Weight of vehicles when stopping

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SUMMARY

Both a 2000 kg vehicle and a 13000 kg vehicle can utilize the same skid-to-stop formula, demonstrating that weight does not significantly affect the stopping speed when using the formula provided. The calculations show that both vehicles, despite their weight difference, achieve a stopping speed of approximately 61.14 km/h when the skid distance is 21 meters and the drag factor is 0.7. The kinetic energy approach confirms that the stopping speed remains consistent across different vehicle weights under these conditions.

PREREQUISITES
  • Understanding of basic physics principles, specifically kinetic energy.
  • Familiarity with the skid-to-stop formula and its variables.
  • Knowledge of drag factors and their impact on vehicle dynamics.
  • Ability to perform calculations involving square roots and basic algebra.
NEXT STEPS
  • Explore the derivation of the skid-to-stop formula in detail.
  • Research the effects of different drag factors on stopping distances.
  • Learn about the relationship between vehicle weight and braking distance in various conditions.
  • Investigate advanced vehicle dynamics and their implications for safety and performance.
USEFUL FOR

Students in physics or engineering, automotive safety researchers, and professionals involved in vehicle dynamics and accident reconstruction will benefit from this discussion.

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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.
 
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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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