Examples of Graph Interp.

Ch 2B Notes

(All velocities in m/s, all time in s)

1. When is the greatest instantaneous velocity?

• From t = 3 to t = 5

• Furthest from x axis

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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

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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

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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

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5. Find Ave Accel. from t = 5 to t = 8 s

2 4 6 8 10

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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

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Point 1 (5,6)

Point 2 (8,0)

m = Δy / Δx = (y2 – y1) /

(x2 – x1)= (0 - 6) / (8 - 5)= -6 / 3 = -2.0aave = -2.0 m/s2

6. Find Average Acceleration from t = 0 to t = 4

2 4 6 8 10

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6. Find Average Acceleration from t = 0 to t = 4

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m = Δy / Δx

= (y2 – y1) /

(x2 – x1)

= (6-0) / (4-0)

= 6/4 = 1.5

aave = 1.5 m/s2

Point 2 (4,6)

Point 1 (0,0)

7. Find Instantaneous Accel. at t = 7 s

• Sketch tangent line at point

• Find two points on tangent line

• Find slope

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Point 1 (4,6)8

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Point 2 (9.6,0)

m = Δy / Δx = (y2 – y1) /

(x2 – x1)= (0 - 6) / (9.6 - 4)= -6 / 5.6 = -1.07 m/s2

a7 = -1.07 m/s2

8. Find Displacement from t = 0 to t = 5 s

• Chop time increments into chunks

• Find area of each chunk

• Sum areas together

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AΔ = 1/2bh = ½ (3)(6) = 9 m

A□ = lw = (2)(6) = 12 m

Atotal = 9 m + 12 m = 21 m

9. Find Displacement from t = 6 to t = 10 s

2 4 6 8 10

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•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□ = AΔ1 + AΔ2

= ½(2)(4.4) + ½(2)(-8)

= 4.4 m – 8.0 m =

- 4.6 m

10. Find Average Velocity from t = 3 to t = 6

2 4 6 8 10

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1.Find point on curve at beginning and end of interval

2.Find slope between these pointsm = Δy/Δx

= (y2-y1) / (x2-x1)

m = (8-15) / (6-3)

m = -7/3 = -2.33 m/s

Point 1

(3s, 15m) Point 2

(6s, 8m)

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

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Point #1

(3s,19m)

Point #2

(8s,0m)

m = Δy / Δx = (y2 – y1) / (x2 – x1)= (0-19) / (8-3) = -19/5= -3.8 m/s = v6

12. Find change in velocity from t = 5 to t = 10 s

2 4 6 8 10

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•Create geometric shapes that are estimates of the actual area beneath curve•Find area of each chunk•Sum areas together, watching signs

AΔ1 = 1/2bh = ½ (3)(5) = 7.5 m/sAΔ2 = 1/2bh = 1/2(2)(-8) = -8 m/sAtotal = 7.5 m + (-8 m) = -0.5 m/s