Presentation on theme: "Chapter 2: One-Dimensional Kinematics Outline 2-1Position, Distance, and Displacement 2-2Average Speed and Velocity 2-3Instantaneous Velocity 2-4Acceleration."— Presentation transcript:
Chapter 2: One-Dimensional Kinematics Outline 2-1Position, Distance, and Displacement 2-2Average Speed and Velocity 2-3Instantaneous Velocity 2-4Acceleration 2-5Motion with Constant Acceleration 2-6Applications of the Equations of Motion 2-7Freely Falling Objects
What we will do today: The most important things to understand in Chapter 2. How to use this information to solve problems.
TA’s Information Name: Tatiana Brusentsova Email: Office hours
Major Concepts Position, distance and displacement Speed and velocity –Average –Instantaneous –Constant Acceleration –Average –Instantaneous –Constant Graphs of position versus time, velocity versus time, and acceleration versus time Equations of motion with constant acceleration Free fall Warning Areas: Speed vs. Velocity Distance vs. Displacement Constant Velocity vs. Constant Acceleration
How the physics is described, step by step. Constant speed: Constant acceleration:
Constant speed, acceleration. Constant Speed t, sec x, M tt xx Slope = v Constant Acceleration t, sec v, m/s tt vv Slope = a Special case of constant a.
Figure 2-26 Problem 2-21 Find the average velocity for each segment of the “walk”, and for the total “walk”. Now, compare DISPLACEMENT and DISTANCE. Compare, average VELOCITY and SPEED.
Figure 2-28 Problem 2-32 What is the average acceleration for each segment, what is the average acceleration for the whole motorcycle ride?
Constant Acceleration: The “master” 1D formula: Uses:
Table 2-4 Constant-Acceleration Equations of Motion Variables RelatedEquationNumber Velocity, time, acceleration v = v 0 + at2-7 Initial, final, and average velocity v av = ½(v 0 + v)2-9 Position, time, velocity x = x 0 + ½(v 0 + v)t2-10 Position, time, acceleration x = x 0 + v 0 t + ½ at 2 2-11 Velocity, position, acceleration v 2 = v 0 2 + 2a(x – x 0 ) = v 0 2 + 2a x 2-12
These equations are related x = x 0 + v 0 t + ½ at 2 v = v 0 + at Use: or a=(v-v 0 )/t Then: x = x 0 + v 0 t + ½ at 2 = x 0 + v 0 t + ½(v-v 0 )t 2 /t = x 0 +½(v+v 0 )t = x 0 +v avg t Everything is derived from definitions of average and constant acceleration and velocity. (the “Master equation”)
Figure 2-29 Problem 2-33 CAREFUL! What is the displacement for each segment of the graph shown? Problem solving: 1.Draw a picture (given) 2.What are they asking for? 3.What is given? 4.What are the mathematical relationships?
The formula that most correctly describes the graph of Drag Racing motion is: 1. v 2 = v 0 2 + 2a(x – x 0 ) 2. x = x 0 + ½ at 2 3. x = x 0 + ½(v 0 + v)t 4. v = v 0 + at
Figure 2-16 Velocity as a Function of Position for the Ranger in Example 2-8 Notice that ½ the speed is lost in last ¼ of stopping distance. CONSTANT DECELERATION v 2 = v 0 2 + 2a(x – x 0 )
Hit the Brakes! V o =11.4 m/s a=-3.8m/s 2 Name that graph! t t x
Name that graph! (in right order) 1. Velocity, position, acceleration 2. Position, acceleration, velocity 3. Velocity, acceleration, velocity V o =11.4 m/s a=-3.8m/s 2 t xt
Constant acceleration problem: A vehicle is traveling at an initial speed v. It brakes with constant acceleration a (which can be negative). What is the total time to come to a stop? What is the total distance traveled? What is the shape of position vs. time? What is the shape of velocity vs. time? V 0 =11.4 m/s a=-3.8 m/s^s t, sec m (x) m/s (v)
Objects in “free-fall” Acceleration is gravity, a = -g = -9.8 m/s ^2 Q: A ball is dropped from the top of the library. At the same time, a ball is thrown from the ground so that it just reaches the top of the library. Which of the following statements is true? Draw picture. What is asked? What is given? What are relationships? Solve for unknown.
Help, I’m in free fall! 1. The dropped ball reaches the ground sooner, because it picks up speed as it drops. 2. The thrown ball reaches the top sooner, because it started with higher speed. 3. They reach the endpoints at the same time. 4. Can’t answer—depends on initial speed of ball. This problem can be quickly solved by looking at the time symmetry of the situation. Let’s do this the “long way”, just to see how it’s done.
P 2-27: The pop fly. A baseball is hit straight upwards. The round-trip time in the air is 4.5 seconds. How high did the ball go? What was the initial speed? Draw a picture. Assign “knowns” to variables. Assign variable names to “unknowns”. Work backwards from “knowns” to “unknowns”.
P. 2-91: Seagull flying upwards with seashell, which drops. Seagull flying upward 5.2 m/s, drops shell at 12.5 m. What is acceleration at time of drop? What is maximum height? When does it return to 12.5 m? What is its speed at 12.5m?
P. 2-94. Ball thrown vertically with initial speed v. Ball one thrown upwards with speed v. Reaches maximum height h, when ball two is thrown with same speed. Draw x vs. t for each ball. At what height do balls cross paths?