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| Oct5-07, 05:43 PM | #1 |
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derivation
1. The problem statement, all variables and given/known data
Is it possible that a vehicle weighing 2000 kg and a vehicle weighing 13000 kg can both use the same skid to stop formula?? Does the difference in weight not come into account? 2. Relevant equations Where : S = speed in km/h µ = .75 d = skid distance S = 15.9 *( sgr ( µ d )) 3. The attempt at a solution I know how to do the above equation. It is the derivation I do not know how to do. in order to prove or disprove the weight effect on the speed. |
| Oct6-07, 04:31 PM | #2 |
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Thanks, but I proved this one myself, with the use of kinetic energy. But if someone would like to check this over to make sure I am correct I would appreciate it.
13000 kg vehicle leaves skid marks 21 meters long, the drag factor is .7 13000 * 21 * .7 = 191100 joules V = sqr ((2gKE)/w) V = sqr ((2* 9.81*191100))/13000) V = sqr (3749382/13000) V = sqr 288.414 V = 16.98 m/s S = 16.98 / .2777 S = 61.14 km/h 2000 kg vehicle leaves skid marks 21 meters long, the drag factor is .7 2000 * 21 * .7 = 29400 joules V = sqr ((2gKE)/w) V = sqr ((2* 9.81*29400))/2000) V = sqr (576828/2000) V = sqr 288.414 V = 16.98 m/s S = 16.98 / .2777 S = 61.14 km/h Check skid to stop S = sqr (254 µ d )) S = sqr (254 * .7 * 21)) S = sqr 3733.8 S = 61.10 km/h |
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