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The Georgia Police Academy

The Georgia Police Academy. Traffic Accident Reconstruction Level IV. Background and Purpose.

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The Georgia Police Academy

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  1. The Georgia Police Academy • Traffic Accident Reconstruction Level IV

  2. Background and Purpose • In serious traffic accidents questions continually arise regarding positions of vehicles and or pedestrians prior to the collision. In order to fully analyze a traffic accident, these time/distance relationships must be established. Traffic Accident Reconstruction Level IV, introduces the students to twelve motion equations that will enhance their investigative abilities. Students who successfully complete this course will be able to establish pre-impact vehicle to vehicle or vehicle to pedestrian relationships.

  3. Time/Distance Equations • Terminal Performance Objective • Given a simulated traffic accident, students will accurately calculate vehicle to vehicle and vehicle to roadway relationships using motion equations.

  4. Time/Distance Equations • Enabling Objectives • Demonstrate the ability to apply three acceleration equations to traffic accident reconstruction problems • Demonstrate the ability to apply three initial velocity equations to traffic accident reconstruction problems • Demonstrate the ability to apply two end velocity equations to traffic accident reconstruction problems

  5. Time/Distance Equations • Enabling Objectives • Demonstrate the ability to apply three distance equations to traffic accident reconstruction problems • Demonstrate the ability to apply one time equation to traffic accident reconstruction problems

  6. Time/Distance • Three Basic Equations • The equations we will use during this course originate from three equations. • a = Ve - Vi/t • d = Vit + 1/2 at² • Ve² = Vi² + 2ad

  7. Time/Distance • a = acceleration in ft/sec/sec • t = time in seconds • Vi = initial velocity in ft/sec • Ve = end velocity in ft/sec

  8. Time/Distance • By using algebraic manipulation each of these equations can be solved for the variables in the equations. The equations on your equation sheet provide the basis to solve many time-distance problems in accident reconstruction.

  9. Time/Distance • Definitions • Understanding the relationships of the variables acceleration (a), time (t), distance (d), initial velocity (Vi), and end velocity (Ve), is essential when addressing accident reconstruction problems.

  10. Time/Distance • Distance is a linear measurement from some point. For reconstruction problems distance is measured relative to a coordinate system fixed on earth. For example, from the point where a vehicle started to accelerate or decelerate. Distance is a scalar quantity, having magnitude only. In U.S.A. units the dimension for distance is always feet.

  11. Time/Distance • Time is measured in seconds. In some cases it may be useful to know the distance that could be traveled during an hour. • Velocity Initial and End is a rate of change of distance with respect to time. Thus velocity has the units of distance per time. The values used are ft/sec. If a vehicle is traveling at constant velocity, then v = d/t.

  12. Time/Distance • By rearranging this equation the other two equations for constant velocity can be seen. They are d = vt and t = d/v. • These equations can only be used if the velocity is constant. If the velocity is not constant the answer will be wrong.

  13. Time/Distance • When velocity changes from one to another, the first velocity is initial velocity (Vi) and the second velocity is called end velocity (Ve). The change in velocity takes place over a time period, t. By definition, velocity is a vector quantity. In most time-distance analysis only the magnitude of the velocity is of interest.

  14. Time/Distance • Acceleration is the rate change of velocity with respect to time. Because velocity is a vector quantity, this rate change of velocity can be either in magnitude or direction. Think of an object speeding up or slowing down as an example of a change of magnitude. An object being deflected from its path by some side force and undergoing a lateral acceleration is an example of a change in direction. The units of acceleration are in feet per second per second (ft/sec²).

  15. Time/Distance • Solving Time/Distance Problems • You have been given an equation sheet to use to solve time, distance, acceleration and velocity problems. There are more equations that can be used than those listed. It is recommended that you limit your confusion by using only those equations on the equation sheet.

  16. Time/Distance • Drag factor, f, will be given to you in many problems. The first thing you should do is change drag factor to acceleration. From On-Scene Level II we know drag factor is related to acceleration by the following equations:

  17. Time/Distance • a = fg • f = a/g • The quantity, g, is the acceleration of gravity. The acceleration rate of gravity is 32.2 ft/sec/sec. Drag factor does not have a positive or negative sign associated with it. You must remember, to add the negative sign in cases where you have deceleration (decreasing velocity).

  18. Time/Distance • Velocity units will be in ft/sec. In some cases you will have to convert from/to mi/hr to/from ft/sec. Remember S = 15/22 x Velocity • V = 22/15 x Speed

  19. Time/Distance • Analysis procedure is made easier if you use the equations provided in the student manual. The most common question asked by investigators is: Which equation do I use? This is not a difficult question to answer. First of all, ask yourself "What do I want to know?" This may be velocity, time, acceleration factor, distance, or whatever else is important.

  20. Time/Distance • Look for an equation that isolates this desired quantity on the left side of the equals sign. Next, ask yourself what information is already known. You must be thorough here, especially if there is a change in velocity. You must find an equation that has those things that you know or can get on the right side of the equals sign. You can then solve the equation.

  21. Time/Distance • It is always a good idea to write the equation you are going to use on your worksheet. Once you have done the arithmetic to get the answer check to see if your answer makes sense. For example, if you just calculated the time it takes to accelerate from a stop across two lanes of traffic and your answer is 44 seconds, you clearly have made an error. Check your work.

  22. Time/Distance • Equations • Acceleration • The equation sheet in the student manual lists three equations for determining acceleration. • a = Ve - Vi/t • a = 2d - 2Vit/t² • a = Ve² - Vi²/2d

  23. Time/Distance • Calculate the acceleration of a vehicle which accelerates from a stop to 30 ft/sec in 6.5 seconds. • Given: Vi = 0 ft/sec • Ve = 30 ft/sec • t = 6.5 seconds

  24. Time/Distance • Solution: • a = Ve -Vi/t • a = 30 - 0/6.5 • a = 4.6 ft/sec/sec

  25. Time/Distance • Calculate the acceleration of a vehicle which decelerates over a distance of 75 feet from a velocity (speed) of 60 mph for 5 seconds. • Given: d = 75 ft. • Vi = 60 x 22/15 = 88 ft/sec • t = 5 seconds

  26. Time/Distance • Solution: • a = [2d - 2Vit]/t² • a = [2 (75) - 2 (88) (5)]/5² • a = [150 - 880]/25 • a = -730/25 • a = -29.2 ft/sec/sec Remember the negative sign means the vehicle is slowing.

  27. Time/Distance • Calculate the acceleration of a vehicle which decelerates from 88 ft/sec to 44 ft/sec over a distance of 130 feet. • Given: Vi = 88 ft/sec • Ve = 44 ft/sec • d = 130 feet

  28. Time/Distance • Solution: • a = Ve² -Vi²/2d • a = 44² - 88²/2 (130) • a = 1936 - 7744/260 • a = -5808/260 • a = -22.3 ft/sec/sec

  29. Time/Distance • Initial Velocity • The equation sheet in the student manual lists three equations for determining initial velocity. • Vi = Ve - at • Vi = d/t - at/2 • Vi = %Ve² - 2ad

  30. Time/Distance • Calculate the initial velocity (speed) of a vehicle which accelerates to 30 ft/sec for 5 seconds at 4 ft/sec/sec. • Given: Ve = 30 ft/sec • t = 5 sec • a = 4 ft/sec/sec

  31. Time/Distance • Solution: • Vi = Ve - at • Vi = 30 - (4) (5) • Vi = 30 - 20 • Vi = 10 ft/sec

  32. Time/Distance • Calculate the initial velocity of a vehicle which has slowed over a distance of 120 feet with a drag factor equal to .75 for 2.5 seconds. • Given: d = 120 feet • a = fg = (.75) (-32.2) = -24.2 ft/sec/sec • t = 2.5 sec

  33. Time/Distance • Solution: • Vi = d/t - at/2 • Vi = 120/2.5 - (-24.2) (2.5)/2 • Vi = 48 - (-30.2) • Vi = 48 + 30.2 • Vi = 78.2 ft/sec

  34. Time/Distance • Calculate the initial velocity of a vehicle which decelerates to a stop over a distance of 150 feet with a drag factor of .80. • Given: Ve = 0 ft/sec • a = fg = (.80) (-32.2) = -25.8 ft/sec/sec • d = 150 feet

  35. Time/Distance • Solution: • Vi = %Ve² - 2ad • Vi = %0² - 2 (-25.76) (150) • Vi = %0 - (-7728) • Vi = %7728 • Vi = 87.90 ft/sec

  36. Time/Distance • End Velocity • Ve = Vi + at • Ve = %Vi² + 2ad

  37. Time/Distance • Calculate the end velocity of a vehicle which accelerates from an initial velocity of 5 ft/sec at a rate of 4 ft/sec/sec after 2 seconds, after 3 seconds. • Given: Part 1Part 2 • Vi = 5 ft/sec Vi = 5 ft/sec • a = 4 ft/sec/sec a = 4 ft/sec/sec • t = 2 sec t = 3 sec

  38. Time/Distance • Solution: • Ve = Vi + at Ve = Vi + at • Ve = 5 + (4) (2) Ve = 5 + (4) (3) • Ve = 5 + 8 Ve = 5 + 12 • Ve = 13 ft/sec Ve = 17 ft/sec

  39. Time/Distance • Calculate the end velocity of a vehicle which slows from an initial velocity of 90 ft/sec over a distance of 125 feet at a rate of 20 ft/sec/sec. • Given: Vi = 90 ft/sec • d = 125 feet • a = -20 ft/sec/sec

  40. Time/Distance • Solution: • Ve = %Vi² + 2ad • Ve = %90² + 2 (-20) (125) • Ve = %8100 + (-5000) • Ve = %8100 - 5000 • Ve = %3100 • Ve = 55.7 ft/sec

  41. Time/Distance • The equation sheet in the student manual lists three equations for determining distance. • d = Vit + 1/2 at² • d = Ve² - Vi²/2a • d = t (Vi + Ve)/2

  42. Time/Distance • Remember, all of the equations given on your equation sheet, apply whether a vehicle accelerates to or from any velocity (not just to/from a stop). Equation (9) would also apply if you do not have acceleration. If acceleration equals zero, then constant velocity occurs. For Equation (9), the term 1/2 at² drops out because a is zero.

  43. Time/Distance • Thus, Equation (9) becomes d = Vit. Equation (11), can be used in situations where there is a constant velocity. With constant velocity the initial velocity equals the end velocity, Vi = Ve. Therefore equation (11) becomes d = t(Vi + Vi)/2

  44. Time/Distance • Vi + Vi = 2Vi • d = t(2Vi)/2 • Cancel the 2's leaving: • d = Vit

  45. Time/Distance • This equation only applies in cases where the velocity is constant. Calculate the distance needed for a vehicle to decelerate from 100 ft/sec for 2.3 seconds at a rate of 20 ft/sec/sec. • Given: Vi = 100 ft/sec • t = 2.3 seconds • a = -20 ft/sec/sec

  46. Time/Distance • Solution: • d = Vit + 1/2 at² • d = 100 (2.3) + 1/2 (-20) (2.3)² • d = 230 + (-52.9) • d = 230 - 52.9 • d = 177 feet

  47. Time/Distance • Calculate the distance needed to stop for a vehicle which decelerates from a velocity (speed) of 60 mph with a drag factor of .5. • Given: Ve = 0 • Vi = 22/15 x 60 = 88 ft/sec • a = fg = (.5) (-32.2) = -16.1 ft/sec/sec

  48. Time/Distance • Solution: • d = Ve² - Vi²/2a • d = 0²/2 - 88²/2 (-16.1) • d = 0 - (+7744)/-32.2 • d = -7744/-32.2 • d = 240 feet

  49. Time/Distance • A car falls off a cliff. Its initial vertical velocity equals zero. Its vertical velocity after 2 seconds is 64.4 ft/sec. Calculate the vertical distance it traveled. • Given: t = 2 seconds • Ve = 64.4 ft/sec • Vi = 0

  50. Time/Distance • Solution: • d = t (Vi + Ve)/2 • d = 2 (0 + 64.4)/2 • d = 128.8/2 • d = 64.4 feet

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