1 / 46

Acceleration

Solve acceleration problems involving distance, velocity, and time with given values. Learn how to calculate distances traveled by cars starting from rest and decelerating, determining acceleration rates, and more. This physics concept helps understand motion in vehicles and objects. Improve your problem-solving skills with these practical examples and formulas. Enhance your knowledge of acceleration dynamics through real-world scenarios.

stephanie-snider
Télécharger la présentation

Acceleration

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Acceleration • A car starts from rest and accelerates at a rate of 6.1 m/s2for 7 seconds. Calculate how far it traveled. • V1 = • V2 = • A = • D = • T =

  2. Acceleration • A car starts from rest and accelerates at a rate of 6.1 m/s2for 7 seconds. Calculate how far it traveled. • V1 = 0 m/s

  3. Acceleration • A car starts from rest and accelerates at a rate of 6.1 m/s2for 7 seconds. Calculate how far it traveled. • V1 = 0 m/s • V2 = no

  4. Acceleration • A car starts from rest and accelerates at a rate of 6.1 m/s2for 7 seconds. Calculate how far it traveled. • V1 = 0 m/s • V2 = no • A = 6.1 m/s2

  5. Acceleration • A car starts from rest and accelerates at a rate of 6.1 m/s2for 7 seconds. Calculate how far it traveled. • V1 = 0 m/s • V2 = no • A = 6.1 m/s2 • D = ?

  6. Acceleration • A car starts from rest and accelerates at a rate of 6.1 m/s2for 7 seconds. Calculate how far it traveled. • V1 = 0 m/s • V2 = no • A = 6.1 m/s2 • D = ? • T = 7 sec.

  7. Solution #1 • D = V1t + at2 2

  8. Solution #1 • D = ½ at2 • D = (6.1 m/s2)(7 sec)2 2

  9. Solution #1 • D = ½ at2 • D = (6.1 m/s2)(7 sec)2 2 • D = 150 m

  10. Acceleration #2 • What was the initial velocity of a car that accelerates at a rate of – 8.0 m/s2 and comes to rest in 484 m? • V1= • V2= • A = • D = • T =

  11. Acceleration #2 • What was the initial velocity of a car that accelerates at a rate of – 8.0 m/s2 and comes to rest in 484 m? • V1 = ?

  12. Acceleration #2 • What was the initial velocity of a car that accelerates at a rate of – 8.0 m/s2 and comes to rest in 484 m? • V1 = ? • V2 = 0 m/s

  13. Acceleration #2 • What was the initial velocity of a car that accelerates at a rate of – 8.0 m/s2 and comes to rest in 484 m? • V1 = ? • V2 = 0 m/s • A = -8.0 m/s2

  14. Acceleration #2 • What was the initial velocity of a car that accelerates at a rate of – 8.0 m/s2 and comes to rest in 484 m? • V1 = ? • V2 = 0 m/s • A = -8.0 m/s2 • D = 484 m

  15. Acceleration #2 • What was the initial velocity of a car that accelerates at a rate of – 8.0 m/s2 and comes to rest in 484 m? • V1 = ? • V2 = 0 m/s • A = -8.0 m/s2 • D = 484 m • T = no

  16. Solution #2 • Use equation # 7 • V22 = V12 + 2ad

  17. Solution #2 • Use equation # 7 • V22 = V12 + 2ad • 0 m/s = V12+ 2(-8.0m/s)( 484 m)

  18. Solution #2 • Use equation # 7 • V22 = V12 + 2ad • 0 m/s = V12+ 2(-8.0m/s2)( 484 m) • 7744 m2/s2 = V12

  19. Solution #2 • Use equation # 7 • V22 = V12 + 2ad • 0 m/s = V12+ 2(-8.0m/s2)( 484 m) • 7744 m2/s2 = V12 • V1= 88 m/s

  20. Acceleration #3 • What is the acceleration of a car that was going at 25.6 m/s and slows to 8.5 m/s in 4.6 sec.? • V1= • V2= • A= • D= • T=

  21. Acceleration #3 • What is the acceleration of a car that was going at 25.6 m/s and slows to 8.5 m/s in 4.6 sec.? • V1 = 25.6 m/s

  22. Acceleration #3 • What is the acceleration of a car that was going at 25.6 m/s and slows to 8.5 m/s in 4.6 sec.? • V1 = 25.6 m/s • V2 = 8.5 m/s

  23. Acceleration #3 • What is the acceleration of a car that was going at 25.6 m/s and slows to 8.5 m/s in 4.6 sec.? • V1 = 25.6 m/s • V2 = 8.5 m/s • A = ?

  24. Acceleration #3 • What is the acceleration of a car that was going at 25.6 m/s and slows to 8.5 m/s in 4.6 sec. • V1 = 25.6 m/s • V2 = 8.5 m/s • A = ? • D = no

  25. Acceleration #3 • What is the acceleration of a car that was going at 25.6 m/s and slows to 8.5 m/s in 4.6 sec. • V1 = 25.6 m/s • V2 = 8.5 m/s • A = ? • D = no • T = 4.6 sec

  26. Solution #3 • Use Equation #2 • A = V2- V1 t

  27. Solution #3 • A = V2- V1 t A = 8.5 m/s – 25.6 m/s 4.6 sec

  28. Solution #3 • A = V2- V1 t A = 8.5 m/s – 25.6 m/s 4.6 sec A = -3.7 m/s2

  29. Acceleration #4 • A car traveling at 14m/s encounters a patch of ice and takes 5.0 sec to stop. How far did it skid? • V1 = • V2= • A = • D = • T =

  30. Acceleration #4 • A car traveling at 14m/s encounters a patch of ice and takes 5.0 sec to stop. How far did it skid ? • V 1 = 14 m/s

  31. Acceleration #4 • A car traveling at 14m/s encounters a patch of ice and takes 5.0 sec to stop. How far did it skid ? • V 1 = 14 m/s • V2 = 0 m/s

  32. Acceleration #4 • A car traveling at 14m/s encounters a patch of ice and takes 5.0 sec to stop. How far did it skid ? • V 1 = 14 m/s • V2 = 0 m/s • A = no

  33. Acceleration #4 • A car traveling at 14m/s encounters a patch of ice and takes 5.0 sec to stop. How far did it skid ? • V 1 = 14 m/s • V2 = 0 m/s • A = no • D = ?

  34. Acceleration #4 • A car traveling at 14m/s encounters a patch of ice and takes 5.0 sec to stop. How far did it skid ? • V 1 = 14 m/s • V2 = 0 m/s • A = no • D = ? • T = 5.0 sec

  35. Solution #4 • Use equation # 4 • D = (V2+V1) t 2

  36. Solution #4 • Use equation # 4 • D = (V2+V1) t 2 D = (14m/s + 0 m/s)5 sec 2

  37. Solution #4 • Use equation # 4 • D = (V2+V1) t 2 D = (14m/s + 0 m/s)5 sec 2 D = 35 m

  38. Acceleration #5 • An object traveling at 16 m/s accelerates at a constant rate of 4.0 m/s2over 50 meters. What is it’s final velocity? • V1= • V2= • a= • d= • t=

  39. Acceleration #5 • An object traveling at 16 m/s accelerates at a constant rate of 4.0 m/s2over 50 meters. What is it’s final velocity? • V1= 16 m/s

  40. Acceleration #5 • An object traveling at 16 m/s accelerates at a constant rate of 4.0 m/s2over 50 meters. What is it’s final velocity? • V1= 16 m/s • V2 = ?

  41. Acceleration #5 • An object traveling at 16 m/s accelerates at a constant rate of 4.0 m/s2over 50 meters. What is it’s final velocity? • V1= 16 m/s • V2 = ? • A = 4.0 m/s2

  42. Acceleration #5 • An object traveling at 16 m/s accelerates at a constant rate of 4.0 m/s2over 50 meters. What is it’s final velocity? • V1= 16 m/s • V2 = ? • A = 4.0 m/s2 • D = 50 m

  43. Acceleration #5 • An object traveling at 16 m/s accelerates at a constant rate of 4.0 m/s2over 50 meters. What is it’s final velocity? • V1= 16 m/s • V2 = ? • A = 4.0 m/s2 • D = 50 m • T = no

  44. Solution #5 • Use equation #7 • V22= V12 + 2ad

  45. Solution #5 • Use equation #7 • V22= V12 + 2ad • V22 = 16 m/s + 2(4.0m/s2)(50m)

  46. Solution #5 • Use equation #7 • V22= V12 + 2ad • V22 = 16 m/s + 2(40.0m/s2)(50m) • V2 = 26 m/s

More Related