{Recall that 88 ft = 60 mi so 88 ft x h = 1 s h s 60 mi General Atomics Sciences Education Foundation All rights reserved.

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1 Simulation Tool Development 1. a. Develop a scale model car tat sows te distance traveled in 1 s at speeds from 10 to 70 mp in increments of 10 mp. Use a scale of 1 in = 100 ft. Include a marker on eac scale model car tat sows te 160 ft distance tat is illuminated by a car's eadligts. b. Develop a scale model car tat sows te distance traveled in 1 s at speeds from 10 to 70 m/s in increments of 10 m/s. Use a scale of 1 cm = 10 m. 2. Te stopping distance for a car is equal to te distance it travels during your reaction time (reaction distance) plus te distance it travels wile te brakes are applied (braking distance). Te typical deceleration of a car is 17 ft/s 2 and a typical reaction time is 1.5 s. Recall te equation for calculating stopping distance is x = v 2 /2a, were v is te initial speed of te car, and te equation for calculating te distance traveled at constant speed is x = vt. Calculate te reaction distance, braking distance, and stopping distance for a car traveling at te following speeds. a. 10 mp b. 20 mp c. 30 mp d. 40 mp e. 50 mp f. 60 mp g. 70 mp. Make a table summarizing te above data. i. Make a plot summarizing te above data. j. Calculate te braking time for te car for eac of te cases a-g. Recall tat for te case of constant acceleration, t = v, a were v is te initial speed and te deceleration is a. Generate a table and a grap for te reaction time, braking time, and total stopping time for eac speed. k. Generate a scale model car and map (or use car D and map A) to simulate tese situations. Te stopping distance for a car is equal to te distance it travels during your reaction time (reaction distance) plus te distance it travels wile te brakes are applied (braking distance). Te typical deceleration (a) of a car is 17 ft/s 2 and a typical reaction time (t) is 1.5 s. Calculate te reaction distance (vt), braking distance (v 2 /(2a)), and total stopping distance (vt + v 2 /(2a)) for a car traveling at te following speeds (v). (a) 10 mp Reaction distance = speed x reaction time (recall tat speed is a constant during te reaction time) Reaction distance = 10 mi x 1.5 s x 88ft x = 22 ft {Recall tat 88 ft = 60 mi so 88 ft x = 1 s

2 Tis is a simple and useful conversion to remember. Or te student could use te previously found speed in ft/s as is done in te braking distance calculation below} (During tis time, te deceleration is constant) = ft 2 x s 2 = 6.4 ft s 2 x 2 x 17ft So stopping distance = reaction distance + braking distance = 22 ft ft = 28.4 ft. (b) 20 mp Reaction distance = 20 mi x 1.5 s x 88ft x = 44 ft = ft 2 x s 2 = 25.2 ft s 2 x 2 x 17ft So stopping distance = reaction distance + braking distance = 44 ft ft = 69.2 ft. (c) 30 mp Reaction distance = 30 mi x 1.5 s x 88ft x = 66 ft = 44 2 ft 2 x s 2 = 56.9 ft s 2 x 2 x 17ft So stopping distance = reaction distance + braking distance = 66 ft ft = ft

3 (d) 40 mp Reaction distance = 40 mi x 1.5 s x 88ft x = 88 ft = ft 2 x s 2 = ft s 2 x 2 x 17ft So stopping distance = reaction distance + braking distance = 88 ft ft = ft. (e) 50 mp Reaction distance = 50 mi x 1.5 s x 88ft x = 110 ft = ft 2 x s 2 = 158 ft s 2 x 2 x 17ft So stopping distance = reaction distance + braking distance = 110 ft ft = 268 ft. (f) 60 mp Reaction distance = 60 mi x 1.5 s x 88ft x = 132 ft = 88 2 ft 2 x s 2 = 228 ft s 2 x 2 x 17ft So stopping distance = reaction distance + braking distance = 132 ft ft = 360 ft

4 (g) 70 mp Reaction distance = 70 mi x 1.5 s x 88ft x = 154 ft = ft 2 x s 2 = 310 ft s 2 x 2 x 17ft So stopping distance = reaction distance + braking distance = 154 ft ft = 464 ft

5 () Speed (mp) Reaction distance (ft) Braking distance (ft) Total stopping distance (ft) (i) 500 Stopping distance vs Speed Reaction distance (ft) distance (ft) Braking distance (ft) Total stopping distance (ft) Stopping Speed (mp)

6 i. 10 mp = 14.7 ft so braking time t= v = 14.7 ft s 2 = 0.9 s s a s 17 ft 20 mp = 29.3 ft so braking time t= v = 29.3 ft s 2 = 1.7 s s a s 17 ft 30 mp = 44 ft so braking time t= v = 44 ft s 2 = 2.6 s s a s 17 ft 40 mp = 58.7 ft so braking time t= v = 58.7 ft s 2 = 3.5 s s a s 17 ft 50 mp = 73.3 ft so braking time t= v = 73.3 ft s 2 = 4.3 s s a s 17 ft 60 mp = 88 ft so braking time t= v = 88 ft s 2 = 5.2 s s a s 17 ft 70 mp = 103 ft so braking time t= v = 103 ft s 2 = 6.1 s s a s 17 ft

7 Speed (mp) Reaction time (s) Braking time (s) Total stopping time (s) Reaction times vs speed Speed (mp) reaction time (s) braking time (s) total stopping time (s) Develop a scale model car tat sows te reaction distance, braking distance, and total stopping distance for speeds for 10 to 70 mp in increments of 10 mp. Include te reaction time, braking time, and total stopping time in te scale model car simulation tool. Also include a marker tat sows te 160 ft distance tat is illuminated by a car's eadligts. Use te scale of 1 in = 100 ft

8 3. Redo te calculations of problem 2 assuming tat you are drunk and your reaction time is twice as long (3 s instead of 1.5 s). Generate a drunk driver scale model and map (or use car E and map A) to simulate tese situations. Te reaction distance will be double te distances calculated in problem 5 because te reaction distance is directly proportional to te reaction time. Te braking distance will remain te same because we are assuming it is not affected. Speed (mp) Reaction distance (ft) Braking distance (ft) Total stopping distance (ft) Speed vs Stopping Distance Reaction distance (ft) Braking distance (ft) Total stopping distance (ft) 500 Distance (ft) Speed (mp)

9 Speed (mp) Reaction time (s) Braking time (s) Total stopping time (s) Drunk Driver Reaction Times vs Speed Speed (mp) reaction time (s) braking time (s) total stopping time ( s ) Develop a scale model car tat sows te reaction distance, braking distance, and total stopping distance for speeds for 10 to 70 mp in increments of 10 mp. Include te reaction time, braking time, and total stopping time in te scale model car simulation tool. Also include a marker tat sows te 160 ft distance tat is illuminated by a car's eadligts. Use te scale of 1 in = 100 ft

10 4. Redo te calculations of problem 2 assuming tat te road is wet. Te deceleration of a car on a wet road is apparently 75-95% of te dry road value (see Appendix A) - assuming tat tere is no ydroplaning. So assume tat te wet road deceleration is 13 ft/s 2 instead of te 17ft/s 2 tat we ave using for a dry road. (a) 10 mp Reaction distance = speed x reaction time (recall tat speed is a constant during te reaction time) Reaction distance = 10 mi x 1.5 s x 88ft x = 22 ft = ft 2 x s 2 = 8.3 ft s 2 x 2 x 13ft So stopping distance = reaction distance + braking distance = 22 ft ft = 20.3 ft. (b) 20 mp Reaction distance = 20 mi x 1.5 s x 88ft x = 44 ft = ft 2 x s 2 = 33.0 ft s 2 x 2 x 13ft So stopping distance = reaction distance + braking distance = 44 ft ft = 77.0 ft. (c) 30 mp Reaction distance = 30 mi x 1.5 s x 88ft x = 66 ft = 44 2 ft 2 x s 2 = 74.5 ft s 2 x 2 x 13ft So stopping distance = reaction distance + braking distance = 66 ft ft = ft

11 (d) 40 mp Reaction distance = 40 mi x 1.5 s x 88ft x = 88 ft = ft 2 x s 2 = ft s 2 x 2 x 13ft So stopping distance = reaction distance + braking distance = 88 ft ft = ft. (e) 50 mp Reaction distance = 50 mi x 1.5 s x 88ft x = 110 ft = ft 2 x s 2 = ft s 2 x 2 x 13ft So stopping distance = reaction distance + braking distance = 110 ft ft = ft. (f) 60 mp Reaction distance = 60 mi x 1.5 s x 88ft x = 132 ft = 88 2 ft 2 x s 2 = ft s 2 x 2 x 13ft So stopping distance = reaction distance + braking distance = 132 ft ft = ft

12 (g) 70 mp Reaction distance = 70 mi x 1.5 s x 88ft x = 154 ft = ft 2 x s 2 = ft s 2 x 2 x 13ft So stopping distance = reaction distance + braking distance = 154 ft ft = ft

13 Speed (mp) Reaction distance (ft) Braking distance (ft) Total stopping distance (ft) Stopping distances vs speed Reaction distance (ft) Braking distance (ft) Total stopping distance (ft) Speed (mp)

14 10 mp = 14.7 ft so braking time t= v = 14.7 ft s 2 = 1.1 s s a s 13 ft 20 mp = 29.3 ft so braking time t= v = 29.3 ft s 2 = 2.3 s s a s 13 ft 30 mp = 44 ft so braking time t= v = 44 ft s 2 = 3.4 s s a s 13 ft 40 mp = 58.7 ft so braking time t= v = 58.7 ft s 2 = 4.5 s s a s 13 ft 50 mp = 73.3 ft so braking time t= v = 73.3 ft s 2 = 5.6 s s a s 13 ft 60 mp = 88 ft so braking time t= v = 88 ft s 2 = 6.8 s s a s 13 ft 70 mp = 103 ft so braking time t= v = 103 ft s 2 = 7.9 s s a s 13 ft Speed (mp) Reaction time (s) Braking time (s) Total stopping time (s)

15 Reaction Times vs Speed Reaction time (s) Braking time (s) Total stopping time (s) Speed (mp)

16 5. Compare te total stopping distance and total reaction time for a driver on a dry road, a driver on a wet road and a drunk driver on a dry road. Use te results of te previous problems. Sow te result in a table for speeds from 10 to 70 mp in increments of 10 mp. Plot te different stopping distances vs speed. Wat are te implications of tese tabulated results for safe driving? Speed (mp) Dry Stopping Distance (ft) Dry Stopping Time (s) Wet Stopping Distance (ft) Wet Stopping Time (s) Drunk Stopping Distance (ft) Drunk Stopping Time (s) Te stopping distances increase dramatically as te speed increases. Te stopping distances for driving on wet roads and for drunk drivers are muc greater tan tat for dry roads and sober drivers. Te implications are: slow down wen te roads are wet (drive about 10 mp slower tan your typical speed) because te stopping distances are greater and don t drive if drunk. Stopping Distances vs Speed Dry stopping distance (ft) Wet stopping distance (ft) Drunk stopping distance (ft) Stopping distance (ft) Speed (mp)