The Georgia Police Academy Traffic Accident Reconstruction Level IV

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.

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.

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

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

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

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

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.

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.

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.

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.

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.

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.

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).

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.

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:

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).

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

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.

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.

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.

Time/Distance Acceleration 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

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

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

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

Time/Distance Solution: a = [2d - 2Vit]/t² a = -29.2 ft/sec/sec Remember the negative sign means the vehicle is slowing.

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

Time/Distance Solution: a = Ve² -Vi²/2d a = 44² - 88²/2 (130) a = -22.3 ft/sec/sec

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

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

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

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

Time/Distance Solution: Vi = d/t - at/2 Vi = 120/2.5 - (-24.2) (2.5)/2 Vi = 78.2 ft/sec

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

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

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

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 1 Part 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

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

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

Time/Distance Solution: Ve = %Vi² + 2ad Ve = %90² + 2 (-20) (125) Ve = 55.7 ft/sec

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

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.

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

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

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

Time/Distance Solution: d = Vit + 1/2 at² d = 177 feet

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

Time/Distance Solution: d = Ve² - Vi²/2a d = 0²/2 - 88²/2 (-16.1) d = 240 feet

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

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

Time/Distance The equation sheet in the student manual lists one equation for determining time. t = Ve - Vi/a Remember if you have a case of constant velocity, the equation t = d/v applies. Do not use this equation if you have acceleration, because you will obtain the wrong answer.

Time/Distance Calculate the time required for a vehicle to slow from 60 mph to 30 mph with a drag factor of .70. Given: Vi = 22/15 x 60 = 88 ft/sec Ve = 22/15 x 30 = 44 ft/sec a = fg = (.70) (-32.2) = -22.5 ft/sec/sec

Time/Distance Solution: t = Ve - Vi/a t = 44 - 88/-22.5 t = -44/-22.5 t = 1.96 seconds

Time/Distance Calculate the time required for a pedestrian to walk a distance of 20 feet at a velocity of 4 ft/sec. Given: V = 4 ft/sec (constant velocity) d = 20 feet

Time/Distance Solution: t = d/v t = 20 /4 t = 5 seconds

Time/Distance Traffic Accident Reconstruction Problems Up to this point you have been given simple examples using the 12 equations listed on your equation sheet. We are now going to attempt problems which are more complicated. Remember, analyze what is known. Decide what you want to know. Write down the equation needed. The solutions given to the problems are only a solution. There may be several ways to solve a problem.

Time/Distance Problem 1 A vehicle accelerates from a stop over a distance of 72 feet in 6 seconds.

Time/Distance Question 1 – What is the vehicle’s velocity after eight seconds of acceleration from a stop if the acceleration is uniform over the whole eight seconds? Question 2 – What is the distance traveled in the first eight seconds?

Time/Distance Question 3 - What is the vehicle's velocity (speed) in miles per hour after the first three seconds of acceleration? Question 4 - How much time does it take to accelerate over the first 36 feet?

Time/Distance Question 1 Solution Given: The statement of Question 1 only gives two values Vi = 0 and t = 6 seconds. Remember, unless it is a constant velocity condition, it's always necessary to know three variables before you can work the problem. Therefore, it is necessary to look at the first part of the problem statement, where we are given:

Time/Distance Vi = 0 ft/sec d = 72 feet t = 6 seconds From these three values, acceleration can be calculated; this is the same acceleration needed to solve Question 1.

Time/Distance a = 2d - 2 Vit/t² a = 2 (72) - 2 (0) (6)/6² a = 4 ft/sec/sec Now Question 1 can be answered, because three values are now known:

Time/Distance Given: a = 4 ft/sec/sec Vi = 0 ft/sec t = 8 seconds

Time/Distance Ve = Vi + at Ve = 0 + (4) (8) Ve = 32 ft/sec

Time/Distance Given: a = 4 ft/sec/sec Vi = 0 ft/sec t = 8 seconds Question 2 Solution Given: a = 4 ft/sec/sec Vi = 0 ft/sec t = 8 seconds

Time/Distance d = Vit + 1/2at² d = (0) (8) + 1/2 (4) (8²) d = 128 feet

Time/Distance Given: Vi = 0 ft/sec a = 4 ft/sec/sec t = 3 seconds Question 3 Solution Given: Vi = 0 ft/sec a = 4 ft/sec/sec t = 3 seconds

Time/Distance Ve = Vi + at Ve = 0 + (4) (3) Ve = 12 ft/sec 15/22 x 12 = 8.18 mph

Time/Distance Given: d = 36 feet Vi = 0 ft/sec a = 4 ft/sec/sec Question 4 Solution Given: d = 36 feet Vi = 0 ft/sec a = 4 ft/sec/sec

Time/Distance The equation for time is: t = Ve - Vi/a Ve is not known. Because three variables are known Ve can be calculated. Ve = %Vi² + 2ad Ve = %0² + (2) (4) (36) Ve = 17 ft/sec

Time Distance Now t can be solved for: t = Ve - Vi/a t = 17 - 0/4 t = 4.25 seconds

Time/Distance Problem 2 A vehicle skids to a stop over two surfaces. The first surface has a drag factor of .85 and the second surface has a drag factor of .60. The vehicle skidded 85 feet over the first surface and 45 feet over the second surface.

Time/Distance Problem 2 Question 1 - What is the vehicle's velocity (speed) when the skidding first occurred? Question 2 - How much time did it take the vehicle to skid over both surfaces? Question 3 - What is the vehicle's velocity after skidding 100 feet?

Time/Distance Question 1 Solution In this case the problem must be worked "backward". Start from the final position of the vehicle to calculate its velocity when it first enters the second surface. Then three quantities are known. Given: Ve = 0 a = fg = -.60 (32.2) = -19.3 ft/sec/sec d = 45 feet

Time/Distance Vi = %Ve² - 2ad Vi = %0² - 2 (-19.3) (45) Vi = 41.7 ft/sec

Time/Distance The velocity, 41.7 ft/sec, is the velocity where the vehicle starts to skid on the second surface. It is also the end velocity after skidding on the first surface. Thus, on the first surface these quantities are now known. Given: Ve = 41.7 ft/sec d = 85 feet a = fg = -.85 (32.2) = -27.4 ft/sec/sec

Time/Distance Using the same equation as before, but now over the first surface, allows the initial velocity at first skidding to be calculated.

Time/Distance Vi = %Ve² - 2ad Vi = %41.7 - 2 (-27.4) (85) Vi = 79.98 ft/sec 15/22 x 79.98 = 54.53 mph

Time/Distance Question 2 Solution Equation 12 is used to solve this problem. It must be used for both surfaces and then the two value of time are added. Given: Surface 1 Vi = 80 ft/sec Ve = 41.7 ft/sec a = -27.4 ft/sec/sec

Time/Distance t = Ve - Vi/a t = 41.7 - 80/-27.4 t = -38.3/-27.4 t = 1.39 seconds

Time/Distance Given: Surface 2 Vi = 41.7 ft/sec Ve = 0 a = -19.3 ft/sec/sec

Time/Distance t = Ve - Vi/a t = 0 - 41.7/-19.3 t = -41.7/-19.3 t = 2.16 seconds total time = 1.39 + 2.16 = 3.55 seconds

Time/Distance Question 3 Solution A simple approach to answering this question is to calculate the end velocity knowing the initial velocity at the beginning of the second surface. For this approach distance equals 15 feet, because an additional 15 feet are required to reach a total skidding distance of 100 feet. Given: Vi = 41.7 ft/sec a = -19.3 ft/sec d = 15 feet

Time/Distance Ve = %Vi² + 2ad Ve = %41.7² + 2 (-19.3) (15) Ve = 34 ft/sec

Time/Distance Problem 3 A car skids for 130 feet and hits a pedestrian. The car continues to skid another 90 feet before coming to a stop. These distances apply to the vehicle's center of mass. The total skidding distance is 220 feet. The vehicle's drag factor during skidding was .85. The pedestrian walked from the pavement edge northbound a distance of 14 feet and was then struck by the car. The pedestrian's walking velocity was 4 ft/sec.

Time/Distance Question 1 - What was the initial velocity of the car? Question 2 - What was the vehicle's velocity when the pedestrian was struck? Question 3 - How far was the car from the pedestrian when the pedestrian first stepped onto the pavement?

Time/Distance Given: Ve = 0 a = fg = .85 (-32.2) = -27.4 ft/sec/sec Question 1 Solution Given: Ve = 0 a = fg = .85 (-32.2) = -27.4 ft/sec/sec d = 220 feet

Time/Distance Vi = %Ve² - 2ad Vi = %0² - 2 (-27.4) (220) Vi = 109.79 ft/sec

Time/Distance Question 2 Solution The car has 90 feet left to skid to a stop when the pedestrian is hit. Thus the following variables apply. Ve = 0 ft/sec a = -27.4 ft/sec/sec d = 90 feet

Time/Distance Vi = %Ve² - 2ad Vi = %0² - 2 (-27.4) (90) Vi = 70.22 ft/sec

Time/Distance Question 3 Solution The third question is more difficult than the other two questions. The question asks for the relative position of the car to the first contact position of the pedestrian when the pedestrian stepped onto the pavement. The time the pedestrian walks to the collision point is the same time the car travels to the collision point. Therefore, calculate the time for the pedestrian to walk to the collision point given a constant velocity of 4 ft/sec and a distance of 14 feet. Given:

Time/Distance d = 14 feet V = 4 ft/sec

Time/Distance t = d/v t = 14/4 t = 3.5 seconds

Time/Distance Now determine where the car was 3.5 seconds before the collision. First, calculate the time it took to skid the 130 feet before the collision: Given: Vi = 109.79 ft/sec Ve = 70.22 ft/sec a = -27.4 ft/sec/sec

Time/Distance t = Ve - Vi/a t = 70.22 - 109.79/-27.4 t = -39.57/-27.4 t = 1.44 seconds

Time/Distance Of the 3.5 seconds that the pedestrian is walking toward impact, the car is skidding for 1.44 seconds. During the remaining time of 2.06 seconds, the car can be assumed to be traveling at the velocity calculated at the beginning of the skid (109.79 ft/sec). Therefore, the distance traveled for 2.06 seconds is. Given: V = 109.79 ft/sec t = 2.06 seconds

Time/Distance d = vt d = 109.79 (2.06) d = 226.16 feet

Time/Distance The distance the car is from its first contact position, when the pedestrian first steps onto the pavement, is the sum of 226.16 feet plus 130 feet (distance skidded before collision). total distance = 226.16 + 130 total distance = 356.16 feet

Time/Distance Problem 4 A car accelerates from a stop across the intersection. An eastbound vehicle travels at a constant speed of 55 mph.

Time/Distance Question 1 - If the northbound vehicle accelerates at 4 ft/sec/sec, how much time is required for the car to get halfway across the east-west street and completely across the street? Question 2 - If the eastbound vehicle was 150 feet from the northbound vehicle when the northbound vehicle started to move, what would be the required acceleration of the northbound car if the eastbound car continued traveling at a constant speed of 55 mph.

Time/Distance Question 1 Solution The distance the northbound car must travel to clear half the distance across the road is the sum of the length of the car plus the distance from the car front to the project road edge plus half the width of the road. d = 15 + 5 + 12 = 32 feet

Time/Distance The other variables that are known are the initial velocity and the acceleration. Given: .d = 32 feet .Vi = 0 ft/sec .a = 4 ft/sec/sec

Time/Distance First, calculate the end velocity; then use that measurement to calculate the time. Ve = %Vi² + 2ad Ve = %0² + 2 (4) (32) Ve = %256 Ve = 16 ft/sec

Time/Distance t = Ve - Vi/a t = 16 - 0/4 t = 4 seconds to travel halfway across the street

Time/Distance To travel completely across the street the car must travel an additional 12 feet, or a total of 44 feet. d = 44 feet Vi = 0 ft/sec a = 4 ft/sec/sec

Time/Distance Ve = %Vi² + 2ad Ve = %0² + 2 (4) (44) Ve = %352 Ve = 18.76 ft/sec

Time/Distance t = Ve - Vi/a t = 18.76 - 0/4 t = 4.69 seconds

Time/Distance Question 2 Solution The time required for the eastbound vehicle to travel 150 feet at a constant speed of 55 mph can be calculated. Given: V = 55 mph 22/15x 55 = 80.66 d = 150 feet

Time/Distance t = d/v t = 150/80.66 t = 1.85 seconds

Time/Distance The northbound car has started to move when the eastbound car is 1.85 seconds from their possible collision point. The northbound car must travel 32 feet (see Question 1 Solution). The northbound car's initial velocity is zero. Given: Vi = 0 ft/sec d = 32 feet t = 1.85 seconds

Time/Distance a = 2d - 2Vit/t² a = [(2) (32) - (2) (0) (1.85)]/1.85² .a = 18.71 ft/sec/sec

Time/Distance Question 2 Solution This required acceleration greatly exceeds the normal acceleration of passenger cars. f = a/g = f = 18.71/32.3 = f = .58

Time Distance Equation Derivations The velocity of an object in motion is the rate of motion. The rate of motion is the distance traveled with respect to a certain period of time. The definition of velocity is represented by the equation: (1) V = d/t