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Understanding Acceleration in One Dimension: Key Concepts and Practice Problems

Explore the concept of acceleration in one dimension, focusing on average acceleration, velocity, and time. Learn how to calculate acceleration with given initial and final velocities, and understand its significance as a vector quantity. This resource includes practical examples, graphs, and derived equations to reinforce learning. From a car accelerating from 4 m/s to 20 m/s to analyzing bugs in graph representations, engage with diverse practice problems that deepen your understanding of motion dynamics.

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Understanding Acceleration in One Dimension: Key Concepts and Practice Problems

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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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