1. Kinematics Graphical Analysis of Motion

2. Goal 2: Build an understanding of linear motion. Objectives – Be able to: 2.04 Using graphical and mathematical tools, design and conduct investigations of linear motion and the relationships among:  Position.  Average velocity.  Instantaneous velocity.  Acceleration.  Time.

3. Graphical Analysis of Motion Three questions can be asked for motion: How far? How fast? How quickly did “how fast” change?

4. Graphical Analysis of Motion To answer these three questions, there three methods of graphical analysis:  INSPECTION  SLOPE  AREA UNDER THE CURVE

5. Graphical Analysis of Motion Inspection Inspection means using ordered pairs … and the definitions of displacement, velocity, acceleration, and time to analyze motion … On x-t, v-t, and a-t graphs.

6. Graphical Analysis of Motion Inspection Given the x-t graph, describe the object’s 5 kinds of motion.

7. Graphical Analysis of Motion Inspection Given the x-t graph, describe the object’s 5 kinds of motion. At t = 0, the object was at the origin. From t = 0-5 s, the object moved with uniform motion away from the origin.

8. Graphical Analysis of Motion Inspection Given the x-t graph, describe the object’s 5 kinds of motion: From t = 5-10 s, the object was at rest at x = +30 m.

9. Graphical Analysis of Motion Inspection Given the x-t graph, describe the object’s 5 kinds of motion: From t = 10-20 s, the object moved toward the origin with uniform motion.

10. Graphical Analysis of Motion Inspection Given the x-t graph, describe the object’s 5 kinds of motion: From t = 20-25 s, the object was at rest at x = +10 m.

11. Graphical Analysis of Motion Inspection Given the x-t graph, describe the object’s 5 kinds of motion: From t = 20-25 s, the object was at rest at x = +10 m.

12. Graphical Analysis of Motion Inspection Given the x-t graph, describe the object’s 5 kinds of motion: From t = 25-30 s, the object moved away from the origin at constant speed.

13. Graphical Analysis of Motion Inspection What was the object’s position at t = 15 s? Answer: x = +20 m When was the object at x = +25 m? Answer t = 4 s and … t = 12.5 s

14. Graphical Analysis of Motion SLOPE Velocity of the object can be determined from the graph’s data using slope: What is the velocity from t = 0 - 5 s? Δx (+30 – 0) m v = ---- = ---------------- = +6 m/s Δt (5 – 0) s Important fact: The ratio that defines average velocity also defines the slope of the x-t graph.

15. Graphical Analysis of Motion SLOPE Velocity of the object can be determined from the graph’s data using slope: What is the velocity from t = 5 - 10 s? Δx (+30 – +30) m v = ---- = ------------------- = +0 m/s Δt (10 – 5) s What is the velocity from t = 10 - 15 s? Δx (+20 – +30) m v = ---- = -------------------- = -2 m/s Δt (15 – 10) s

16. Graphical Analysis of Motion SLOPE Velocity of the object can be determined from the graph’s data using slope: What is the velocity from t = 0-30 s? Δx (+20 – 0) m v = ---- = ------------------- Δt (30 – 0) s v = +.67 m/s

17. Graphical Analysis of Motion Inspection Now consider an object moving in a straight line on the y-axis in such a way that its velocity v as a function of time is given on the following v-t graph: What does the graph tell us? At time t = 0, the object’s velocity was 0. Over the first two seconds, its velocity increased steadily to +2 m/s. From t = 2 – 8 s, the object maintained a constant speed of +2 m/s.

18. Graphical Analysis of Motion Inspection Now consider an object moving in a straight line on the y-axis in such a way that its velocity v as a function of time is given on the following v-t graph: What does the graph tell us? From t = 8 – 10 s, the object slowed steadily to a stop. From 10 – 14 s, the object was at rest. From 14 – 16 s, the object steaadily inceased speed in the negative (opposite, or backward) direction.

19. Graphical Analysis of Motion Inspection Now consider an object moving in a straight line on the y-axis in such a way that its velocity v as a function of time is given on the following v-t graph: What does the graph tell us? From 16 – 18 s, the object maintained a constant speed (uniform motion) of -2 m/s (away from the origin). From 18 – 20 s, the object steadily decreased speed and came to a halt.

20. Graphical Analysis of Motion SLOPE What can we ask about this motion? First, the fact that the velocity changed from t = 0 to t = 2 s tells us that the object accelerated. The acceleration for this time was: Δv (+2 – 0) m/s a = ---- = ---------------- = +1 m/s 2 Δt (2 – 0) s Therefore, the slope of a velocity time graph gives the acceleration.

21. Graphical Analysis of Motion SLOPE What was the acceleration from time t = 2 s to t = 8 s? Δv (+2 – +2) m/s a = ---- = ---------------- = 0 m/s 2 Δt (8 – 2) s What was the acceleration from time t = 8 s to t = 10 s? Δv (0 – +2) m/s a = ---- = ---------------- = -1 m/s 2 Δt (10 – 8) s

22. Graphical Analysis of Motion Another question can be asked when velocity-time graph is given: How far did the object travel for a given time interval? Let’s figure out the displacement of the object from time t = 2 s to time t = 8 s. During this interval, the velocity was a constant +2 m/s, so the displacement was Δy = vΔt = (+2 m/s)(6 s) = +12 m. 12 m

23. Graphical Analysis of Motion AREA UNDER THE CURVE Geometrically, we’ve determined the area between the graph and the horizontal axis. After all, the area of a rectangle is base X height and, for the shaded rectangle shown, the base is Δt, and the height is v. So base x height equals Δt X v, which is displacment. 12 m

24. Graphical Analysis of Motion AREA UNDER THE CURVE Given a velocity- time graph, the area between the graph and the t axis equals the object’s displacement. 12 m

25. Graphical Analysis of Motion AREA UNDER THE CURVE What was the displacement of the object from time t = 0 s to time t = 2 s? During this interval, the velocity was accelerating, and covering larger distances with each time interval. This shape is a triangle. Area Δ = (½)(base)(height) = (½)(2 s)(+2 m/s) = +2 m.

26. Graphical Analysis of Motion AREA UNDER THE CURVE Therefore, what is the total displacement from t = 0 – 8 s? Answer: Add the triangle and rectangle areas: +2 m + (12 m) = +14 m 14 m

27. Graphical Analysis of Motion AREA UNDER THE CURVE What was the displacement of the object from time t = 14 s to time t = 16 s? This triangle area is the same area as t = 0 – 2 s, but with negative sign … -2 m. The minus sign indicates negative displacement … Remember, the object was speeding up in the negative direction!

28. Graphical Analysis of Motion AREA UNDER THE CURVE On a v-t graph, Regions below the horizontal axis are negative displacements (since the object’s velocity is negative, its displacement must be negative). Regions above the horizontal axis are positive displacements.

29. Graphical Analysis of Motion Three questions can be asked for motion: How far? How fast? How quickly did “how fast” change?

30. Graphical Analysis of Motion To answer these three questions, there three methods of graphical analysis:  INSPECTION  SLOPE  AREA UNDER THE CURVE

31. Goal 2: Build an understanding of linear motion. Objectives – Be able to: 2.04 Using graphical and mathematical tools, design and conduct investigations of linear motion and the relationships among:  Position.  Average velocity.  Instantaneous velocity.  Acceleration.  Time.