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Motion in One Dimension 2.2

Motion in One Dimension 2.2. Ch 2.2. Acceleration. Avg acceleration = change in velocity time needed for change a = v = v f – v i t t f – t i. Acceleration. variable unit Acceleration a m/s 2

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Motion in One Dimension 2.2

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  1. Motion in One Dimension 2.2 Ch 2.2

  2. Acceleration • Avg acceleration = change in velocity time needed for change a = v = vf – vi t tf – ti

  3. Acceleration variableunit Acceleration am/s2 Velocity vm/s (final or initial) Time ts

  4. Acceleration is a vector • Acceleration has direction and magnitude • Velocity time graph has a slope of acceleration • See figure 10 on page 50 • Check out Table 3 on page 51 • Conceptual Challenge #2 pg 50

  5. Acceleration A car is traveling at 4 m/s and accelerates to 20 m/s in 4 seconds. What is the acceleration? A = 20m/s – 4 m/s = + 4 m/s2 4 s

  6. Acceleration Time 0 s 1s 2s 3s 4s Accel +4 +4 +4 +4 Velocity 4 8 12 16 What would the graph look like?

  7. Deceleration A car is traveling at 10 m/s and accelerates to 0 m/s in 5 seconds. What is the acceleration? A = 0m/s – 10 m/s = - 2 m/s2 5 s Deceleration- negative acceleration (slowing down) Practice: page 49 1-5 Practice: graphing sheet

  8. Acceleration Time 0 s 1s 2s 3s 4s 5s Accel -2 -2 -2 -2 -2 Velocity 10m/s 8 6 4 2 0 What would the graph look like?

  9. Fundamental Equations V (avg) = 1/2 (vf + Vi) (Used mainly for deriving) V = Dx = xf - xi Dt Dt a = vf – virearranged vf = vi + at t

  10. Derived Equations • Derived equations - equations formed from other equations • Dx = ½ (vi + vf) t • Dx = vi t + ½ at2 • Vf2 = vi2 + 2a Dx

  11. DerivingDx = ½ (vi + vf) t Put V = Dx and V = 1/2 (vf + Vi) Dt together Dx = VDt usesubstitution Dx = 1/2 (vf + Vi)Dt

  12. DerivingDx = vi t + ½ at2 Put Dx = 1/2 (vf + Vi) Dtand vf = vi + a Dttogether Dx = 1/2 (vf + vi) Dt Use substitution Dx = 1/2 (vi + aDt + vi) Dt Dx = ½ (2 vi + a Dt) Dt Dx = ½Dt (2 vi) + ½Dt (a Dt) Dx = Vi Dt + ½ a Dt 2

  13. DerivingVf2 = vi2 + 2a Dx v = vf + vi and t = vf - vi 2 a Dx = vD t Dx = vf + vi multiplied vf - vi 2 a Dx = vf2 – vi2 rearrange and get 2a Vf2 = vi2 + 2a Dx

  14. Practice Problems • Find the acceleration of an amusement park ride that falls from rest to a speed of 28 m /s in 3.0 s. a = vf – vi =28m/s - 0 m/s tf – ti 3.0 s a = 9.3 m/s2

  15. Practice Problems • A bicyclist accelerates from 5.0 m /s to 16 m /s in 8.0 s. Assuming uniform acceleration, what distance does the bicyclist travel during this time interval. Dx = ½ (vi + vf) t = ½ (5.0 m/s + 16 m/s) 8.0 s = 84 m

  16. Practice Problems • An aircraft has a landing speed of 83.9 m /s. The landing area of an aircraft carrier is 195 m long. What is the minimum uniform acceleration required for a safe landing? a = Vf2 - vi2 = (0)2 - (83.9m/s)2 2 Dx 2(195 m) = - 18.0 m/s2

  17. Practice Problems • An electron is accelerated uniformly from rest in an accelerator at 4.5 X 107 m/s2 over a distance of 95 km. Assuming constant acceleration, what is the final velocity of the electron? Vf2 = vi2 + 2a Dx Vf2 = (0)2 + 2(4.5 X 107 m/s2) 95,000 m Take the square root of each side Vf = 2.9 X 106 m/s

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