with the positive direction of y upward. Free-Fall acceleration is negative – that is, downward on the y-axis, toward Earth’s center – and so it has the value –g in the equations.

Example 2: A pitcher tosses a baseball up along a y-axis with an initial speed of 12 m/s. What is the ball’s maximum height above its release point? How long does the ball take to reach its maximum height? How long does the ball take to reach a point 5.0 m above its release point.

Example 2: A pitcher tosses a baseball up along a y-axis with an initial speed of 12 m/s. What is the ball’s maximum height above its release point? STEP 1: ???? STEP 1: Determine what we know and what the problem is asking for. Problem is asking for the maximum height, so we know we need to have the final height 𝑦 𝑓 in the equation we choose. We are given: v i =12 m s g=9.8 m s 2 v f =0 m/s y 𝑖 =0 m

Example 2: A pitcher tosses a baseball up along a y-axis with an initial speed of 12 m/s. What is the ball’s maximum height above its release point? STEP 2: ???? STEP 2: Choose an equation. We need an equation with y f , y 𝑖 , v i , v f , and 𝑔. We do not know time, so we need to find an equation without time. v f 2 = v i 2 +2g ( y f − y i )

Example 2: A pitcher tosses a baseball up along a y-axis with an initial speed of 12 m/s. What is the ball’s maximum height above its release point? STEP 3: ???? STEP 3: Manipulate the equation to get what we are solving for. v f 2 = v i 2 +2𝑔 ( y f − y i ) v f 2 − v i 2 =2𝑔 ( y f − y i ) y f − y i = v f 2 − v i 2 2𝑔

Example 2: A pitcher tosses a baseball up along a y-axis with an initial speed of 12 m/s. What is the ball’s maximum height above its release point? STEP 4: ???? STEP 4: Plug and Chug y f − y i = v f 2 − v i 2 2𝑔 y f = 7.3 m

Example 2: A pitcher tosses a baseball up along a y-axis with an initial speed of 12 m/s. (B) How long does the ball take to reach its maximum height? What equation would you use? At this point, you can use any equation as long as it has time in it but the easiest one would be: v f = v i +gt Answer: t = 1.2 s

Example 2: A pitcher tosses a baseball up along a y-axis with an initial speed of 12 m/s. (C) How long does the ball take to reach a point 5.0 m above its release point? There are two separate times when the ball passes the five meter mark so we can solve for both. If you will notice, the below equations looks like the quadratic equation. y f − y i = v i t+ 1 2 𝑔 t 2 Move all of the variables to the same side of equation so that it all equals 0, and use the quadratic formula.

Example 2: A pitcher tosses a baseball up along a y-axis with an initial speed of 12 m/s. (C) How long does the ball take to reach a point 5.0 m above its release point? y f − y i = v i t+ 1 2 𝑔 t 2 t = 0.53 s and t = 1.9 s

Check Point 2: A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon. 4) 1.29 s

Exit Ticket 2: 5) What does uniform acceleration mean? What doe we mean by free-fall acceleration? Why does a feather fall slower than a bowling ball on earth? Who discovered that all objects fall at the same acceleration?