Invetigation #6: Driving Safety Solution 1. How far can a car travelling at 60 mi/h go in: (a) 0.1? 60 mi x 1 h x 1 min x 5280 ft = 88 ft h 60 min 60 1 mi x = v x t = 88 ft x 0.1 = 8.8 ft (b) 0.5? x = v x t = 88 ft x 0.5 = 44 ft (c) 1? x = v x t = 88 ft x 1.0 = 88 ft (d) 2? x = v x t = 88 ft x 2.0 = 176 ft (e) 5? x = v x t = 88 ft x 5.0 = 440 ft - 84 -
2. Travelling at 60 mi/h, (a) Etimate how far you will travel when you turn around to talk to a friend in the back eat. What are the afety implication of thi? Talking briefly to a friend in the back eat take about 2. In 2, you will travel 176 feet (ee 1 (d)). Thi i a long ditance o that driving condition might be dratically different compared to when you lat looked at the road, poibly leading to a crah. It i bet not to take your eye off of the road (b) Etimate how far you will travel when you earch for a CD in the glove compartment. What are the afety implication of thi? Reaching and finding a CD in the glove compartment take about 4. In thi time, you will travel the following ditance: x = v x t = 88 ft x 4.0 = 352 ft Thi i a long ditance o that driving condition might be dratically different compared to when you lat looked at the road, poibly leading to a crah. It i bet not to take your eye off of the road. (c) Etimate how far you will travel when you turn to the ide to ee if the pace next to you i clear for paing. What are the afety implication of thi? Looking to the ide take about 1. In 1, you will travel 88 feet (ee 1 (c)). Thi i a long ditance o that driving condition might be dratically different compared to when you lat looked at the road, poibly leading to a crah. It i bet to minimize the time you take your eye off of the road and it i prudent to leave much more than 88 feet between you and the car in front of you. (d) The low beam of your headlight will allow you to ee about 160 feet in front of you at night. How long doe it take your car to travel thi ditance? What are the afety implication of thi? t = x = 160 ft = 1.8 v 88 ft You will travel the ditance illuminated by your headlight in only 1.8. If an object i een in your headlight, you have at mot 1.8 to avoid it if you tay travelling at 60 mph. - 85 -
(e) Mot driver need about 1.5 to react to a new ituation. How far will your car travel in thi time interval? What are the afety implication of thi? x = v x t = 88 ft x 1.5 = 132 ft Your car will travel 132 feet in an emergency ituation before you even have a chance to hit the brake! Teacher' Note: There i an intereting dicuion of reaction time in the book Traffic Safety and the Driver. If you are anticipating an event, reaction time can be a hort a 0.15. While driving, your reaction time i divided into a perception reaction time ("I need to brake") and a movement reaction time (movement of your foot). For driver focuing on the car ahead of them, an average reaction time i 1.6. For driver encountering an unexpected obtacle around a blind curve, an average reaction time i cloer to 2.5. (f) After the brake are applied to a car traveling at 60 mi/h, the car need about 227 feet to top. Conidering reaction time (about 1.5 ) and braking ditance, how long a ditance will it take to top if you ee a problem up ahead at night? Why doe thi how the danger of driving at night? The car will travel 132 ft while you are reacting to the ituation (ee 2e above) and then it will take 227 feet to top. So to top will require a total topping ditance of 132 ft + 227 ft = 359 ft. Since your headlight illuminate only the road 160 feet in front of you, at 60 mph, you will be unable to top in time if an obtruction uddenly appear on the road in front of you at night. Activity Suggetion: Have your tudent tand near a toplight. They can determine a typical reaction time by meauring the time it take for a car to begin moving after a toplight turn from red to green. - 86 -
3. How far will a drunk driver' car travel before it top if it wa traveling at 60 mi/hr? What are the afety implication of thi? (Recall that the reaction time of a drunk driver i doubled compared to that of a ober driver o the reaction time of a drunk driver i about 3.0 ) The reaction time i 3.0 o the ditance the car travel at 60 mph in 3.0 i: x = v x t = 88 ft x 3.0 = 264 ft The braking ditance i, a in 2 above, 227 ft. So the total braking ditance i 264 ft + 227 ft = 491 ft. The topping ditance of a drunk driver i 491 ft when traveling at 60 mph. Thi i (491ft - 359 ft) = 132 ft more than the topping ditance for a ober driver. A drunk driver react more lowly than a ober driver, o their car will travel a longer ditance before it top. If there i a problem in the road le than the topping ditance of 491 feet, the drunk driver will hit it. Since the reaction time of a driver on a cell phone i comparable to that of a drunk driver, the above analyi alo hold for them. - 87 -
4. On a 2-lane road (one lane in each direction), you (car 1) decide to pa a car (car 2) in front of you that i traveling at 50 mi/hr. You ee another car (car 3) coming toward you from the other direction that i traveling at 50 mi/hr. When you are jut behind car 2, you intantly accelerate to 60 mi/hr and move into the other lane. You pa car 2 then move back into your original lane, and intantly decelerate back to 50 mph. Aume that it take 2 to travel into the paing lane and 2 to move back into your original lane. Alo aume that you will pa back into your lane when you are two complete car length in front of the car you are paing. Each car i 5 m long. The oncoming car i 0.25 mile away from car 1. a. Will car 1 uccefully pa car 2 without hitting the oncoming car 3? a. If o, how many econd later would your car and the oncoming car be at the ame poition on the road? What leon did you learn from thi? 0.25 mile A 1 2 The tarting ituation. All car are traveling at 50 mph. 3 B 1 30 ft It take car 1 two to move into the paing lane. Then car 1 intantly accelerate to 60mph 2 C 30 ft 1 Car 1 move 30 feet in front of car 2. Then car 1 intantly decelerate to 50 mph 2 D 30 ft 2 1 It take car 1 two to move back into it original lane. Car 1 i again traveling at 50 mph. - 88 -
a. Car 1 mut travel 60 feet more than car 2 to be able to pa it. Car 1 i travelling 10 mph (15 ft/) fater than car 2 while it i in the paing lane. So car 1 will take the following amount of time to pa car 2. t = x = 60 ft = 4 v 15 ft So car 1 take 2 to move into the paing lane (at 50 mph), 4 to pa car 1 (at 60 mph), and then 2 more to move back into it original lane (at 50 mph), the total time i 8. In 8, car 1 will travel for 2 at 50 mph, then for 4 at 60 mph, then 2 for 50 mph, o the total ditance traveled by car 1 i: x = 73.3 ft x 2 + 88 ft x 4 + 73.3 ft x 2 = 645 ft In 8, car 3 will travel the following ditance (toward car 1) x = 73.3 ft x 8 = 586 ft So at the end of 8 car 1 and car 2 are 645 ft + 586 ft = 1231 ft cloer. The car tarted 0.25 mile apart, which i 1320 ft (recall that 1 mile = 5280 ft). When car 1 i back in it original lane after paing car 2, car 1 will be only 89 feet from car 3. Since their relative peed of approach i 50 mph + 50 mph = 100 mph = 147 ft/, the afety margin for paing car 2 wa le than 1. b. Even though car 1 tarted to pa car 2 when car 3 wa 0.25 mile away, car 1 and car 3 mied colliding at 50 mph by le than 1. So you mut allow much more ditance and time for paing than i obviou. Excerpt from Traffic Safety and the Driver (p. 118): "It i found that while driver make reliable etimate of the ditance to the oncoming car, they are inenitive to it peed... The inability of driver to etimate oncoming peed lead them to decline afe paing opportunitie when the oncoming car i travelling lower than expected, and to initiate unafe paing maneuver when the oncoming car i travelling fater than expected." Note: A an extenion, tudent could alo plot the poition v. time and the velocity (or peed) v. time for each of the three car. - 89 -
5. Suppoe you a traffic engineer working for the California Department of Tranportation. Your job i to et the timing of traffic light. You are alo to only ue metric unit. How long hould traffic light be yellow for the following peed: 45 km/hr, 65 km/hr, 85 km/hr, and 105 km/hr? Ue the reult of problem 2 in the imulation tool development ection. Firt tranlate the metric peed to Englih unit peed. a. 45 km x 1 mi = 28 mph h 1.6 km b. 65 km x 1 mi = 41 mph h 1.6 km c. 85 km x 1 mi = 53 mph h 1.6 km d. 105 km x 1 mi = 66 mph h 1.6 km Uing the table of total topping time in problem 2 in the imulation tool development ection to etimate the total topping time at the 4 peed. Speed (km/h) Total topping time () from problem 2 in imulation tool development ection 45 3.9 3.1 65 5.0 3.9 85 6.0 4.9 105 7.2 5.8 California traffic manual uggeted time for yellow light The yellow light time uggeted by the California traffic manual eem to be about 1 econd le than would be expected. Let' explore thi further in the activitie below. Suggeted tudent activity: Have your tudent meaure the duration of yellow light. For a certain peed limit, are they alway et the ame? How are they et at different peed limit? - 90 -
e. What i the ditance traveled by car during California department of tranportation yellow light time if it tay at it initial peed. At 28 mph: 28mi x 88ft hr x 3.1 = 41.1 ft x 3.1 = 127 ft hr 60 mi At 41 mph: 41mi x 88ft hr x 3.9 = 60.1 ft x 3.9 = 234 ft hr 60 mi At 53 mph: 53mi x 88ft hr x 4.9 = 77.7 ft x 4.9 = 381 ft hr 60 mi At 66 mph: 66mi x 88ft hr x 5.8 = 96.8 ft x 5.8 = 561 ft hr 60 mi f. What i the topping ditance at the above 4 peed: Recall that the otal topping ditance = reaction ditance + braking ditance = v (1.5 ) + v 2 2a At 28mph: Total topping ditance = 41.1 ft x 1.5 + (41.1 ft) 2 2 = 111 ft 2 (2)(17 ft) At 41mph: Total topping ditance = 60.1 ft x 1.5 + (60.1 ft) 2 2 = 197 ft 2 (2)(17 ft) At 53mph: Total topping ditance = 77.7 ft x 1.5 + (77.7 ft) 2 2 = 294 ft 2 (2)(17 ft) At 66mph: Total topping ditance = 96.8 ft x 1.5 + (96.8 ft) 2 2 = 373 ft 2 (2)(17 ft) - 91 -
g. Calculate the difference between the contant peed ditance and the total topping ditance determined in e and f above. At 28mph: Contant peed ditance - Total topping ditance =127 ft - 111 ft = 16 ft At 41mph: Contant peed ditance - Total topping ditance =235 ft - 197 ft = 38 ft At 53mph: Contant peed ditance - Total topping ditance =381 ft - 294 ft = 87 ft At 66mph: Contant peed ditance - Total topping ditance =561 ft - 373 ft = 188 ft h. Determine the time it take to travel the ditance determined in g. Thi i the amount of time you have to decide to brake o that you can top before the light. Time to travel contant peed ditance - total topping ditance At 28mph: 16 ft = 0.4 41.1 ft So at 28mph, you have 0.4 of afety margin to decide to brake. At 41mph: 38 ft = 0.6 60.1 ft So at 41mph, you have 0.6 of afety margin to decide to brake. At 53mph: 87 ft = 1.1 77.7 ft So at 53mph, you have 1.1 of afety margin to decide to brake. - 92 -
At 66mph: 188 ft =1.9 96.8 ft So at 66mph, you have 1.9 of afety margin to decide to brake. i. Draw a diagram howing the ditance traveled at 66 mph, the total topping ditance at 66 mph, and the afety margin in ditance and time you have to decide to brake. A B C 561 ft - ditance traveled at 66 mph in 5.8 Safety margin for braking: 188 ft or 1.9 at 66 mph 373 ft - total topping ditance at 66 mph Signal i yellow for 5.8 at 66 mph At point A, you cannot make it through the light traveling at 66 mph o you hould brake and top. You hould know from experience that your car can eaily and afely brake at thi ditance from the light. At point B, you can either drive at the ame peed or brake. You will pa the light before it turn red if you continue traveling at the peed limit. You alo have enough time to afely brake if you decide to top before the light. So at thi critical ditance from the light, either deciion will be a afe one. At point C, you do not have enough ditance to brake before the light. You hould know from experience that your car cannot afely brake at thi ditance from the light. You hould continue traveling at the ame peed - you will eaily pa the light before it turn red. From the San Diego Union Tribune 10/6/01, page B1. Headline: "Davi ign bill on red-light camera; yellow to be timed." "Gov. Gray Davi igned a bill yeterday requiring that traffic light with camera how the yellow caution light for a reaonable amount of time before taking photo of red-light violator. The amount of time for triggering the automatic camera would be determined by the California Department of Tranportation, There have been allegation that ome traffic light in San Diego and other area were et to witch too wiftly from green to red. SB 667 by Sen. Steve Peace require that traffic light with camera comply with the Caltran traffic manual for minimum yellow-light interval. The minimum depend on the peed limit. For example, at 25 mph the yellow light mut be on for at leat 3 econd. At 45 mph, the interval mut be at leat 4.3 econd." - 93 -
j. Redo the diagram of part i auming that you are peeding: you are traveling at 66 mph in a zone where the peed limit i 53 mph. At 53 mph, the ignal i yellow for 4.9. Ditance = 66mph x 88ft/ x 4.9 = 474 ft; 66 mph x 88ft/ = 96.8 ft/ 60 mph 60 mph Safety margin time = (474ft - 373ft)/(96.8ft/) = 1.04 474 ft - ditance traveled at 66 mph in 5.8 Safety margin for braking: 101 ft or 1.04 at 66 mph 373 ft - total topping ditance at 66 mph Signal i yellow for 4.9 at 53 mph - 94 -
k. Redo the diagram of part i auming that you are peeding: you are traveling at 66mph in a zone where the peed limit i 41 mph. At 41 mph, the ignal i yellow for 3.9. What do thee reult indicate about the danger of peeding. Ditance = 66mph x 88ft/ x 3.9 = 378 ft; 66 mph x 88ft/ = 96.8 ft/ 60 mph 60 mph Safety margin time = (378ft - 373ft)/(96.8ft/) = 0.05 If the car i in the hatched in area hown in the diagram below, it mut continue through the light. If it i in the area hown by the arrow, it mut top in order to not run the red light. There i no area B a hown in part i of thi problem ince the afety margin for making the correct deciion i now only 5 ft or 0.05 when traveling at 66 mph. Near the tranition region D, the driver mut make an immediate correct deciion a to whether or not they hould hit the brake or maintain their peed. D Safety margin for braking: 5ft or 0.05 at 66 mph 378 ft - ditance traveled at 66 mph in 3.9. 373 ft - total topping ditance at 66 mph Signal i yellow for 3.9 at 41mph - 95 -
6. A general rule of thumb taught to driver i to leave one car length of ditance between car for every 10 mph. Aume that a typical car length i 15 feet or 3 meter. Doe thi make ene? Why or why not? Conider 3 different cae. a. You are traveling at 60 mph and uddenly the traffic in front of you low to 50 mph. The ituation i a hown below: A B 90 ft. Firt calculate how long it take car A to brake from 60 mph to 50 mph. It take a driver about 1.5 to react, o car A travel the following ditance: x = vt = 88 ft x 1.5 = 132 ft. Next determine the time it take a car to low down from 60 mph to 50 mph at a contant rate of deceleration of 17 ft/ 2. Previouly, we found that 50 mph = 73.3 ft/. Since a = v f v I, then t = v f v I = (73.3 ft 88 ft) 2 t a x (-17 ft) = 0.86 The average peed of car A during thi time i v f + v I = 81 ft. 2 So car A travel x = vt = 81 ft x 0.86 = 70 ft a it decelerate to 50 mph. So the total time it take car A to low down to 50 mph = 1.5 + 0.86 = 2.36. The total ditance it travel i 132 ft + 70 ft = 202 ft. In thi time interval, car B will travel x = vt = 73.3 ft x 2.36 = 182 ft. When both car are traveling at 50 mph, car A will have traveled (202 ft 182 ft) = 20 feet more than car B. Since thi ditance i le than 90 ft, car A will not come too cloe to car B car A will end up 70 feet behind car B. - 96 -
b. You are traveling at 60 mph and uddenly the traffic in front of you low to 40 mph. Firt calculate how long it take car A to brake from 60 mph to 40 mph. It take a driver about 1.5 to react, o car A travel the following ditance: x = vt = 88 ft x 1.5 = 132 ft. Next determine the time it take a car to low down from 60 mph to 40 mph at a contant rate of deceleration of 17 ft/ 2. Previouly, we found that 40 mph = 49 ft/. Since a = v f v I, then t = v f v I = (49 ft 88 ft) 2 t a x (-17 ft) = 2.3 The average peed of car A during thi time i v f + v I = 68.5 ft. 2 So car A travel x = vt = 68.5 ft x 2.2 = 151 ft a it decelerate to 40 mph. So the total time it take car A to low down to 50 mph = 1.5 + 2.3 = 3.8. The total ditance it travel i 132 ft + 151 ft = 283 ft. In thi time interval, car B will travel x = vt = 49 ft x 3.8 = 186 ft. When both car are traveling at 40 mph, car A will have traveled (283 ft 186 ft) = 97 feet more than car B. Since thi ditance i more than 90 ft, car A will hit car B. c. You are travelling at 60 mph and all of a udden you become aware that the traffic i topped in front of you when the traffic i only 6 car length in front of you. The total topping ditance at 60 mph i 360 ft (ee problem 5f), much more than the ditance of 6 car length of 90 ft. So you will hit the car in front of you. It take you 1.5 to react to the ituation. During thi time, your car travel 132 ft, o you will till be traveling at 60 mph when you hit the car in front of you which will probably reult in eriou injury or death. In concluion, a long a you keep your relative peed to within about 15 mph of the traffic in front of you, you will be able to react in time. 7. Another way to judge appropriate pacing between car when driving i to conider what i called "headway." Headway i defined a the elaped time between the front of the lead vehicle paing a point on the roadway and the front of the following vehicle paing - 97 -
the ame point. Mot driving manual recommend a headway of at leat 2. How doe a headway of 2- compare to the rule of thumb that you hould leave 1 car length between the front of your car and the back of the car in front of you for every 10-mph of peed? Hint 1: Aume that each car length i 14.7 ft long. Recall that 10mph = 14.7 ft/. Hint 2: Aume two car are moving at the ame contant peed, one behind the other. Call the peed v. At time t = 0 the front of the lead vehicle pae a given point on the highway. Two econd later the front of the econd vehicle pae that ame point. The total ditance between the front of the two vehicle i therefore (v)(2 ). Aume two car are moving at the ame contant peed, one behind the other. Call the peed v. At time t = 0 the front of the lead vehicle pae a given point on the highway. Two econd later the front of the econd vehicle pae that ame point. The total ditance between the front of the two vehicle i therefore (v)(2 ). Thu the ditance D between the back of the firt car and the front of the econd i (v) (2 ) L, where L i the length of the car. Car 1 Car 2 D (v)(2 ) L L The length of car vary, typically ranging around 14-15 ft. For implicity, take L = 14.7 ft. Thi give u D = (v)(2 ) - L = (v)(2 ) - 14.7ft For v = 10mph, v = 14.7 ft/ o D = (14.7 ft)(2 ) - 14.7 ft = 14.7 ft = 1 car length - 98 -
For v = 60mph, v = 88 ft/ o D = (88ft) (2 ) - 14.7 ft = 161.3 ft 161.3 ft x 1 car length = 11 car length 14.7 ft Thu, the ditance between car varie from 1 car length for every 10mph (at low peed) to a bit le than 2 car length for every 10mph (at high peed), which i in accord with the rough rule of thumb but i better becaue it allow for greater pacing at higher peed. Not only i a headway of 2 econd a afer rule of thumb, you can more readily meaure headway on the road. Etimating the ditance between vehicle i more difficult than noting the time between car paing the ame point on the road. Excerpt from Traffic Safety and the Driver (p. 314-316): "... driver who are following other vehicle do o with an average headway of 1.32 econd; that i, the average headway i coniderably horter than the recommended minimum.... Why do driver chooe to follow o cloely? It eem to me that it become largely a driving habit, rather than reaoned conciou behavior.... Following at a headway of 2.0 econd intead of 0.5 econd mean you will arrive 1.5 econd later, auming that no vehicle cut in front of you.... Even if a few vehicle do cut into the gap in front... thi add only about 2 econd per uch incident to the overall trip time. Driver probably object to other vehicle cutting in front of them not becaue it delay them a couple of econd, but becaue it i interpreted a ome ort of peronal affront, an aault on manhood or womanhood. If detached rationality cannot dipel uch feeling, comfort might be ought in the confident expectation that the offending driver i likely to be experiencing more than the average crah rate of one per 10 year. Let uch driver have their fun - they are paying a high price for it; recapture your two econd by walking fater to your vehicle." - 99 -
8. You are driving at night to a friend houe. You make a harp right hand turn onto another treet and uddenly, 75 feet in front of you, you ee omeone croing the treet. Baed on the reult of problem 5, what i the maximum peed you could be traveling at and not hit the peron. Baed on problem 5, the maximum peed i about 20 mph, ince at 20 mph, the total topping ditance i about 70 ft. How fat hould you be driving when making a turn onto a treet? You peed hould be uch that your topping ditance i le than the ditance your light are illuminating in front of you. When making a turn, that ditance may be a little a 25 feet, o you hould not make turn at night at peed higher than about 10 mph. - 100 -
9. Headlight illuminate the road up to 160 feet in front of you. If you are on a road with top ign, what i the fatet peed you can drive and till top afely at night? The total topping ditance mut be le than or equal to 160 feet we will aume it i equal to 160 ft o that your car will come to a complete top right at the top ign. If you are initially travelling at a peed v, then the reaction ditance i vt and the braking ditance i v 2 /(2a). So the total topping ditance (d total ) i: d total = vt + v 2. 2a d total = 160 ft, the braking deceleration a = 17 ft, and the reaction time t = 1.5. 2 The equation that mut be olved i a quadratic equation: v 2 + vt - d total = 0 or 0.029 2 v 2 + 1.5 v - 160ft = 0 2a ft Uing the quadratic formula v = -1.5 +/- qr {1.5 2 2-4 *.029 2 /ft * (-160ft)} 2 * 0.029 2 /ft Only the + ign yield a phyically meaningful olution, o: v = 52.8 ft or 36 mph. Perhap thi i why the peed limit on country road, where you may encounter a top ign without warning, i often 35 mph. If you were drunk and traveling at thi peed, your reaction time would double to 3. So your reaction ditance would be: Reaction ditance = vt = 52.7 ft x 3 = 158.1 ft. Your braking ditance i till v 2 = 52.7 2 ft 2 2 = 81.7 ft 2a 2 2 x 17 ft So the total braking ditance = 151.8 ft + 81.7 ft = 239.8 ft. Thi ditance i greater than you can ee uing your light (160 ft), o you would not be able to top before the top ign. You will be traveling through the interection, poibly cauing a crah. - 101 -
9A. You are traveling on a freeway with a mall amount of traffic. After about a hour of driving, you notice that there appear to be no car traveling at your peed - all car eem to be either paing you or you are paing them. Why i thi? The reaon i that car traveling at the ame peed a you never get cloer to you or farther from you - they tay the ame ditance from you. So you will never ee the car traveling at your peed. 9B. Smiley face want to walk from the idewalk at point A to the idewalk at point B at night. She need to cro the treet. Where i the afet place for Smiley to cro and why? Smiley hould cro at the dahed line ince it give her the mot time to react to a car entering Walk St. from either Top St. or Bottom St. Top St. B Walk St. Bottom St. A - 102 -
10. A fire engine i traveling at 25 m/ directly toward a bu tation on it way to a fire. At it cloet approach it pae right next to the tation. Starting at 400 m before the tation, it end out a very hort blat of ound every 100 m. It top ending thee meage when it i 400 m pat the tation. Sound travel at 330 m/. If you are tanding at the bu tation, determine the time interval between ucceive blat of ound. Calculate and compare (uing a table and a chart) how the time interval change when the fire engine i approaching you veru when it i moving away from you. At time t=0, the firt ound i ent out. At thi point, the fire engine i 400 meter away, o the ound take t = d/v = 400 /330 = 1.2. So the firt ound appear at the tation at t = 1.2. After the fire engine ha traveled 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 4. The ound mut now travel 300 m, which take t = d/v = 300 /330 = 0.9. So thi ound appear at the tation at t = 4.9. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 8. The ound mut now travel 200 m, which take t = d/v = 200 /330 = 0.6. So thi ound appear at the tation at t = 8.6. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 12. The ound mut now travel 100 m, which take t = d/v = 100 /330= 0.3. So thi ound appear at the tation at t = 12.3. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. It i now at the tation. So thi ound i ent out at t = 16. The ound mut now travel 0 m, which take t = d/v = 0 /330 = 0. So thi ound appear at the tation at t = 16. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. It i now 100 m pat the tation. So thi ound i ent out at t = 20. The ound mut now travel 100 m, which take t = d/v = 100 /330 = 0.3. So thi ound appear at the tation at t = 20.3. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 24. The ound mut now travel 200 m, which take t = d/v = 200 /330 = 0.6. So thi ound appear at the tation at t = 24.6. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 28. The ound mut now travel 300 m, which take t = d/v = 300 /330 = 0.9. So thi ound appear at the tation at t = 28.9. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 32. The ound mut now travel 400 m, which take t = d/v = 400 /330 = 1.2. So thi ound appear at the tation at t = 33.2. - 103 -
Make a table ummarizing the data. Ditance of fire engine from tation (m) Time ignal arrived at tation () -400 1.2-300 4.9 3.7-200 8.6 3.7-100 12.3 3.7 0 16.0 3.7 100 20.3 4.3 200 24.6 4.3 300 28.9 4.3 400 33.2 4.3 Time between ucceive ignal () So the time between ucceive blat of ound i le a the fire engine i approaching the tation compared with the time between ucceive blat of ound a the fire engine travel away from the tation. Thi i the origin of the Doppler effect. Note that there i a difference between the fire engine approaching the tation and receding from the tation. A the fire engine i approaching the tation: it end out a ignal it travel toward the tation it end out the next ignal when it i cloer to the tation A the fire engine i receding from the tation: it end out a ignal it travel away from the tation it end out the next ignal when it i farther from the tation 4.4 Time v Poition Time Bewteen Succeive Sound () 4.3 4.2 4.1 4 3.9 3.8 3.7 3.6-400 -300-200 -100 0 100 200 300 400 500 Poition of Fire Engine (m) - 104 -
10A. A fire engine i traveling at 25 m/ on it way to a fire. At it cloet approach it pae 100m from a bu tation. Starting at 400 m before the tation, it end out a very hort blat of ound every 100 m. It top ending thee meage when it i 400 m pat the tation. Sound travel at 330 m/. If you are tanding at the bu tation, determine the time interval between ucceive blat of ound. Calculate and compare (uing a table and a chart) how the time interval change when the fire engine i approaching you veru when it i moving away from you. Bu tation 100m -400m 0m 400m Fire engine motion At time t=0, the firt ound i ent out. At thi point, the fire engine i (400 2 +100 2 ) 0.5 m = 412.3 m away, o the ound take t = d/v = 412.3/330 = 1.249. So the firt ound appear at the tation at t = 1.249. After the fire engine ha traveled 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 4. The ound mut now travel (300 2 +100 2 ) 0.5 m = 316.2 m, which take t = d/v = 316.2 /330 = 0.958. So thi ound appear at the tation at t = 4.958. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 8. The ound mut now travel (200 2 +100 2 ) 0.5 m = 223.6 m, which take t = d/v = 223.6 /330 = 0.678. So thi ound appear at the tation at t = 8.678. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 12. The ound mut now travel (100 2 +100 2 ) 0.5 m = 141.4 m, which take t = d/v = 141.4 /330= 0.428. So thi ound appear at the tation at t = 12.428. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. It i now at it cloet approach to the bu tation. So thi ound i ent out at t = 16. The ound mut now travel 100 m, which take t = d/v = 100 /330 = 0.303. So thi ound appear at the tation at t = 16.303. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 20. The ound mut now - 105 -
travel (100 2 +100 2 ) 0.5 m = 141.4 m, which take t = d/v = 141.4 /330= 0.428. So thi ound appear at the tation at t = 20.428. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 24. The ound mut now travel (200 2 +100 2 ) 0.5 m = 223.6 m, which take t = d/v = 223.6 /330 = 0.678. So thi ound appear at the tation at t = 24.678. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 28. The ound mut now travel (300 2 +100 2 ) 0.5 m = 316.2 m, which take t = d/v = 316.2 /330 = 0.958. So thi ound appear at the tation at t = 28.958. After the fire engine ha traveled another 100 m (which take 4 at 25 m/), the fire engine end out another ound. So thi ound i ent out at t = 32. The ound mut now travel (400 2 +100 2 ) 0.5 m = 412.3 m, o the ound take t = d/v = 412.3/330 = 1.249. So thi ound appear at the tation at t = 33.249. Make a table ummarizing the data. Ditance of fire engine from tation (m) Time ignal arrived at tation () -400 1.249-300 4.958 3.709-200 8.678 3.720-100 12.428 3.750 0 16.303 3.875 100 20.428 4.125 200 24.678 4.250 300 28.958 4.280 400 33.249 4.291 Time between ucceive ignal () - 106 -
So the time between ucceive blat of ound lowly increae a the fire engine i approaching the bu tation, then increae rapidly a it pae it cloet approach to the bu tation, then lowly increae again a it travel away from the bu tation. Thi i the origin of the Doppler effect. 4.4 Time v. Poition Time Between Succeive Sound () 4.3 4.2 4.1 4 3.9 3.8 3.7 3.6-400 -300-200 -100 0 100 200 300 400 500 Poition of fire engine (m) - 107 -
11. Jack and Jill each drive their vehicle 10,000 mile per year. Jack' vehicle ha a fuel economy of 10 mile per gallon, Jill' 30 mile per gallon. a. How much fuel doe each of them ue in a year? Jack: 10,000 mi x 1 gallon = 1000 gallon year 10 mi year Jill: 10,000 mi x 1 gallon = 333 gallon year 30 mi year b. How much fuel doe the Jack and Jill houehold ue in a year? The Jack and Jill houehold ue 1333 gallon per year. c. How far do they travel in a year? The Jack and Jill houehold travel a total of 20,000 mile. d. What i their average houehold fuel economy? I it the average of Jack' fuel economy and Jill' fuel economy? The Jack and Jill houehold average fuel economy i given by the total mile driven divided by the total fuel ued, o Houehold average fuel economy = 20,000 mile = 15 mile. 1333 gallon gallon Note that thi i NOT the average of Jack' fuel economy and Jill' fuel economy, which would be 20 mile per gallon. e. What would their average houehold fuel economy be if Jill' vehicle got 100 mile per gallon? Jill: 10,000 mi x 1 gallon = 100 gallon year 100 mi year So the total houehold would ue 1100 gallon, therefore: Houehold average fuel economy = 20,000 mile = 18.2 mile. 1100 gallon gallon - 108 -
f. What would their average houehold fuel economy be if Jill' vehicle got 1000 mile per gallon? Jill: 10,000 mi x 1 gallon = 10 gallon year 1000 mi year So the total houehold would ue 1010 gallon, therefore: Houehold average fuel economy = 20,000 mile = 19.8 mile. 1010 gallon gallon g. What would their average houehold fuel economy be if Jill' vehicle got 10,000 mile per gallon? Jill: 10,000 mi x 1 gallon = 1 gallon year 10,000 mi year So the total houehold would ue 1001 gallon, therefore: Houehold average fuel economy = 20,000 mile = 19.98 mile. 1001 gallon gallon h. What would their average houehold fuel economy be if Jack' vehicle got 30 mile per gallon, the ame a Jill' original vehicle? Jack: 10,000 mi x 1 gallon = 333 gallon year 30 mi year So the total houehold would ue 666 gallon, therefore: Houehold average fuel economy = 20,000 mile = 30 mile. 666 gallon gallon i. If you were in charge of making policy to reduce fuel conumption, what would you do? It i far better to try to improve the fuel economy of the wort vehicle than the bet vehicle. Therefore, you might try to make a minimum fuel economy tandard or, a Congre ha done, mandate a required average fuel economy for all vehicle old by a car manufacturer. - 109 -
12. Conider the following ditribution of dot on the line below. Let' call the dot "galaxie" and let' call the line "the univere." Suppoe that adjacent galaxie are all located a ditance of L apart from each other in the univere. At a time T later, the univere ha expanded a factor of two o that now all of the adjacent galaxie are a ditance of 2L apart. a. Suppoe you are living in galaxy A. How fat doe it appear that galaxie B, C, and D are receding from you? From the perpective of galaxy A, galaxy B traveled a ditance of L in time T o it peed of receion i L/T. From the perpective of galaxy A, galaxy C traveled a ditance of 2L in time T o it peed of receion i 2L/T. From the perpective of galaxy A, galaxy D traveled a ditance of 3L in time T o it peed of receion i 3L/T. b. I there a correlation between the ditance the galaxy i located from you and the peed with which it i receding from you. What i that relationhip? The further away the galaxy i from you, the fater it appear to be moving. The relationhip (known a the Hubble Contant in atronomy) i that the peed of receion i L/T for every ditance L the galaxy i located from you. Thee data are ummarized in the table below. Galaxy Original ditance of galaxy from galaxy A Ditance traveled by galaxy in time T a oberved by galaxy A Speed of receion of galaxy a oberved by galaxy A Speed of receion of galaxy a oberved by galaxy A divided by original ditance of galaxy from galaxy A B L L L/T 1/T {or (L/T)/L} C 2L 2L 2L/T 1/T {or (2L/T)/(2L)} D 3L 3L 3L/T 1/T {or (3L/T)/(3L)} - 110 -
c. Do all galaxie ee the ame thing happening? Ye. To a peron on any galaxy, it appear that all the other galaxie are moving away from them and the peed of receion i proportional to the ditance from your galaxy. A B C D L A B C 2L D - 111 -
13. From the Aociated Pre dated 11/4/02: Long Beach Nearly 200 car and big-rig truck collided in two incident on the fogbound Long Beach Freeway early yeterday, injuring dozen of people, nine critically, and cloing the highway for hour CHP officer Joeph Pace aid viibility wa down to about 50 feet in heavy fog when the chain reaction crahe began jut before 7 a.m Some motorit etimated car were moving at 25 to 35 mph. From the Aociated Pre dated 11/5/02: Lo Angele The chain-reaction crahe that piled up nearly 200 car on the Long Beach Freeway likely could have been avoided if driver had imply lowed down when they hit foggy condition, California Highway Patrol officer aid yeterday. The crahe, which left a five-mile ection of the freeway looking like an auto junkyard, hut down the highway for 11 hour Sunday. Eight people uffered critical or eriou injurie in the accident, which took place within minute. Motorit reported driving into fog o thick it reminded ome of being on an airliner a it travel into the cloud. In that weather condition, we re ure if driver had dratically reduced their peed, thi could have been avoided, aid California Highway Patrol Officer Lui Mendoza. a. What i the total breaking ditance for a car traveling at 25 mph? b. What i the total breaking ditance for a car traveling at 35 mph? c. How do thee total breaking ditance compare to the 50 foot viibility of that day? Why did the accident occur? d. What maximum peed hould the car have been traveling at if the viibility wa only 50 feet. a. 25 mi x 88ft hr = 36.7 ft hr 60 mi Reaction ditance = 36.7 ft x 1.5 = 55 ft Braking ditance = (initial peed) 2 2 x deceleration = 36.7 2 ft 2 x 2 = 39.6 ft 2 x 2 x 17ft So topping ditance = reaction ditance + braking ditance = 55 ft + 39.6 ft = 94.6 ft. - 112 -
b. a. 35 mi x 88ft hr = 51.3 ft hr 60 mi Reaction ditance = 51.3 ft x 1.5 = 77 ft Braking ditance = (initial peed) 2 2 x deceleration = 51.3 2 ft 2 x 2 = 77.4 ft 2 x 2 x 17ft So topping ditance = reaction ditance + braking ditance = 77 ft + 77.4 ft = 154.4 ft. c. The total topping ditance for car traveling at 25 and 35 mile per hour i 94 and 154 feet repectively, much greater than the viibility of 50 feet. So the car could not top in time to avoid a crah. d. The total topping ditance i equal to 50 ft. If you are initially traveling at a peed v, then the reaction ditance i vt and the braking ditance i v 2 /(2a).. So the total topping ditance (d total ) i: d total = vt + v 2. 2a d total = 50 ft, the braking deceleration a = 17 ft, and the reaction time t = 1.5. 2 The equation that mut be olved i a quadratic equation: v 2 + vt - d total = 0 or 0.029 2 v 2 + 1.5 v - 50ft = 0 2a ft Uing the quadratic formula v = -1.5 +/- qr {1.5 2 2-4 *.029 2 /ft * (-50ft)} 2 * 0.029 2 /ft Only the + ign yield a phyically meaningful olution, o: v = 23 ft or 16 mph. So if the viibility wa only 50 feet, car hould have been traveling at a maximum peed of 16 mph. - 113 -
14. From the January/February 2003 iue of the AAA magazine entitled A Glaring Concern: HID headlight: boon to vehicular afety or blight on the automotive landcape? Neverthele, the viual improvement high-intenity dicharge (HID) lamp provide i dramatic: A motorit uing HID can ee about 330 feet in front of the vehicle, compared with 190 feet with tandard halogen lighting almot a 75 percent improvement. One of the problem with halogen lighting i that motorit often overdrive their headlight they re unable to ee people, animal, or object until it too late. The data are tartling: Under perfect condition (dry pavement, mechanically ound vehicle, antilock brake, alert and killed driver) at 45 mph, it take 170 feet to top your car; at 50 mph, 205 feet; and at 60 mph, 282 feet. So, omewhere between 45 and 50 mph, you ve overdriven halogen headlight. a. What reaction time/breaking time i being aumed for the perfect condition driver dicued above? I thi a fair dicuion? b. Dicu how HID can improve driving afety given your work o far in thi unit. Conider the average driver with an average car and average reaction time. - 114 -