Stopping Distance

[MolaKule]

The simplest equation for stopping distance of a vehicle is

Workdue to friction = - umgd = - 1/2 mV^2.

u - friction coefficient, m is mass of the vehicle (kg), v is the vehicle's speed (m/s), and g is gravitational acceleration = 9.8m/s^2.

So solving for stopping distance d = V^2/(2ug).

A car that was moving at 50 mph is stopping on a dry, flat surface in which the tire/surface friction coefficient is u = 0.7 without skidding.

What is the stopping distance in feet?

 

[skyactiv]

I feel dumb, I know that is kinematics which is above the geometry and algebra I took in high school and couldn't solve that to save my life.

 

[Leo99]

I'm getting 119 feet.

 

[MolaKule]

I feel dumb, I know that is kinematics which is above the geometry and algebra I took in high school and couldn't solve that to save my life.

No problem. Convert v = 50mph to meters per second m/s, square it, and then divide by 2 X u X g. Multiply the result (meters) by 3.281 to get feet.

 

[AZjeff]

119.45

Not familiar with the ^ symbol.

 

[Leo99]

119.45

Not familiar with the ^ symbol.

Exponent. 10^2 is 10 squared.

 

[MolaKule]

119.45

Not familiar with the ^ symbol.

Here it means square.the value to the left, as in vxv.

 

[XCIDMigs]

That is not really the simply way to do that....

The standard distance to skid to stop formula used in crash recon math is...

D=S^2/30f so D=50^/30(0.7) = 2500/21 = 119.04 ft


in your example you have replaced the accepted coef of friction value with "u" for some reason but the math works out the same # without any need to convert to metric system or anything else, if you want an english unit answer seems easier...

 

[Gyro Gearloose]

That is not really the simply way to do that....

The standard distance to skid to stop formula used in crash recon math is...

D=S^2/30f so D=50^/30(0.7) = 2500/21 = 119.04 ft


in your example you have replaced the accepted coef of friction value with "u" for some reason but the math works out the same # without any need to convert to metric system or anything else, if you want an english unit answer seems easier...

Mola's equations are the classical physics teacher version with mixed units of measure for extra bonus. (Typical physics teacher trick)

The crash recon formula is a partially precalculated formula where terms have been combined and the conversion from mph to feet/sec have been rolled into the number used for g. That simplifies the math (and reduces opportunity for errors) for the user, but it limits the formula to only being valid for mph speed and feet stopping distance. Canada would need a completely different g number for their crash recon formula, which I presume would need to be in kph/meters.

 

[MolaKule]

That is not really the simply way to do that....

The standard distance to skid to stop formula used in crash recon math is...

D=S^2/30f so D=50^/30(0.7) = 2500/21 = 119.04 ft


in your example you have replaced the accepted coef of friction value with "u" for some reason but the math works out the same # without any need to convert to metric system or anything else, if you want an english unit answer seems easier...

Gyro Gearloose is correct.

In today's physics courses we use the metric system exclusively and as an exercise we state English units in the problem so the student can gain experience in making conversions.

In terms of friction coefficients u is the accepted symbol for that coefficient in modern-day mechanics, physics, and Tribology literature.

I fail to see your point.

 

[Hohn]

The simplest equation for stopping distance of a vehicle is

Workdue to friction = - umgd = - 1/2 mV^2.

u - friction coefficient, m is mass of the vehicle (kg), v is the vehicle's speed (m/s), and g is gravitational acceleration = 9.8m/s^2.

So solving for stopping distance d = V^2/(2ug).

A car that was moving at 50 mph is stopping on a dry, flat surface in which the tire/surface friction coefficient is u = 0.7 without skidding.

What is the stopping distance in feet?

Work is force times distance. So umgd breaks down as just the braking force (umg) times distance. The braking force further breaks down as essentially the max friction, which is Mu time the normal force (MG, mass time gravity).

50mph= 22.352 m/sec. Round to 22.4m/s

D= 22.4m/s^2/ (2*0.7*9.8)
D=501.76 m^2/s^2 / 13.72m/s^2
-- seconds squared cancel, meters squared is reduced to just meters.
D=501.76/13.72 meters
D=36.57m or ~119.98 feet.

Call it 120 feet if you're a sig fig pedant. The tedium of English units is really just worse the more comfortable you get with regular work in Metric.

"Metric or Standard?"
My reply: METRIC *IS* THE STANDARD.

 

[MolaKule]

There are three main reasons for QOTD:

Educate

Review automotive mechanics, lubricant chemistry, and physics

Have some fun.

Off topic complaints about units and other minute are neither welcome nor warranted.

 

[NMP]

As an accident investigator I always used drag factor rather than coefficient of friction. The difference is drag factor corrects for slope/ grade whereas coefficient of friction does not.

 

[MolaKule]

See https://cfm-calculator.com/calculator.php?utm_source=/physics/Stopping-Distance-Calculator.php

and includes the reaction time of the driver.

which shows it is a bit more complex. I am using typical physics text equations to keep things simple.

The opposing aerodynamic drag force FAd is about - 278N.

So if anyone wants to determine how the FAd drag force would affect the stopping distance, please calculate away.

 

[Surestick]

As an accident investigator I always used drag factor rather than coefficient of friction. The difference is drag factor corrects for slope/ grade whereas coefficient of friction does not.

Respect for what you guys do. I've seen reports that correlate very closely to the actual speeds of vehicles involved when they're available (dash cam or other sources).
Also the info that can be obtained from a modern car's computer(s) system post-collision is amazing.

 

[MolaKule]

The NHTSB ran experiments and tests to correlate actual physical crashes with the mathematics to provide crash investigators with more accurate software to quickly make determinations.

This SAE document, J2505, Measurement of Vehicle-Roadway Frictional Drag, presents the drag factor calculations in Section 5:

and is uploadable as a PDF file.

PREPARED BY THE SAE ACCIDENT INVESTIGATION AND RECONSTRUCTION PRACTICES COMMITTEE

 

[NMP]

Reaction time is somewhat of a crapshoot. University of Michigan had a professor who spent a great deal of research time coming to the conclusion "about a second and a half". Except for reduced visibility, distractions, alcohol, drugs, fatigue, unexpected information, you name it.
Regardless of the accuracy of the physics there is always a certain unpredictability of the seat / steering wheel interface.

 

[MolaKule]

Except for reduced visibility, distractions, alcohol, drugs, fatigue, unexpected information, you name it.
Regardless of the accuracy of the physics there is always a certain unpredictability of the seat / steering wheel interface.

Human factors is a major topic in many incidents, whether it be terrestrial vehicles or flying vehicles.

See https://cfm-calculator.com/calculator.php?utm_source=/physics/Stopping-Distance-Calculator.php

and includes the reaction time of the driver.

This is why many advanced SD calculators include a Delta Reaction Time factor to graph the expected versus "on-scene" observations. 1.5 seconds is simply a global average for unimpeded response times.

 

[NMP]

"On scene observations"? As in 'you were lucky enough to see the whole thing'? The reason Dr. Olsen came up with his conclusion (as I understand it) is there are just too many variables in perception / reaction. For example: Driver 1 is on a straight road with no view obstructions. Every mile there is an intersecting farm road controlled by a stop sign. Driver 1 observes Driver 2 approaching the intersection. Driver 2 is slowing down. Driver 1 anticipates driver 2 will stop and then shifts his attention away from driver 2. Driver 2 fails to stop and driver 1 has to realize that the situation has changed.

It's the sort of thing that creates overtime for cops and employment for expert witnesses. I was both and I'm happy to say that today, NMP (Not my problem).

 

[MolaKule]

"On scene observations"? As in 'you were lucky enough to see the whole thing'?

Of course not. Thr reference is to post crash observations verses the expected values from the SD software based on J2505 equations.