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ongzoe01

Science

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55 mph ≈ 80.7 ft/s 

trucks: 

Ek = ½mv² = ½ * (8730lb / 32.2 ft/s²) * (80.7ft/s)² = 882 100 ft·lb 

worst case friction: Ffw = µmg = 0.55 * 8730lb = 4802 lb 

→ stopping distance d = Ek / Ffw = 184 ft 

best case friction: Ffb = 0.941 * 8730lb = 8215 lb 

→ stopping distance d = Ek / Ffb = 107 ft 

bugs: 

Ek = ½ * (1190lb / 32.2ft/s²) * (80.7ft/s)² = 120 240 ft·lb 

worst case friction: Ffw = 0.55 * 1190lb = 655 lb 

→ stopping distance d = 184 ft 

best case friction: Ffb = 0.941 * 1190lb = 1120 lb 

→ stopping distance d = 107 ft 

Given that the maximum allowable distance is 155 ft, we've got to reduce the maximum allowable Ek of the vehicles, and it appears not to matter which one we analyze. 

worst case friction for bug over 155 ft entails Work = 655lb * 155ft = 101 525 ft·lb

This corresponds to Ek = 101 525 ft·lb = ½ * (1190lb / 32.2ft/s²) * v² 

→ v ≈ 74 ft/s ≈ 50 mph ← maximum desired speed limit


Please let me know if you need any clarification. I'm always happy to answer your questions.


Anonymous
Very useful material for studying!

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