Phys 201 Fall 2009 Thursday, September 10, 2009 & Tuesday, September 15, Chapter 2: Mo>on in One Dimension

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1 Phys 201 Fall 2009 Thursday, September 10, 2009 & Tuesday, September 15, 2009 Chapter 2: Mo>on in One Dimension

2 The first exam will be 5:45pm 7pm on Tuesday, September 29 Please let us know as soon as possible (and certainly by 9/15) if (1) you are enrolled in another class that meets at that time, or (2) you have a McBurney VISA and need accommodation for a disability so that we can make the appropriate accommodations.

3 Displacement: change in posi>on 2008 by W.H. Freeman and Company

4 Average velocity and speed Average velocity = net displacement total time = s Δt Average speed = total distance total time = s Δt

5 Distance and displacement of a dog (p 29) total distance traveled = 35 ft net displacement = 5 ft 2008 by W.H. Freeman and Company

6 Average velocity and speed Average velocity = net displacement total time elapsed Average speed = total distance traveled total time elapsed Velocity includes direction. (It is fundamental for the laws of motion; speed is not.)

7 The average velocity for a >me interval (t 1, t 2 ) is the slope of the straight line connec>ng the points (t 1,x 1 ) and (t 2,x 2 ) on a graph of x versus t. The average velocity of particle between times t 1 and t 2 is greater than the average velocity of particle between times t 1 and t by W.H. Freeman and Company

8 Average velocity and average speed of a dog (p 31) Average velocity = (net displacement)/(total time) = (5 m)/(2.5 s) = 2 m/s Average speed = (total distance traveled)/(total time) = (35 m)/(2.5 s) = 14 m/s 2008 by W.H. Freeman and Company

9 The instantaneous velocity along x, v x, is the limit of the ra>o Δx/Δt as Δt approaches zero.

10 Approaching the limit Δt 0 to find the instantaneous velocity 2008 by W.H. Freeman and Company

11 Instantaneous velocity at a given >me is the deriva>ve of the x versus t curve by W.H. Freeman and Company

12 Displacement versus >me when x = 5t 2, with x in meters and t in seconds (p 34). velocity = derivative of displacement dx/dt = 10 t m/s by W.H. Freeman and Company

13 Accelera>on Average acceleration along x = net change in velocity along x total time = Δv x Δt Instantaneous acceleration along x: = derivative of v x -versus-t curve.

14 Instantaneous acceleration along x is first derivative of v x with respect to time t, and second derivative of x with respect to t:

15 A 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 m/s to 28 m/s in 1.5 s. Which of the following is correct? v a y 1. v>0, a>0 2. v>0, a<0 3. v<0, a>0 4. v<0, a<0

16 A 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 m/s to 28 m/s in 1.5 s. Which of the following is correct? v a y 1. v>0, a>0 2. v>0, a<0 3. v<0, a>0 4. v<0, a<0 correct a = Δv Δt = v v f i Δt = 28 m/ s ( 16 m/s) 1.5 s = 8 m/s 2 end 98

17 Parachute opens During a later part of the fall, after the parachute has opened, her speed decreases from 48 m/s to 26 m/s in 11 s. Which of the following is correct? v a y 1. v>0, a>0 2. v>0, a<0 3. v<0, a>0 4. v<0, a<0

18 Parachute opens During a later part of the fall, after the parachute has opened, her speed decreases from 48 m/s to 26 m/s in 11 s. Which of the following is correct? v a y 1. v>0, a>0 2. v>0, a<0 3. v<0, a>0 4. v<0, a<0 correct If speed is increasing, v and a are in same direction. If speed is decreasing, v and a are in opposite directions. a = Δv Δt = v f v i Δt = 26 m/ s ( 48 m/s) 11 s = 2 m/s 2

19 Special case: constant accelera>on If accelera>on is constant, velocity increases linearly with >me by W.H. Freeman and Company

20 Mo>on with constant accelera>on When acceleration is constant, the average velocity is the velocity halfway through the time interval (THIS IS NOT ALWAYS TRUE IF THE ACCELERATION IS NOT CONSTANT!) 2008 by W.H. Freeman and Company

21 Mo>on with constant accelera>on: find velocity and posi>on versus >me Consider a particle that at time t 0 has velocity along x of v x0 and is at position x 0. If the acceleration a x is constant, the velocity at time t is: v x (t) = v x0 + a x (t-t 0 ) and the particle s position at time t, x(t), is x(t) = x 0 + v x0 (t-t 0 ) + (1/2)a x (t-t 0 ) 2.

22 Mo>on with constant accelera>on: find velocity versus posi>on Recall velocity is: v x (t) = v x0 + a x (t-t 0 ) t-t 0 = (v x (t)-v x0 )/a x. Since the particle s position at time t, x(t), is x(t) = x 0 +v x0 (t-t 0 )+(1/2)a x (t-t 0 ) 2, this means x(t) = x 0 + v x0 (v x (t)-v x0 )/a x + (1/2)a x ((v x (t)-v x0 )/a x ) 2 x(t)-x 0 = v x0 (v x (t)-v x0 )/a x + (1/2)(v x (t)-v x0 ) 2 /a x. Multiply both sides of equation by 2a x and simplifying yields 2a x (x(t)-x 0 ) = v x (t) 2 - v x0 2.

23 Free fall: If no air, all objects in free fall with same ini>al velocity move iden>cally. acceleration due to gravity = g 9.81 m/s 2 32 ft/s 2, directed downwards (same for all objects) The acceleration results in a widening of the spaces between images by W.H. Freeman and Company gravity demos

24 What is x f? (Ex. 2 10, stopping distance) 2008 by W.H. Freeman and Company

25 What is x f? (Ex. 2 10, stopping distance) Let Δt be the time It takes the car to come to a stop: (Δt = Δv/a x = (-15 m/s/(-5.0 m/s 2 )). Distance traveled in 3.0 s = v 0 Δt + (1/2) a x (Δt) by W.H. Freeman and Company = (15 m/s)(3.0 s) + (1/2) (-5.0 m/s 2 )(3.0 s) 2 = 45 m m = 23 m (since we are keeping 2 significant figures)

26 Example 2 13, The Flying Cap (p 44) A cap is tossed in the air. It starts at height y 0 with a velocity v 0y =14.7 m/s upward. How long does it take for the cap to reach its highest point? How high does the cap go? 2008 by W.H. Freeman and Company

27 Example 2 13, The Flying Cap (p 44) A cap is tossed in the air. It starts at height y 0 with a velocity v 0y =14.7 m/s upward. How long does it take for the cap to reach its highest point? Highest point is when the velocity is zero. v y (t) = v 0y - a y t f t f = (v y (t)-v 0y )/a y = (14.7 m/s)/(-9.81 m/s2) = 1.50 s How high does the cap go? The cap goes distance v 0y t f + (1/2)a y t 2 f = (14.7 m/s)(1.50 s) + (1/2)(-9.81 m/s 2 )(1.50 s) 2 = 11.0 m by W.H. Freeman and Company

28 Example 2 13; Height and velocity of the flying cap 2008 by W.H. Freeman and Company

29 Geang velocity from accelera>on by integra>on (The only tricky part is putting in the limits of integration.)

30 Geang posi>on from velocity by integra>on (The only tricky part is putting in the limits of integration.)

31 The position is the integral of the velocity (the limit of the sum of areas of the rectangles) by W.H. Freeman and Company

32 The displacement is given by the area under the v x versus t curve 2008 by W.H. Freeman and Company

33 Problem 2.31 Reginald and Josie go the same distance. What is rela>onship between v J and v max? v J 2008 by W.H. Freeman and Company

34 Problem 2.31 Reginald and Josie go the same distance. What is rela>onship between v J and v max? v J Distance traveled is area under v-versus-t curve by W.H. Freeman and Company For Josie, this area is v J t f. For Reginald, this area is t max v max /2 + (t f -t max )v max /2. The areas are equal when v max =2v J.

35 Problem write algebraic expressions for x(t), v x (t), and a x (t) 2008 by W.H. Freeman and Company

36 Problem write algebraic expressions for x(t), v x (t), and a x (t) Reading from graph, v x (t) = v 0 + At, with v 0 =50 m/s and A=-10 m/s 2. Acceleration is derivative of velocity: a x (t)=a=-10m/s 2. Position is integral of velocity: x(t) = x(0)+v 0 t + ½At 2 = 50t - 5t by W.H. Freeman and Company

37 Mo>on with >me varying accelera>on An object undergoes a linearly increasing acceleration a(t) = 10 t m/s 2 along a straight line. It starts at the origin x(0)=0 with a velocity of 2 m/s. Where is its position at time t=4 seconds?

38 Mo>on with >me varying accelera>on An object undergoes a linearly increasing acceleration a(t) = 10 t m/s 2 along a straight line. It starts at the origin x(0)=0 with a velocity of 2 m/s. What is its position at time t=4 seconds? In this problem the acceleration is NOT constant. So must do the integrals explicitly. a(t) = 10t v(t) = a(t)dt = 10 t2 + v 0 = 5t 2 + v 0 2 x(t) = v(t)dt = 5 t3 + v 0 t + x 0 3 x(4) = = 114.7m

39 Which graph of v versus t best describes the mo>on of a par>cle with posi>ve velocity and nega>ve accelera>on?

40 Which graph of v versus t best describes the mo>on of a par>cle with posi>ve velocity and nega>ve accelera>on?

41 A car accelerates uniformly from a velocity of 10 km/h to 30 km/h in one minute. Which graph best describes the motion of the car?

42 A car accelerates uniformly from a velocity of 10 km/h to 30 km/h in one minute. Which graph best describes the motion of the car?

43 The accelera>on of a vehicle is given by a(t) = At where A is a constant. Its velocity as a func>on of >me is (v o is a constant) correct. is the above answers of None E. 3 ) ( D. ) ( C. ) ( B. 2 ) ( A o o o o v t A t v v At t v v At t v v t A t v + = + = + = + =

44 The accelera>on of a vehicle is given by a(t) = At where A is a constant. Its velocity as a func>on of >me is (v o is a constant) correct. is the above answers of None E. 3 ) ( D. ) ( C. ) ( B. 2 ) ( A o o o o v t A t v v At t v v At t v v t A t v + = + = + = + =

45 Concept ques>on: velocity and accelera>on A ball is thrown vertically upward. At the very top of its trajectory, which of the following statements is true: a. velocity is zero and acceleration is zero b. velocity is not zero and acceleration is zero c. velocity is zero and acceleration is not zero

46 Concept ques>on: velocity and accelera>on A ball is thrown vertically upward. At the very top of its trajectory, which of the following statements is true: a. velocity is zero and acceleration is zero b. velocity is not zero and acceleration is zero c. velocity is zero and acceleration is not zero correct At the top of the path, the velocity of the ball is zero, but the accelera>on is not zero. The velocity at the top is changing, and the accelera>on is the rate at which velocity changes. Accelera>on is the change in velocity. Just because the velocity is zero does not mean that it is not changing. Accelera>on is not zero since it is due to gravity and is always a downward poin>ng vector.

stopping distance : Physics 201, Spring 2011 Lect. 4 Example 2 13, The Flying Cap (p 44) Example 2 13, The Flying Cap (p 44) 1/19/11

stopping distance : Physics 201, Spring 2011 Lect. 4 Example 2 13, The Flying Cap (p 44) Example 2 13, The Flying Cap (p 44) 1/19/11 Physics 201, Spring 2011 Lect. 4 Chapter 2: Mo;on in One Dimension (examples and applica;ons) stopping distance : You were driving on a country road at an instantaneous velocity of 55 mph, east. You suddenly