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Edit Comment Close Premium member Presentation Transcript Physics 201: Lecture 2: Physics 201: Lecture 2 Kinematics One dimensional motion Equations of motion for constant accelerationMotion in 1 dimension: Motion in 1 dimension Motion: change in position with time. In 1-D, we usually write position as x(t). Also called trajectory. Since it’s in 1-D, all we need to indicate direction is + or . Displacement in a time t = t2 - t1 is x = x(t2) - x(t1) = x2 - x1 t x t1 t2 x t x1 x2 some particle’s trajectory in 1-DAverage Velocity: Average Velocity t x t1 t2 x x1 x2 trajectory Velocity v is the “rate of change of position” Sign of v tells the direction the object is moving Magnitude of v is the speed Average velocity vav in the time t = t2 - t1 is: t Vav = slope of line connecting x1 and x2.Instantaneous Velocity: Consider the limit t1 t2 Instantaneous velocity v is defined as: Instantaneous Velocity t x t1 t2 x x1 x2 t so v(t2) = slope of line tangent to path at t2.Acceleration: Acceleration Acceleration a is the “rate of change of velocity” Average acceleration aav in the time t = t2 - t1 is: And instantaneous acceleration a is defined as: usingAcceleration: Acceleration Constant velocity Zero acceleration Constant acceleration in the same direction as v Increasing velocity Constant acceleration opposite of v Decreasing velocitySkydiver Jumps Out: Skydiver Jumps Out A skydiver is falling straight down, along the negative y direction. During the initial part of the fall, her speed increases from 16 to 28 m/s in 1.5 s. Which of the following is correct? 1) v>0, a>0 2) v>0, a<0 3) v<0, a>0 4) v<0, a<0Parachute Opens: Parachute Opens During a later part of the fall, after the parachute has opened, her speed decreases from 48 to 26 m/s in 11 s. Which of the following is correct? 1) v>0, a>0 2) v>0, a<0 3) v<0, a>0 4) v<0, a<0 If speed is increasing, v and a are in same direction. If speed is decreasing, v and a are in opposite direction.Question: Question When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path? (a) Both v = 0 and a = 0. (b) v 0, but a = 0. (c) v = 0, but a 0. Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero. Since the velocity is continually changing there must be some acceleration. In fact the acceleration is caused by gravity (g = 9.81 m/s2). correctSlide10: A ball is thrown straight up in the air and returns to its initial position. For the time the ball is in the air, which of the following statements is true? 1 - Both average acceleration and average velocity are zero. 2 - Average acceleration is zero but average velocity is not zero. 3 - Average velocity is zero but average acceleration is not zero. 4 - Neither average acceleration nor average velocity are zero. Question Free fall: acceleration is constant (-g) Initial position = final position: x=0 avg vel = x/ t = 0 More 1-D kinematics: More 1-D kinematics We saw that v = dx / dt In “calculus” language we would write dx = v dt, which we can integrate to obtain: Graphically, this is adding up lots of small rectangles: v(t) t + +...+ = displacement1-D Motion with constant acceleration: Calculus: Also recall that Since a is constant, we can integrate this using the above rule to find: Similarly, since we can integrate again to get: 1-D Motion with constant accelerationUseful Formula: Useful Formula Eliminating t: Solving for t: Simplifying:Alternatively - (Calculus): Alternatively - (Calculus) (chain rule) orEquations for Constant Acceleration: Equations for Constant Acceleration x = v0t + 1/2 at2 (parabolic) v = at (linear) v2 = v02 + 2a x (independent of time) Kinematics of long-range relationships: Kinematics of long-range relationships He makes a weekly round trip, in an airplane, to visit Her in California (about 5000 miles) Assume that he does not move around much otherwise His average velocity and average speed are zero We need to know how fast his airplane flies His average velocity and average speed are ~30 mph His average speed is about 30 mph Instantaneous speed is magnitude of instantaneous velocity Average speed is NOT magnitude of average velocity Rather it is total distance traveled divided by the elapsed timeWhat is his time-averaged position?: What is his time-averaged position? Near Madison, Wisconsin Near Stanford, California Near Denver, Colorado Near Toronto, CanadaQuestion : Question If the average velocity of a car during a trip along a straight road is positive, is it possible for the instantaneous velocity at some time during the trip to be negative? 1 - Yes 2 - No As long as the net distance traveled over the given time was positive, the average velocity will be positive- regardless of whether the car went in reverse at any point during that time. I could have forgotten something at home and had to turn around, but eventually I reached my destination away from my starting point. Question: If the velocity of some object is not zero, can its acceleration ever be zero ? 1 - Yes 2 - No Question if something is moving with constant velocity, it's acceleration equals zero Acceleration is a change in velocity. Therefore, if something has constant velocity, its acceleration is nil.Slide20: Is it possible for an object to have a positive velocity at the same time as it has a negative acceleration? 1 - Yes 2 - No Question if the car was moving in a positive direction and at the same time decreasing its speed then its velocity is positive and its acceleration is negative. positive velocity means you are moving forward. negative acceleration while the velocity is positive means you are slowing down. You can move forward and slow down at the same time.Slide21: An object undergoes a linearly increasing acceleration a=10 t m/s2 along a straight line. Initially, it starts at the origin x(0)=0 and has a velocity 2 m/s. How far does it travel from the origin in 4 seconds (origin is at t=0)? 1 - 328 m 2 - 221.3 m 3 - 114.7 m 4 - 98.0 m Question The problem states that acceleration is NOT constant. You do not have the permission to view this presentation. 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1DMotion Denise Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINTLite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 1086 Category: Travel/ Places.. License: All Rights Reserved Like it (3) Dislike it (0) Added: March 11, 2008 This Presentation is Public Favorites: 1 Presentation Description No description available. Comments Posting comment... By: phteach (29 month(s) ago) great presentation..would like to use some of it for a lesson in physics. can i download it? Saving..... Post Reply Close Saving..... Edit Comment Close By: madhavmantri (37 month(s) ago) its really a nice presentation.....can u allow it to download?? Saving..... Post Reply Close Saving..... Edit Comment Close Premium member Presentation Transcript Physics 201: Lecture 2: Physics 201: Lecture 2 Kinematics One dimensional motion Equations of motion for constant accelerationMotion in 1 dimension: Motion in 1 dimension Motion: change in position with time. In 1-D, we usually write position as x(t). Also called trajectory. Since it’s in 1-D, all we need to indicate direction is + or . Displacement in a time t = t2 - t1 is x = x(t2) - x(t1) = x2 - x1 t x t1 t2 x t x1 x2 some particle’s trajectory in 1-DAverage Velocity: Average Velocity t x t1 t2 x x1 x2 trajectory Velocity v is the “rate of change of position” Sign of v tells the direction the object is moving Magnitude of v is the speed Average velocity vav in the time t = t2 - t1 is: t Vav = slope of line connecting x1 and x2.Instantaneous Velocity: Consider the limit t1 t2 Instantaneous velocity v is defined as: Instantaneous Velocity t x t1 t2 x x1 x2 t so v(t2) = slope of line tangent to path at t2.Acceleration: Acceleration Acceleration a is the “rate of change of velocity” Average acceleration aav in the time t = t2 - t1 is: And instantaneous acceleration a is defined as: usingAcceleration: Acceleration Constant velocity Zero acceleration Constant acceleration in the same direction as v Increasing velocity Constant acceleration opposite of v Decreasing velocitySkydiver Jumps Out: Skydiver Jumps Out A skydiver is falling straight down, along the negative y direction. During the initial part of the fall, her speed increases from 16 to 28 m/s in 1.5 s. Which of the following is correct? 1) v>0, a>0 2) v>0, a<0 3) v<0, a>0 4) v<0, a<0Parachute Opens: Parachute Opens During a later part of the fall, after the parachute has opened, her speed decreases from 48 to 26 m/s in 11 s. Which of the following is correct? 1) v>0, a>0 2) v>0, a<0 3) v<0, a>0 4) v<0, a<0 If speed is increasing, v and a are in same direction. If speed is decreasing, v and a are in opposite direction.Question: Question When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path? (a) Both v = 0 and a = 0. (b) v 0, but a = 0. (c) v = 0, but a 0. Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero. Since the velocity is continually changing there must be some acceleration. In fact the acceleration is caused by gravity (g = 9.81 m/s2). correctSlide10: A ball is thrown straight up in the air and returns to its initial position. For the time the ball is in the air, which of the following statements is true? 1 - Both average acceleration and average velocity are zero. 2 - Average acceleration is zero but average velocity is not zero. 3 - Average velocity is zero but average acceleration is not zero. 4 - Neither average acceleration nor average velocity are zero. Question Free fall: acceleration is constant (-g) Initial position = final position: x=0 avg vel = x/ t = 0 More 1-D kinematics: More 1-D kinematics We saw that v = dx / dt In “calculus” language we would write dx = v dt, which we can integrate to obtain: Graphically, this is adding up lots of small rectangles: v(t) t + +...+ = displacement1-D Motion with constant acceleration: Calculus: Also recall that Since a is constant, we can integrate this using the above rule to find: Similarly, since we can integrate again to get: 1-D Motion with constant accelerationUseful Formula: Useful Formula Eliminating t: Solving for t: Simplifying:Alternatively - (Calculus): Alternatively - (Calculus) (chain rule) orEquations for Constant Acceleration: Equations for Constant Acceleration x = v0t + 1/2 at2 (parabolic) v = at (linear) v2 = v02 + 2a x (independent of time) Kinematics of long-range relationships: Kinematics of long-range relationships He makes a weekly round trip, in an airplane, to visit Her in California (about 5000 miles) Assume that he does not move around much otherwise His average velocity and average speed are zero We need to know how fast his airplane flies His average velocity and average speed are ~30 mph His average speed is about 30 mph Instantaneous speed is magnitude of instantaneous velocity Average speed is NOT magnitude of average velocity Rather it is total distance traveled divided by the elapsed timeWhat is his time-averaged position?: What is his time-averaged position? Near Madison, Wisconsin Near Stanford, California Near Denver, Colorado Near Toronto, CanadaQuestion : Question If the average velocity of a car during a trip along a straight road is positive, is it possible for the instantaneous velocity at some time during the trip to be negative? 1 - Yes 2 - No As long as the net distance traveled over the given time was positive, the average velocity will be positive- regardless of whether the car went in reverse at any point during that time. I could have forgotten something at home and had to turn around, but eventually I reached my destination away from my starting point. Question: If the velocity of some object is not zero, can its acceleration ever be zero ? 1 - Yes 2 - No Question if something is moving with constant velocity, it's acceleration equals zero Acceleration is a change in velocity. Therefore, if something has constant velocity, its acceleration is nil.Slide20: Is it possible for an object to have a positive velocity at the same time as it has a negative acceleration? 1 - Yes 2 - No Question if the car was moving in a positive direction and at the same time decreasing its speed then its velocity is positive and its acceleration is negative. positive velocity means you are moving forward. negative acceleration while the velocity is positive means you are slowing down. You can move forward and slow down at the same time.Slide21: An object undergoes a linearly increasing acceleration a=10 t m/s2 along a straight line. Initially, it starts at the origin x(0)=0 and has a velocity 2 m/s. How far does it travel from the origin in 4 seconds (origin is at t=0)? 1 - 328 m 2 - 221.3 m 3 - 114.7 m 4 - 98.0 m Question The problem states that acceleration is NOT constant.