Presentation is loading. Please wait.

Presentation is loading. Please wait.

Physics 101: Lecture 6, Pg 1 Lecture 5: Introduction to Physics PHY101 Chapter 2: è Equations of Kinematics for Constant Acceleration in 1 Dim. (2.4, 2.5,

Similar presentations


Presentation on theme: "Physics 101: Lecture 6, Pg 1 Lecture 5: Introduction to Physics PHY101 Chapter 2: è Equations of Kinematics for Constant Acceleration in 1 Dim. (2.4, 2.5,"— Presentation transcript:

1 Physics 101: Lecture 6, Pg 1 Lecture 5: Introduction to Physics PHY101 Chapter 2: è Equations of Kinematics for Constant Acceleration in 1 Dim. (2.4, 2.5, 2.7) è Free Fall (2.6) Chapter 3: è Equations of Kinematics for Constant Acceleration in 2 Dim. (3.1, 3.2)

2 Physics 101: Lecture 6, Pg 2 Summary of concepts from last lecture l position: your coordinates (just x in 1-D) l displacement: x = change of position l velocity: rate of change of position è average : x/ t è instantaneous: slope of x vs. t : lim t->0 x/ t l acceleration: rate of change of velocity è average: v/ t è instantaneous: slope of v vs. t : lim t->0 v/ t

3 Physics 101: Lecture 6, Pg 3 Concept Question l 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 During another part of the trip, 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 correct v a +x

4 Physics 101: Lecture 6, Pg 4 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. Concept Question correct On a circular track, the direction that the car is traveling in is always changing, and since velocity takes into account the direction of travel, the velocity is always changing. Speed, however, is independent of direction and so the speed can stay constant.

5 Physics 101: Lecture 6, Pg 5 Kinematics in One Dimension Constant Acceleration Consider an object which moves from the initial position x 0, at time t 0 with velocity v 0, with constant acceleration along a straight line. How does displacement and velocity of this object change with time ? a av =a = (v-v 0 ) / (t-t 0 ) => v(t) = v 0 + a (t-t 0 ) (1) v av = (x-x 0 ) / (t-t 0 ) = (v+v 0 )/2 => x = x 0 + (t-t 0 ) (v+v 0 )/2 (2) Use Eq. (1) to replace v in Eq.(2): x(t) = x 0 + (t-t 0 ) v 0 + a/2 (t-t 0 ) 2 (3) Use Eq. (1) to replace (t-t 0 ) in Eq.(2): v 2 = v 0 2 + 2 a (x-x 0 ) (4)

6 Physics 101: Lecture 6, Pg 6 Application of Eqs. of Kinematics l A runner accelerates to a velocity of 5.36 m/s due west in 3.00 s. His average velocity is 0.640 m/s 2 due west. What was his velocity when he began accelerating ? [Chapter 2, problem #15] t 0 = 0 s, v= -5.36 m/s, t=3.00 s, a av =-0.640 m/s 2 v 0 = ? m/s a av = (v-v 0 )/(t-t 0 ) => v 0 = v- a av (t-t 0 ) = -3.44 m/s v 0 = 3.44 m/s due west

7 Physics 101: Lecture 6, Pg 7 Application of Eqs. of Kinematics l A drag racer starting from rest, speeds up for 402 m with a=+17 m/s2. A parachute then opens, slowing the car down with a=-6.10 m/s2. How fast is the racer after moving 3.50 x 10 2 m after the parachute opens ? [2-28] 1. Before the parachute opens (car moves +x direction): t 0 = 0 s, v 01 = 0 m/s, x 1 =+402 m, a 1 =+17 m/s 2 2. After the parachute opens: t 0 = 0 s, x 2 =+3.50 x 10 2 m, a 2 =-6.10 m/s 2, v=? m/s v 2 =v 02 2 +2 a 2 x 2 Get v 02 2 from 1.: v 02 =(2 a 1 x 1 ) 1/2 =+117 m/s => v 2 =(v 02 +2 a 2 x 2 ) 1/2 =+96.9 m/s

8 Physics 101: Lecture 6, Pg 8 Free Fall l Free fall is the idealized description of the motion of a downward falling body due to gravity: Air resistance is neglected Acceleration due to gravity is considered to be constant The acceleration due to gravity is always pointing downward with magnitude g=9.80 m/s 2.

9 Physics 101: Lecture 6, Pg 9 Concept Question An object is dropped from rest. If it falls a distance D in time t then how far will it fall in a time 2t ? 1. D/4 2. D/2 3. D 4. 2D 5. 4D Correct x = 1 / 2 at 2 Followup 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

10 Physics 101: Lecture 6, Pg 10 Concept Question 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 The velocity vector changes from moment to moment, buts its acceleration vector does not change. Though the velocity at the top is zero, the acceleration is still constant because the velocity is changing.

11 Physics 101: Lecture 6, Pg 11 Free Fall l A wrecking ball is hanging from rest from a crane when suddenly the cable breaks. The time it takes the ball to fly half way to the ground is 1.2 s. Find the time for the ball to fall from rest all the way to the ground. [2-45] 1. Half way to the ground (-y direction) t 0 = 0 s, v 0 = 0 m/s, t=1.2 s, a=-9.80 m/s 2 Y 1/2 =v 0 t + ½ a t 2 = -7.1 m 2. From rest all the way to the ground, y=2 Y 1/2 t 0 = 0 s, v 0 = 0 m/s, a=-9.80 m/s 2, t= ? s Y=v 0 t + ½ a t 2 = ½ a t 2 => t= (2 y/a) 1/2 =1.7 s

12 Physics 101: Lecture 6, Pg 12 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 Concept Question Correct : v 2 = v 0 2 -2g y

13 Physics 101: Lecture 6, Pg 13 Kinematics in Two Dimensions Constant Acceleration Consider an object which moves in the (x,y) plane from the initial position r 0, at time t 0 with velocity v 0, with constant acceleration. l position: your coordinates (just r=(x,y) in 2-D) l displacement: r = r-r0 change of position l velocity: rate of change of position è average : r/ t è instantaneous: lim t->0 r/ t l acceleration: rate of change of velocity è average: v/ t è instantaneous: lim t->0 v/ t Same concepts as in one dimension ! Equations of kinematics are derived for the x and y components separately. Same equations as in one dimension !


Download ppt "Physics 101: Lecture 6, Pg 1 Lecture 5: Introduction to Physics PHY101 Chapter 2: è Equations of Kinematics for Constant Acceleration in 1 Dim. (2.4, 2.5,"

Similar presentations


Ads by Google