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# Traffic congestion2

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### Traffic congestion2

1. 1. Traffic jam By Hannah B. und Henrike G., Pestalozzi Gymnasium and students of 5b
2. 2. •The development of traffic jam has been watched over time. •The function f shows the traffic's length (in km) from 12:00 to 19:00. •f(t)=-0,1t³+4,5t²-66,3t+322,7 Math problem: Traffic jam
3. 3. Math problem: Traffic jam The graph of the function f for 12<t<19,5
4. 4. Traffic jam in art One task of an art class grade 5 was to draw pictures of a traffic jam on a highway. The art teacher was going to sort these drawings to make a long row of these art work for an exhibition in the classroom. Bye chance Mrs. Schwarze had seen this and was allowed to take pictures from the best sketches. Mrs. Schwarze suggested to put some pictures from these young students between our solution of the task.
5. 5. Traffic jam: tasks a) (1) Calculate how long the traffic congestion is at 1pm. (2)If you want to know how many vehicles are in the traffic congestion at 1pm, you have to make assumptions. Calculate with two reasonable assumptions the number of vehicles at this time. b) Calculate the time where the congestion is the longest as well as its length. c) Calculate how many kilometres the congestion averagely increases between 13:00 and 17:00 per hour. d) For function f apply to 15<t<17 two inequalities: f'(t)>0 and f''(t)<0 Interprete the meaning of these inequalities referring to its application of real life in the task. e) It can be assumed, that the congestion decreases continously from 19:00 with 3,6 km per hour. (1) Calculate the time, when the congestion has completely resolved. (2) The congestion's length after 19:00 is supposed to be shown by the function g. Calculate the term of g.
6. 6. Traffic jam: solutions a) (1) f(13)=-0,1(13)³+4,5(13)²-66,3(13)+322,7 f(13)= 1,6 Answer: The congestion at 13:00 is 1,6 km long. (2) [various possibilities!] length of one vehicle: (L) aprox. 3m space between the vehicles: (S) aprox. 1m length of the congestion: (C) 1600 m number of vehicles: x x= C : (L+S) x=1600:4 x=400 => There are aprox. 400 cars
7. 7. b) You need to find the maximum point => f‘(x)=0 -> -0,3t²+9t-66,3=0  x= 17 => f‘‘(x)<0 - >-0,6t+9≠0  -0,6*17 + 9 < 0 => calculate the y-value f(17)=4,8 Pmax(17|4,8) Answer: At 5pm the traffic jam has a length of 4,8 km. Traffic jam: solutions
8. 8. c) 1. Calculate the y-values for the points 13 and 17 LP=lowest point (13|1,6) HP=highest point (17|4,8) 2. Use the known formula m= (y₂-y₁):(x₂-x₁) m=(4,8-1,6):(17-13) = 0,8 Answer: On an average the traffic jam increases 0,8 km per hour between 1pm and 5pm. Traffic jam: solutions
9. 9. d) The first inaquality f'(t)>0 means that the congestion only becomes longer in this period of time, because the slope is always bigger than 0. The second inequality f''(t)<0 means that the speed of jam development decreases and the lenght of the jam increases in the time between 3pm and 5pm. Traffic jam: solutions
10. 10. e)(1) f(17)=1,6 1,6:3,6= 0,44 0,44*60= 26,4 f) (2) g(x)=mx+b  1,6=19*3,6 +b  g(x)=3,6x+70 Answer: At 19:26 the congestion has completely resolved. Traffic jam: solutions
11. 11. We hope you enjoyed watching. If you have any questions to this exercises you are welcome to ask Henrike G. and Hannah B., e.g. in fb group of MMM, the writers of this presentation. ;) (We had to do this task in the last central exam in grade 9 at the end of last school year. It had been drafted by the ministery of education of North-Rhine Westfalia. We translated the task and wrote down our solutions. Traffic jam

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