1. 11/10/2015Dr. Sasho MacKenzie - HK 376 1 Kinematic Graphs Graphically exploring derivatives and integrals

2. 11/10/2015Dr. Sasho MacKenzie - HK 376 2 Slope X Y (0,0) (4,8) 8 4 Slope = rise =  Y = Y2 – Y1 = 8 – 0 = 8 = 2 run  X X2 – X1 4 – 0 4

3. 11/10/2015Dr. Sasho MacKenzie - HK 376 3 Velocity is the slope of Displacement X Y (0,0) (4,8) 8 4 Average Velocity = rise =  D = D2 – D1 = 8 – 0 = 8 m = 2 m/s run  t t2 – t1 4 – 0 4 s Displacement (m) Time (s)

4. 11/10/2015Dr. Sasho MacKenzie - HK 376 4 1.The displacement graph on the previous slide was a straight line, therefore the slope was 2 at every instant. 2.Which means the velocity at any instant is equal to the average velocity. 3.However if the graph was not straight the instantaneous velocity could not be determined from the average velocity

5. 11/10/2015Dr. Sasho MacKenzie - HK 376 5 Instantaneous Velocity The average velocity over an infinitely small time period. Determined using Calculus The derivative of displacement The slope of the displacement curve

6. 11/10/2015Dr. Sasho MacKenzie - HK 376 6 Instantaneous Acceleration The average acceleration over an infinitely small time period. Determined using Calculus The derivative of velocity The slope of the velocity curve

7. 11/10/2015Dr. Sasho MacKenzie - HK 376 7 Average vs. Instantaneous Average Velocity = rise =  D = D2 – D1 = 8 – 0 = 8 m = 2 m/s run  t t2 – t1 4 – 0 4 s X Y 8 4 Displacement (m) Time (s) (0,0) (4,8) The average velocity does not accurately represent slope at this particular point.

8. 11/10/2015Dr. Sasho MacKenzie - HK 376 8 Derivative The slope of the graph at a single point. Slope of the line tangent to the curve. The limiting value of  D/  t as  t approaches zero.

9. 11/10/2015Dr. Sasho MacKenzie - HK 376 9 Infinitely small time period (dt) dtdt Tangent line Instantaneous Velocity Displacement Time t1t1 t2t2 t3t3

10. 11/10/2015Dr. Sasho MacKenzie - HK 376 10 Velocity from Displacement Displacement Time Velocity > 0 Velocity = 0 Velocity < 0 The graph below shows the vertical displacement of a golf ball starting immediately after it bounces off the floor and ending when it lands again.

11. 11/10/2015Dr. Sasho MacKenzie - HK 376 11 Graph Sketching Differentiation 0 + 0 Displacement Velocity Velocity Acceleration OR Slope 0 0

12. 11/10/2015Dr. Sasho MacKenzie - HK 376 12 Displacement Velocity Velocity Acceleration OR 0 0 0 + 0 0 _ Slope Graph Sketching Differentiation

13. 11/10/2015Dr. Sasho MacKenzie - HK 376 13 Going the other way: area under the curve Area under curve = Height x Base =  Y x  X = 4 x 2 = 8 Y X 4 2 (0,0) (2,4)

14. 11/10/2015Dr. Sasho MacKenzie - HK 376 14 Displacement is the Area Under the Velocity Curve Displacement = V x  t = 4 x 2 = 8 m Velocity (m/s) Time (s) Y X 4 2 (0,0) (2,4)

15. 11/10/2015Dr. Sasho MacKenzie - HK 376 15 What if velocity isn’t a straight line? This would be an over estimate of the area under the curve (2,4) 4 2 (0,0) Velocity Time

16. 11/10/2015Dr. Sasho MacKenzie - HK 376 16 Integration Finding the area under a curve. Uses infinitely small time periods. All the areas under the infinitely small time periods are then summed together. D =  Vdt =  V  t

17. 11/10/2015Dr. Sasho MacKenzie - HK 376 17 Infinitely small time periods Velocity Time D =  Vdt = V 1  t 1 + V 2  t 2 + V 3  t 3 + …… Instantaneous Velocity Infinitely small time period The area under the graph in these infinitely small time periods are summed together.

18. 11/10/2015Dr. Sasho MacKenzie - HK 376 18 Graph Sketching Integration Displacement Velocity Velocity Acceleration OR 0 2 0 Constant slope of 2 tt tt tt Area under curve increases by the same amount for each successive time period (linear increase).

19. 11/10/2015Dr. Sasho MacKenzie - HK 376 19 Graph Sketching Integration Displacement Velocity Velocity Acceleration OR 0 tt tt tt Area under curve increases by a greater amount for each successive time period (exponential increase). 0 Exponential curve

20. 11/10/2015Dr. Sasho MacKenzie - HK 376 20 Indicate on the acceleration graph below the location of the following point(s). Place the letter on the graph. A. Zero velocityB. Zero accelerationC. Max velocity D. Min velocityE. Max accelerationF. Min acceleration 0 1 s2 s3 s 2

21. 11/10/2015Dr. Sasho MacKenzie - HK 376 21 Indicate on the velocity graph below the location of the following point(s). Place the letter on the graph. A. Zero velocityB. Zero accelerationC. Max velocity D. Min velocityE. Max accelerationF. Min acceleration G. Max displacementH. Min displacement 0