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Kinematic Graphs

Kinematic Graphs. Graphically exploring derivatives and integrals. Slope. Y. (4,8). 8. (0,0). X. 4. Slope = rise =  Y = Y2 – Y1 = 8 – 0 = 8 = 2 run  X X2 – X1 4 – 0 4. Velocity is the slope of Displacement. Y. 8. (4,8). Displacement

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Kinematic Graphs

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  1. Kinematic Graphs Graphically exploring derivatives and integrals Dr. Sasho MacKenzie - HK 376

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

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

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

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

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

  10. Displacement Velocity = 0 Velocity > 0 Velocity < 0 Time Velocity from Displacement 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. Dr. Sasho MacKenzie - HK 376

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

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

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

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

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

  17. The area under the graph in these infinitely small time periods are summed together. Velocity Time Instantaneous Velocity Infinitely small time period Infinitely small time periods • D = Vdt = V1t1 + V2t2 + V3t3 + …… Dr. Sasho MacKenzie - HK 376

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

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

  20. Indicate which point(s) [letter(s)] on the acceleration graph below represent(s) the following: (at time = 0 s, velocity = 0) 1. Zero velocity 2. Zero acceleration 3. Max velocity 4. Min. velocity 5. Max acceleration 6. Min. acceleration E F D 2 Acceleration (m/s/s) G C 0 A I B H -1 1 s 2 s 3 s 0 s Time Dr. 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 velocity B. Zero acceleration C. Max velocity D. Min velocity E. Max acceleration F. Min acceleration G. Max displacement H. Min displacement 0 Dr. Sasho MacKenzie - HK 376

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