50 %
50 %
Information about 1DMotion

Published on March 11, 2008

Author: Denise

Source: authorstream.com

Physics 201: Lecture 2:  Physics 201: Lecture 2 Kinematics One dimensional motion Equations of motion for constant acceleration Motion 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-D Average 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: using Acceleration:  Acceleration Constant velocity  Zero acceleration Constant acceleration in the same direction as v  Increasing velocity Constant acceleration opposite of v  Decreasing velocity Skydiver 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<0 Parachute 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).  correct Slide10:  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 + +...+ = displacement 1-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 acceleration Useful Formula:  Useful Formula Eliminating t: Solving for t: Simplifying: Alternatively - (Calculus):  Alternatively - (Calculus) (chain rule) or Equations 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 time What is his time-averaged position?:  What is his time-averaged position? Near Madison, Wisconsin Near Stanford, California Near Denver, Colorado Near Toronto, Canada Question :  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.

Add a comment

Related presentations

Related pages

1DMotion - scribd.com

1DMotion - Download as PDF File (.pdf), Text File (.txt) or read online.
Read more

Practice Test 1 1dmotion - scribd.com

Practice Test 1 1dmotion - Download as PDF File (.pdf), Text File (.txt) or read online. AP exam
Read more

SparkNotes: 1D Motion: How to Cite This SparkNote

SparkNotes Editors. “SparkNote on 1D Motion.” SparkNotes LLC. n.d.. http://www.sparknotes.com/physics/kinematics/1dmotion/ (accessed September 29, 2016 ...
Read more

1-D Kinematics

1-D Kinematics. Lesson 1 - Describing Motion with Words; Introduction; Scalars and Vectors; Distance and Displacement; Speed and Velocity; Acceleration ...
Read more

SparkNotes: 1D Motion

From a general summary to chapter summaries to explanations of famous quotes, the SparkNotes 1D Motion Study Guide has everything you need to ace quizzes ...
Read more

Physlets - High Point University

Motion Diagrams (cart on a track) Describe the motion in words. Compare constant velocity and constant acceleration motion using ...
Read more

1DMotion - lhsscience's library

Share URL: Copy and paste this URL to share content with viewers. For example, paste into email or instant messages
Read more

Physlets - High Point, NC

Home | Courses | Matter and Interactions | Projects | Physlets | Download | Tools | Vita. Introduction. 1-D Motion-- free-fall. The animation below ...
Read more