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Weight of vehicles when stopping

  • Thread starter Probie
  • Start date
  • #1
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Homework Statement



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?


Homework Equations



Where :
S = speed in km/h
µ = .75
d = skid distance

S = 15.9 *( sgr ( µ d ))

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.
 

Answers and Replies

  • #2
24
0
solved myself

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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