NationStates Jolt Archive

I quote my california drivers handbook while talking about the speed limit: "The force of a 60mph crash isn't just twice as great as a 30mph crash, it's four times as great!"

Now, I've heard this before. However, with the semester of physics I took(obviously not much, but still), it seems to me that this just doesn't make sense.

Car traveling at 30mph(48.269kmh).

Assume car weight is 3000kg.

Force is therefore 144807N, as Force = (Mass)(Acceleration)

Doubling the speed to 60mph(96.539kmh) makes the Force = 289617

Ok, so the numbers are a bit off, mostly due to how I calculated kmh... but still, it seems pretty linear to me... Am I screwing up? misinterpreting acceleration?

I quote my california drivers handbook while talking about the speed limit: "The force of a 50mph crash isn't just twice as great as a 30mph crash, it's four times as great!"

Now, I've heard this before. However, with the semester of physics I took(obviously not much, but still), it seems to me that this just doesn't make sense.

Car traveling at 30mph(48.269kmh).

Assume car weight is 3000kg.

Force is therefore 144807N, as Force = (Mass)(Acceleration)

Doubling the speed to 60mph(96.539kmh) makes the Force = 289617

Ok, so the numbers are a bit off, mostly due to how I calculated kmh... but still, it seems pretty linear to me... Am I screwing up? misinterpreting acceleration?
First, they are idiots. When are car increase its speed from 30mph to 50 mph, it isn't going twice as fast. It is going 66.6r% faster.
Second, LG proved I am an idiot for ignoring the crumple zone. Its ability to absorb energy increase at a rate much slower than the speed increases.

Well, I can't be sure as there isn't enough information.

Are they referring to a head-on crash? That'd be a quick and easy answer.

But when calculating the force of the impact of these two cars, they might not be measuring in a perfect system. They might be taking into account crumpling and other ways that cars absorb impact. Afterall, a car will absorb considerably less of a higher speed collision.

But your basic calculation is right. Assuming equal acceleration(the negative acceleration of the sudden stop)of a moving object into a stationary one, the force is linearly proportional to the velocity at the time of impact. In a perfect system.

I knew I would fuck up... its 60mph, not 50.

Second, they make no mention of situation. The line before was "The faster you go, the less time you have to avoid a hazard or accident".

Dunno much aqbout physics either, but I notice you're using acceleration there. A car being driven at 30 mph (or 50mph or 60 or whathaveyou) isn't accelerating.
I assume to accelerate to that speed would take twice as much energy, due to the inertia, momentum stuff. So if the force of the impact uses a formula where that is squared, you would have your 'four times the force' claim.

Actually, the acceleration comes from going from 30mph to 0. That would be the crash. :)

I knew I would fuck up... its 60mph, not 50.

Second, they make no mention of situation. The line before was "The faster you go, the less time you have to avoid a hazard or accident".
Here the line is "The faster you go, the bigger the mess"..
Anyway, I'm too tired for physics heh.

Accelleration isnt part of it. Energy is mass multiplied by velocity squared.

Plug the numbers into that equation and see what you get.

Accelleration isnt part of it. Energy is mass multiplied by velocity squared.

Plug the numbers into that equation and see what you get.

I like you. You're silly. :)

KE = mv²/2
or
KE = 1/2 mv²

where KE is kinetic energy, m is mass, and v is velocity.

I like you. You're silly. :)

I like you too. You are silly for liking me though.

I quote my california drivers handbook while talking about the speed limit: "The force of a 60mph crash isn't just twice as great as a 30mph crash, it's four times as great!"

Now, I've heard this before. However, with the semester of physics I took(obviously not much, but still), it seems to me that this just doesn't make sense.

Car traveling at 30mph(48.269kmh).

Assume car weight is 3000kg.

Force is therefore 144807N, as Force = (Mass)(Acceleration)

Doubling the speed to 60mph(96.539kmh) makes the Force = 289617

Ok, so the numbers are a bit off, mostly due to how I calculated kmh... but still, it seems pretty linear to me... Am I screwing up? misinterpreting acceleration?

Why are using km/hr? The equation F=ma is for metres per second per second. Also 60mph is a velocity not an acceleration. This means you need a time frame for the change in velocity to occur to get the acceleration. The safety features in cars are also ignored and unless you knew much about them it would be hard to say what a change in acceleration would do to force felt by certain parts of the car.

I'll stop now before I start blithering on incomprehensively ...any more.

I like you too. You are silly for liking me though.
http://www.clicksmilies.com/s0105/aetsch/cheeky-smiley-004.gif

I like you too. You are silly for liking me though.
LG does not need another reason for being silly.

I agree with the (1/2)m(v^2) for kinetic energy, giving an exponential increase for the velocity term for KE; I think that when discussing the "force" of the crash, more would need to be known about the elasticity/inelasticity of the collision. I imagine much would be lost to various thermal, audial, and dissipative forces, yes?

Generally, these kinds of things are sometimes modeled with an "impulse" event, where the energy is transmitted from one part to another via the contact, aren't they?

Why are using km/hr? The equation F=ma is for metres per second per second. Also 60mph is a velocity not an acceleration. This means you need a time frame for the change in velocity to occur to get the acceleration. The safety features in cars are also ignored and unless you knew much about them it would be hard to say what a change in acceleration would do to force felt by certain parts of the car.

I'll stop now before I start blithering on incomprehensively ...any more.

Hmm...

Did you actually read my post? I think I made it pretty fucking clear that I was simply trying to figure out the statement with my basic comprehension of physics. I wouldn't have posted this, or at least not in the manner I did, if I thought I was totally correct. I was asking for people to correct me, so I could learn.

I see nothing that warranted you acting like a total jackass and berating me for simply asking for help.

Except using the wrong unit for acceleration, it still looks like my basic assumption was right. The ratio wouldn't change, just the newtons. The speeds were velocities, but as others have said, its the deacceleration of the car(the acceleration the other car is feeling, due to the transferral of energy). The car being hit would be, theoretically in a vacuum, accelerating 60mph the opposite direction. This is all afaik, again.

Safety features? The booklet made a broad assumption for all cars, so as for safety measures, the only thing that comes close to equally applying is what LG said about absorbtion of the force.

Stick around on the forums a bit longer, learn to be a bit more civil, then make posts. It will take you a long way.

I think the point was(in drivers ed), mostly, that a doubled velocity means it takes 4 times as much force to stop your car... not that the impact is 4 times as great.

on another note...

Why are using km/hr? The equation F=ma is for metres per second per second. Also 60mph is a velocity not an acceleration. This means you need a time frame for the change in velocity to occur to get the acceleration. The safety features in cars are also ignored and unless you knew much about them it would be hard to say what a change in acceleration would do to force felt by certain parts of the car.

I'll stop now before I start blithering on incomprehensively ...any more.


Did you actually read my post? I think I made it pretty fucking clear...

He sounds a bit more civil than you, I think. No offense.

I think the point was(in drivers ed), mostly, that a doubled velocity means it takes 4 times as much force to stop your car... not that the impact is 4 times as great.

on another note...





He sounds a bit more civil than you, I think. No offense.

When someone is an asshole to me, I respond in kind usually. I have my limits :P

But how was he an asshole? He was not profane; he accounted for assumptions you failed to make... Granted, I think you were looking for the mathematical explanation rather than the practical aspects of a crash...

But how was he an asshole? He was not profane; he accounted for assumptions you failed to make... Granted, I think you were looking for the mathematical explanation rather than the practical aspects of a crash...

Woah! I just made a huge mistake.

I read what he wrote at the end as: I'd stop now before you start blithering on incomprehensively ...any more.

I apologize profoundly. Its late.

heh, word.

goodnight moon

Why're we going into physics? Simply math works.

30mph + 30mph = 1
60mph + 30mph = 2
60mph + 60mph = 4

That's what they were thinking. A vehicle moving at 60mph and impacting another vehicle moving at 60mph would do twice as much as impacting a vehicle moving at 30mph, and four times as much than if impacting at 30mph. Whether this is right, I don't know or care. I don't see how you can quantify damage.

Why're we going into physics? Simply math works.

30mph + 30mph = 1
60mph + 30mph = 2
60mph + 60mph = 4

That's what they were thinking. A vehicle moving at 60mph and impacting another vehicle moving at 60mph would do twice as much as impacting a vehicle moving at 30mph, and four times as much than if impacting at 30mph. Whether this is right, I don't know or care. I don't see how you can quantify damage.

But how often do you see head on head crashes? Thats also assuming both cars are going twice as fast. The statement is simply that the force of a crash is exponentially greater as you linearly increase your speed.

Though I agree, its not simple math, and not always the same.

But how often do you see head on head crashes? Thats also assuming both cars are going twice as fast. The statement is simply that the force of a crash is exponentially greater as you linearly increase your speed.

Though I agree, its not simple math, and not always the same.

It may be that they are not applying the term "force" in a rigorously precise manner. Perhaps they are confusing force in Newtons with Kinetic Energy in Joules.

Again, I think in Newtonian mechanics, in a collision event (ablative structural effects, thermal and sound energy aspects notwithstanding), force is transmitted between the objects in an impulse event, wherein force is applied over a brief but sharp period, the total work transmitted being the area under that spike (in the graph of Force over time, intervalled from beginning of contact to end).

This is a driving manual, not a physics handbook. They aren't using the mathematical definition of force. They're using force as a synonym for strength, power, energy, etc. In language force, energy and power have very similar meanings, whereas in physics they are precisely defined and there is no room for overlap.

As others have said, they are talking about kinetic energy, where energy is proportional to the square of the car's velocity.

I think they generally look at the breaking distance, which also increases with the square of the velocity. (roughly)
(Which makes sense if you assume constant friction in breaking, all kinetic energy has to be absorbed after all)

This is a driving manual, not a physics handbook. They aren't using the mathematical definition of force. They're using force as a synonym for strength and the like.

As others have said, they are talking about kinetic energy, where energy is proportional to the square of the car's velocity.

I wonder if the point might better have been made by relating fatalities as a function of velocity (if such data exists).

It would seem that if you say there are y deaths at speed Alpha and z deaths at speed Beta, the z/y might be more evocative.

(note, I use only the scalar value of speed and not the possibly more correct vector of velocity, for simplicity)

I think they generally look at the breaking distance, which also increases with the square of the velocity. (roughly)
(Which makes sense if you assume constant friction in breaking, all kinetic energy has to be absorbed after all)

Good point.

While contemporary brakes sometimes vary their action to avoid locking, I think a constant coefficient of friction is reasonable for a short event.

I wonder if the point might better have been made by relating fatalities as a function of velocity (if such data exists).

It would seem that if you say there are y deaths at speed Alpha and z deaths at speed Beta, the z/y might be more evocative.

(note, I use only the scalar value of speed and not the possibly more correct vector of velocity, for simplicity)http://www.safespeed.org.uk/killspeed.gifAt 20 miles an hour there is a 10% chance the child pedestrian will be killed. At 40 miles an hour there is 10% chance the child pedestrian won't be killed.

http://www.safespeed.org.uk/killspeed.gifAt 20 miles an hour there is a 10% chance the child pedestrian will be killed. At 40 miles an hour there is 10% chance the child pedestrian won't be killed.

Well, there you go, then.

Well, there you go, then.I know this infomation from all the lovely UK car accident ads. No blood and guts, but they're definately trying to shock you.

I know this infomation from all the lovely UK car accident ads. No blood and guts, but they're definately trying to shock you.

In Las Vegas, the problem is particularly acute. Accidents occur disproportionately often, and its not just tourists. Our insurance almost doubled when we moved here.

That graph only shows you statistics about when you do hit someone at a certain speed. Btu the chance that happens also increases with speed, because you have less time to react when something props up in front of you.
Any statistics on that?