1. Examples of Graph Interp. Ch 2B Notes (All velocities in m/s, all time in s)

2. 1. When is the greatest instantaneous velocity? From t = 3 to t = 5 Furthest from x axis TIME 8 6 4 2 0 -2 -4 -6 -8 2 4 6 8 10 VELOCITYVELOCITY

3. 2. When is the individual moving in the backwards direction? From t = 8 to t = 10 because v vs. t graph is below x- axis TIME 8 6 4 2 0 -2 -4 -6 -8 2 4 6 8 10 VELOCITYVELOCITY

4. 3. When is acceleration constant? t = 0 to t = 3 t = 3 to t = 5 t = 8 to t = 10 linear intervals have constant slope Constant slope = constant acceleration 2 4 6 8 10 VELOCITYVELOCITY TIME 8 6 4 2 0 -2 -4 -6 -8

5. 4. When is the instantaneous acceleration = 0? From t = 3 to t = 5 because m = 0 At around 6.2 s because slope of tangent line = 0 TIME 8 6 4 2 0 -2 -4 -6 -8 2 4 6 8 10 VELOCITYVELOCITY

6. 5. Find Ave Accel. from t = 5 to t = 8 s 2 4 6 8 10 VELOCITYVELOCITY TIME 8 6 4 2 0 -2 -4 -6 -8

7. 5. Find Ave Accel. from t = 5 to t = 8 s Find starting point and end point of interval on curve Find slope between these points 2 4 6 8 10 VELOCITYVELOCITY TIME 8 6 4 2 0 -2 -4 -6 -8 Point 1 (5,6) Point 2 (8,0) m = Δy / Δx = (y 2 – y 1 ) / (x 2 – x 1 ) = (0 - 6) / (8 - 5) = -6 / 3 = -2.0 a ave = -2.0 m/s 2

8. 6. Find Average Acceleration from t = 0 to t = 4 2 4 6 8 10 VELOCITYVELOCITY TIME 8 6 4 2 0 -2 -4 -6 -8

9. 6. Find Average Acceleration from t = 0 to t = 4 8 6 4 2 0 -2 -4 -6 -8 2 4 6 8 10 VELOCITYVELOCITY TIME m = Δy / Δx = (y 2 – y 1 ) / (x 2 – x 1 ) = (6-0) / (4-0) = 6/4 = 1.5 a ave = 1.5 m/s 2 Point 2 (4,6) Point 1 (0,0)

10. 7. Find Instantaneous Accel. at t = 7 s Sketch tangent line at point Find two points on tangent line Find slope 2 4 6 8 10 VELOCITYVELOCITY TIME Point 1 (4,6) 8 6 4 2 0 -2 -4 -6 -8 Point 2 (9.6,0) m = Δy / Δx = (y 2 – y 1 ) / (x 2 – x 1 ) = (0 - 6) / (9.6 - 4) = -6 / 5.6 = -1.07 m/s 2 a 7 = -1.07 m/s 2

11. 8. Find Displacement from t = 0 to t = 5 s Chop time increments into chunks Find area of each chunk Sum areas together 2 4 6 8 10 VELOCITYVELOCITY TIME 8 6 4 2 0 -2 -4 -6 -8 A Δ = 1/2bh = ½ (3)(6) = 9 m A □ = lw = (2)(6) = 12 m A total = 9 m + 12 m = 21 m

12. 9. Find Displacement from t = 6 to t = 10 s 2 4 6 8 10 VELOCITYVELOCITY TIME 8 6 4 2 0 -2 -4 -6 -8 Create best- estimate shapes of each chunk of area under curve Find area of each chunk (below x-axis is negative) Sum areas together A□ = A Δ1 + A Δ2 = ½(2)(4.4) + ½(2)(-8) = 4.4 m – 8.0 m = - 4.6 m

13. 10. Find Average Velocity from t = 3 to t = 6 2 4 6 8 10 DISPLACEMENTDISPLACEMENT TIME 20 15 10 5 0 -5 -10 -15 -20 1.Find point on curve at beginning and end of interval 2.Find slope between these points m = Δy/Δx = (y 2 -y 1 ) / (x 2 -x 1 ) m = (8-15) / (6-3) m = -7/3 = -2.33 m/s Point 1 (3s, 15m) Point 2 (6s, 8m)

14. 11. Find velocity at t = 6 s Sketch tangent line to curve at time desired Find two points along tangent line Find slope of tangent line using these new points 2 4 6 8 10 TIME 20 15 10 5 0 -5 -10 -15 -20 DISPLACEMENTDISPLACEMENT Point #1 (3s,19m) Point #2 (8s,0m) m = Δy / Δx = (y 2 – y 1 ) / (x 2 – x 1 ) = (0-19) / (8-3) = -19/5 = -3.8 m/s = v 6

15. 12. Find change in velocity from t = 5 to t = 10 s 2 4 6 8 10 ACCELERATIONACCELERATION TIME 8 6 4 2 0 -2 -4 -6 -8 Create geometric shapes that are estimates of the actual area beneath curve Find area of each chunk Sum areas together, watching signs A Δ1 = 1/2bh = ½ (3)(5) = 7.5 m/s A Δ2 = 1/2bh = 1/2(2)(-8) = -8 m/s A total = 7.5 m + (-8 m) = -0.5 m/s