Weight of vehicles when stopping

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Both a 2000 kg and a 13000 kg vehicle can use the same skid to stop formula, as the weight does not affect the stopping speed derived from the skid distance and drag factor. The calculations demonstrate that both vehicles, despite their weight difference, achieve similar speeds of approximately 61.14 km/h when stopping over a skid distance of 21 meters with a drag factor of 0.7. The use of kinetic energy in the calculations confirms that the stopping speed remains consistent across different vehicle weights. The derived equations validate that the weight's impact is negligible in this context. Thus, the stopping formula is applicable regardless of vehicle weight.
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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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