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1/29/03 Physics 103, Spring 2004, U. Wisconsin 1 Physics 103: Lecture 3 Position & Velocity with constant Acceleration l Today’s lecture will be on kinematic.

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Presentation on theme: "1/29/03 Physics 103, Spring 2004, U. Wisconsin 1 Physics 103: Lecture 3 Position & Velocity with constant Acceleration l Today’s lecture will be on kinematic."— Presentation transcript:

1 1/29/03 Physics 103, Spring 2004, U. Wisconsin 1 Physics 103: Lecture 3 Position & Velocity with constant Acceleration l Today’s lecture will be on kinematic equations ç1-d motion with constant acceleration çfree-fall

2 1/29/03 Physics 103, Spring 2004, U. Wisconsin 2 The slope of the curve in the position versus time graph for a particle’s motion is: 1. the particle’s instantaneous speed 2. the particle’s instantaneous acceleration 3. the particle’s average velocity 4. the particle’s instantaneous velocity Review Question?

3 1/29/03 Physics 103, Spring 2004, U. Wisconsin 3 A car’s position on a highway was plotted versus time. It turned out to be a straight line. Which of these statements is most definitely true? 1. Its acceleration is negative 2. Its acceleration is positive 3. Its acceleration is zero 4. Its velocity is zero Review Question?

4 1/29/03 Physics 103, Spring 2004, U. Wisconsin 4 Summary of Concepts (from last lecture) l kinematics: A description of motion l position: coordinates of a point l displacement:  x = change of position l velocity: rate of change of position çaverage :  x/  t çinstantaneous: slope of x vs. t l acceleration: rate of change of velocity çaverage:  v/  t çinstantaneous: slope of v vs. t

5 1/29/03 Physics 103, Spring 2004, U. Wisconsin 5 Velocity and Acceleration A car is moving along the negative x direction. During part of the trip, the 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 4) v<0, a<0  correct v a +x

6 1/29/03 Physics 103, Spring 2004, U. Wisconsin 6 Velocity and Acceleration During another part of the trip, still continuing along the negative x direction, the 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 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.  correct v a +x

7 1/29/03 Physics 103, Spring 2004, U. Wisconsin 7 Which of the following statements is most nearly correct? 1 - A car travels around a circular track with constant velocity. 2 - A car travels around a circular track with constant speed. 3- Both statements are equally correct. Lecture 3, Pre-Flight Q.5 correct Velocity is a vector! Direction changes when going around circle. Speed is the magnitude of velocity.

8 1/29/03 Physics 103, Spring 2004, U. Wisconsin 8 1D Kinematics Equations for Constant Acceleration Demo time!

9 1/29/03 Physics 103, Spring 2004, U. Wisconsin 9 Equations for Constant Acceleration  x = v 0 t + 1/2 at 2 (parabolic)  v = at (linear) v 2 = v 0 2 + 2a  x (independent of time)

10 1/29/03 Physics 103, Spring 2004, U. Wisconsin 10 Lecture 3, Pre-Flight Q.4 An object is dropped from rest. If it falls a distance D in time t then how far will if fall in a time 2t ? 1. D/4 2. D/2 3. D 4. 2D 5. 4D Correct x=1/2 at 2 Follow-up question: If the object has speed v at time t then what is the speed at time 2t ? 1. v/4 2. v/2 3. v 4. 2v 5. 4v Correct v=at

11 1/29/03 Physics 103, Spring 2004, U. Wisconsin 11 A ball is thrown downward (not dropped) from the top of a tower. After being released, its downward acceleration will be: 1. greater than g 2. exactly g 3. smaller than g Throwing Down

12 1/29/03 Physics 103, Spring 2004, U. Wisconsin 12Free-Fall l constant downward acceleration l g: acceleration due to gravity l same for all bodies: g=9.81 m/s 2 l a y = -g = -9.81 m/s 2 x y up down Summary of Free-Fall Equations y = y 0 + v 0y t - 1/2 gt 2 v y = v 0y - gt v y 2 = v 0y 2 - 2g  y

13 1/29/03 Physics 103, Spring 2004, U. Wisconsin 13 Lecture 3, Pre-Flight Q. 1 & 2 A ball is thrown vertically upward. At the very top of its trajectory, which of the following statements is true: 1. velocity is zero and acceleration is zero 2. velocity is not zero and acceleration is zero 3. velocity is zero and acceleration is not zero 4. velocity is not zero and acceleration is not zero correct Acceleration is the change in velocity. Just because the velocity is zero does not mean that it is not changing. At the top of the path, the velocity of the ball is zero,but the acceleration is not zero. The velocity at the top is changing, and the acceleration is the rate at which velocity changes. Acceleration is not zero since it is due to gravity and is always a downward-pointing vector.

14 1/29/03 Physics 103, Spring 2004, U. Wisconsin 14 Dennis and Carmen are standing on the edge of a cliff. Dennis throws a basketball vertically upward, and at the same time Carmen throws a basketball vertically downward with the same initial speed. You are standing below the cliff observing this strange behavior. Whose ball is moving fastest when it hits the ground? 1. Dennis' ball 2. Carmen's ball 3. Same v0v0v0v0 v0v0v0v0 Dennis Carmen H vAvAvAvA vBvBvBvB Lecture 3, Pre-Flight Q.3 Correct : v 2 = v 0 2 -2g  y On the dotted line:  y=0 ==> v 2 = v 0 2 v = ±v 0 When Dennis’s ball returns to dotted line its v = -v 0 Same as Carmen’s

15 1/29/03 Physics 103, Spring 2004, U. Wisconsin 15 Dennis and Carmen are standing on the edge of a cliff. Dennis throws a basketball vertically upward, and at the same time Carmen throws a basketball vertically downward with the same initial speed. You are standing below the cliff observing this strange behavior. Whose ball hits the ground at the base of the cliff first? 1. Dennis' ball 2. Carmen's ball 3. Same Lecture 3, Pre-Flight Q.3 Follow-up correct v0v0v0v0 v0v0v0v0 Dennis Carmen y=y 0 vAvAvAvA vBvBvBvB y=0 Time for Dennis’s ball to return to the dotted line: v = v 0 - g t v = -v 0 t = 2 v 0 / g This is the extra time taken by Dennis’s ball

16 1/29/03 Physics 103, Spring 2004, U. Wisconsin 16 Summary Equations with constant acceleration Free-fall  x = v 0 t + 1/2 at 2  v = at v 2 = v 0 2 + 2a  x  y = v 0y t + 1/2 at 2  v y = -g t v 2 y = v 0y 2 - 2 g  y


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