1. A. Position, Distance, Displacement 1. Position along a number line - 0 + 2. Distance is total length traveled 3. Displacement x = x f - x i B. Average Speed distance traveled / elapsed time C. Average velocity = v x = x/ t Ch2.1 Position, Velocity, and Speed Chapter 2: Motion in One Dimension

2. This question is asking about displacement. What is the change in the distance x? Common misconceptions - Distance traveled is determined by giving the final position or the student incorrectly identifies the initial position as zero. Correct answer

3. Concept Question 2.1 You travel from your house to the grocery store, shop, bring the groceries to your friend’s house and return home. The distance you travel is? A. 0 mi B. 6.4 mi C. 8.5 mi D. 10.7 mi E. 12.8 mi

4. Concept Question 2.2 You travel from your house to the grocery store, shop, bring the groceries to your friend’s house and return home. Your displacement is? A. 0 mi B. 6.4 mi C. 8.5 mi D. 10.7 mi E. 12.8 mi

5. Ch2.1 Position, Velocity, and Speed Concept Question 2.3 Do QQ2.1 p. 22 with clickers Chapter 2: Motion in One Dimension

6. Ways to represent data Equation Table Graph

7. V x (from t = 2 to t = 4) = (16m – 4m) / (4s – 2s) = 6 m/s Slope = rise / run = 12 m / 2 s = 6 m/s Equation Table Graph

8. Concept Question 2.4 The average velocity between t = 1 and t = 2 is? A. -1 m/s B. 0.5 m/s C. 1 m/s D. -0.5 m/s E. 2 m/s

9. Ch2.2 Instantaneous Velocity Chapter 2: Motion in One Dimension A. Instantaneous velocity definition v x = lim x/t t 0 B. Graphical Interpretation V x is the slope of the line tangent to the x vs. t curve at the instant of time in question

10. Look at xt2inst.xls

11. P2.5 (p. 46)

12. A. B. C. D. Concept Question 2.5

13. Ch2.4 Acceleration Chapter 2: Motion in One Dimension A. Average acceleration a x = v/t B. Instantaneous Acceleration a x = lim v/t t 0 a x is the slope of the line tangent to the v x vs. t curve at the instant of time in question

14. Concept Question 2.6 The average acceleration between t = 0 and t = 20 is? A. 1 m/s 2 B. -1 m/s 2 C. 0.5 m/s 2 D. 2 m/s 2 E. 0.2 m/s 2

15. Fig. P2.17, p.51 P2.16 (p. 47)

16. Ch2.5 Motion Diagrams Moving Man Simulation Chapter 2: Motion in One Dimension

17. ABC DE F Concept Question 2.7

18. 2.6 The Particle Under Constant Acceleration Assuming the conditions: t i = 0, t f = t, x(0) = x i and v(0) = v xi and a x is constant.

19. 1.v xf = v xi + a x t 2.x f = x i + (v xi + v xf ) t / 2 3.x f = x i + v xi t + a x t 2 /2 4.v xf 2 = v xi 2 + 2a x (x f – x i ) 2.6 The Particle Under Constant Acceleration P2.20 (p. 47) P2.23 (p. 48)

20. Acceleration occurs when there is a change in velocity. The velocity is constant 1 m/s between 0 and 2 seconds and it is zero from 2 to 4 seconds. The only change is at t = 2 seconds. Common misconception - Student interprets sloping up (or down) on a position graph to mean the object is speeding up (or slowing down). Correct answer

21. Answer [b:b] is correct. Many picked answer [c:d], [c:c], or [c:b]. What is wrong with those choices?

22. Answer b is correct. Some picked answer c.

23. Equations for Constant Acceleration Only 1.v xf = v xi + a x t 2.x f = x i + (v xi + v xf ) t / 2 3.x f = x i + v xi t + a x t 2 /2 4.v xf 2 = v xi 2 + 2a x (x f – x i ) Assuming the conditions: t i = 0, t f = t, x(0) = x i and v(0) = v xi and a x is constant.

24. Strategy 1.Convert units if necessary. 2.Draw a picture that includes given information. 3.Pick coordinate origin. 4.List givens and finds. 5.Select equation based on results of 3 and solve. 6.Check if answer reasonable.

25. 1.v xf = v xi + a x t 2.x f = x i + (v xi + v xf ) t / 2 3.x f = x i + v xi t + a x t 2 /2 4.v xf 2 = v xi 2 + 2a x (x f – x i ) 2.6 The Particle Under Constant Acceleration P2.32 (p. 49)

26. 2.7 Freely Falling Objects

27. CT2.8: You are throwing a ball straight up in the air. At the highest point, the ball’s A.velocity and acceleration are zero. B.velocity is nonzero but its acceleration is zero. C.acceleration is nonzero, but its velocity is zero. D.velocity and acceleration are both nonzero.

28. The ball is continuing to lose velocity during the whole trip. When it turns around at the top, the velocity is momentarily zero, but the velocity is still decreasing as it becomes greater negative. Common misconception - Student concludes that if an object has zero speed, even for an instant, it also has zero acceleration. (This instant may appear at the starting point, ending point, or a turn around point.) Correct answer

29. 2.7 Freely Falling Objects A. Neglect air resistance B. Assume a spherically symmetrical earth C. Assume near Earth’s surface (within several hundred meters) Then the acceleration of gravity is g = 9.8 m/s 2 downward for all objects.

30. 1.v yf = v yi + a y t 2.y f = y i + (v yi + v yf ) t / 2 3.y f = y i + v yi t + a y t 2 /2 4.v yf 2 = v yi 2 + 2a y (y f – y i ) 2.7 Freely Falling Objects P2.41 (p. 49)

31. 19.6 m/s t = 4 s 19.6 m/s t = 0 s The symmetry of throwing a ball upward. 0 m/s t = 2 s

32. A woman is reported to have fallen 144 ft from the 17 th floor of a building, landing on a metal ventilator box, which she crushed to a depth of 18 in. She suffered only minor injuries. Neglecting air resistance, calculate (a) the speed of the woman just before she collided with the ventilator, (b) her average acceleration while in contact with the box, and (c) the time it took to crush the box.

33. Ct2.9: If you drop an object in the absence of air resistance, it accelerates downward at 9.8 m/s 2. If instead you throw it downward, its downward acceleration after release is A.less than 9.8 m/s 2. B.9.8 m/s 2. C.more than 9.8 m/s 2.

34. 1.v yf = v yi + a y t 2.y f = y i + (v yi + v yf ) t / 2 3.y f = y i + v yi t + a y t 2 /2 4.v yf 2 = v yi 2 + 2a y (y f – y i ) 2.7 Freely Falling Objects P2.57 (p. 51)

35. Ct2.10: A person standing at the edge of a cliff throws one ball straight up and another ball straight down at the same initial speed. Neglecting air resistance, the ball to hit the ground below the cliff with the greater speed is the one initially thrown A.upward. B.downward. C.neither—they both hit at the same speed.

36.  x = (v x )(  t) = area of triangle  x = (v x )(  t) = area of trapezoid

37. Statements Connecting Motion Variables that are Always True The instantaneous velocity is the slope of the tangent to the x vs. t curve. The instantaneous acceleration is the slope of the tangent to the v vs. t curve. The change in position is the area under the v vs. t curve. The change in velocity is the area under the a vs. t curve.