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ELI5 : Is the amount of energy or Force needed to Brake a Car from 100km/h to 50 km/h bigger than the amount needed to brake from 50km/h to 0km/h

ELI5 : Is the amount of energy or Force needed to Brake a Car from 100km/h to 50 km/h bigger than the amount needed to brake from 50km/h to 0km/h

phiwong

Yes, it is more energy needed. And, yes the braking force is the same for a given deceleration (assuming no drag) Don't confuse the two - energy is not equal to force. For a given force, the time taken to decelerate from 100 to 50 is the same as 50 to 0. But in that time, the vehicle travels much further in the faster car than the slower car, therefore the energy expended is much higher. Energy (work done) = force \* distance


OrganizationMoney701

Thank you this was the most simple answer


Belzeturtle

I'll just add energy is quadratic in velocity, not exponential.


OrganizationMoney701

But why do we need infinite energy to reach the speed of light if we have mass


Chaosfox_Firemaker

Because the explanation given here is an eli5 simplification. In day to day life, far bellow the speed of light kinetic energy is effectively equal to mass\*velocity^(2) / 2 in actuality its: kinetic energy = m*c^2 / (sqrt[1 - (v/c)^2. ]) -mc^2, a mess by comparison. Which means for any thing with mass, the closer you are to light speed, you need to add even more energy to accelerate by the same amount. also the "your mass increases at relativistic speeds" thing is a bad metaphor some professors use that has pervaded discussion.


crossedstaves

> also the "your mass increases at relativistic speeds" thing is a bad metaphor some professors use that has pervaded discussion. Your momentum increases faster than your velocity in a relativistic setting is the best way I can think of to say it.


aCleverGroupofAnts

But since momentum is mass*velocity, to me it's saying exactly the same thing. Is it incorrect to say that mass increases?


middlegroundnb

I'm no physicist, but I think what we consider "mass" in day-to-day terms (how it is affected by forces like gravity, or invariant mass) doesn't change, the relativistic mass does. Since mass is the measurement of an objects resistance to acceleration, time dilation changes this property to an observer - it's relativistic mass increases.


localPhenomnomnom

I've never seen mass referred that way before. Thank you.


maveric_gamer

That's the best ELI5 I've seen for why relativistic mass isn't the same as "your mass increases", so thank you for that.


crossedstaves

It's not *wrong*, but it can be misleading. Remember of course that we have a mass-energy equivalence, so E=mc^2, we can take the total relativistic energy and divide by c^2 and we have what is often referred to as the relativistic mass. There is also the *rest mass* which is the mass when it's at rest, that's what we conventionally think of as mass. Force is defined as the time derivative of momentum, so it can make things cleaner to think of momentum as decoupled from the idea of mass as we conventionally know it. Thinking in terms of energy and momentum just means we don't have to juggle different ideas of mass. Then general relativity comes in and mass is a nebulous thing without a singular straight definition.


Milleuros

Of course this comes mostly to semantic, but to me `E = mc²` is valid for an object _at rest_. The "m" in there is the rest mass. For objects not at rest, the equation becomes `E = γ mc²` , which includes the kinetic energy.


Pixelated_

I thought the full equation with kinetic energy was E2 = m2c4 + p2c2?


eyekwah2

I was watching a video that was discussing the idea behind the Planck length, and it comes from pushing two particles so close together that force of gravity is equal to the electromagnetic repulsion between these two particles, since force of gravity is affected by distance squared. Beyond where they're equal, gravity actually becomes too strong for even light to escape, meaning you'd create an extremely small black hole where literally nothing can be measured. So it was determined from this that a Planck length should be where the two forces are equal because anything smaller could not be detected in any significant way. Things get really weird at the quantum level.


maveric_gamer

> Things get really weird at the quantum level. In this sentence you have summed up most of my understanding of quantum mechanics.


gcross

To clarify: it's not that a black hole really does form in that scenario. If this were how things actually worked, black holes would be popping into existence all the time. It's that because our current theories would seem to predict this, there clearly must be something wrong with them at that scale, and we haven't figured out what that is yet. The problem is that they work *extremely well* at the scales we have been able to probe through observation and experiment, so we are lacking clues to guide us to the solution.


aCleverGroupofAnts

Ahh I see. Thank you for the explanation, this is very interesting! I loved learning about special relativity, though I clearly have forgotten some of it in the years since college. I should see if there are any online courses in this stuff, I'd love to learn more about general relativity and quantum mechanics (and why those two are somehow at odds?) though I only retain so much from going down wikipedia rabbit holes.


maveric_gamer

I'm no expert, but my understanding is that each of our models of physics works up until a certain point, where the formulae no longer apply. For instance, the Newtonian formula we generally use for measuring kinetic energy works up until we get into significant fractions of the speed of light, and the closer we get the more off our results get, and this is where special relativity comes in to "clean things up" so to speak. Similarly, it seems that things at the subatomic level just don't seem to care about things like relativity, and can and will do things like pop in and out of existence sometimes. (As you may be able to tell, as we get further from Newtonian physics, the less familiar I really am and the less I really grok what is going on with them.)


crossedstaves

One word of advice: if you want to learn quantum mechanics you need to commit to the math. The only way to understand quantum mechanics is to understand the math, trying to understand it in terms of the "kooky counter-intuitive" interpretations and descriptions you sometimes see is not a thing that makes sense. The math is the stuff that works, you have to understand it to be able to understand the situation of the quantum world IMO.


lankymjc

Momentum isn’t actually mass*velocity; it gets multiplied by another figure that is much more complicated to figure out. It’s just that at day-to-day speeds (so not approaching the speed of light) that number is extremely close to 1. So we treat it as 1 and the equation comes out as mass*velocity. Close enough for most things, especially anyone doing physics at a level lower than postgrad.


aCleverGroupofAnts

Ah that makes sense. I know lots of stuff gets weird when you go at speeds that fast, but for some reason I thought momentum was consistent. I almost went into physics, but I definitely had a harder time when we started learning about special relativity and then moving frames of reference. I find it interesting and probably could have stuck with it, but ultimately decided to just stick to math.


Zemedelphos

It also relates to how mass and energy are equivalent. Light has no mass, but it has energy, therefore we can show light must have momentum with E = mc^2 therefore E/c^2 = m And thus the momentum of light could be expressed as p = E/c (since for light, v = c, and E/c^2 * c = E/c) And knowing how large c is shows that the momentum of light is very, very low at practically any value of E you'd see in real life, but still makes sense as to how something like a solar sail could work.


Throwaway-donotjudge

This. Can confirm. Source: I am a janitor at Harvard.


Kretenkobr2

> it gets multiplied by another figure that is much more complicated to figure out. It gets multiplied by gamma, the famous 1/ (sqrt[1 - (v/c)2. ])


Special__Occasions

momentum = m*v is an approximation that works for newtonian mechanics. Relativistic momentum (p) is: p= m*v / sqrt(1-v^2 /c^2 ) Rest mass is constant and relativistic mass is an unnecessary concept.


Ewind42

Momentum is mass\*velocity in non relativistic regimes for massive objects


KamikazeArchon

Momentum = mass \* velocity is *also* a simplification that ignores relativistic factors - for example, photons have zero mass but nonzero momentum.


Chaosfox_Firemaker

Pretty much.


woaily

Also not exactly ELI5, but I like the series expansion of relativistic energy. The first term is mc^2 (rest energy, usually ignored classically), the second term is mv^2 /2 (classical kinetic energy), and everything else is powers of (v/c)^2 that only get big at significant fractions of c.


DCSMU

Maybe its just too early in the morning or something, but I'm having difficulty seeing how mc^(2)(1/(1-v^(2)/c^(2) )^(1/2) – 1) simplifies to mv^(2)/2 when velocity is tiny compared to c. (I see how it simplifies to E=mc^2 , but not KE=mv^(2)/2.) Can you clear this up for me and others that may be having a hard time following along? Edit:fixed the superscripts and typos


zebediah49

So, pulling out the mc^2 for later, and using u=v/c... f(u)=((1-u^(2))^(-1/2) -1) Let's take a Maclaurin expansion: f(u) = f(0) + f'(0)u + ... f(u) = 0 + [u/(1 - u^(2))^(3/2)](0)u + ... f(u) = 0 + 0u + ... Not enough terms, so let's go for the next one f(u) = 0 + 0u + f''(0)u^(2)/2 + ... f(u) = 0 + [(3 u^(2))/(1 - u^(2))^(5/2) + 1/(1 - u^(2))^(3/2)](0) u^(2)/2 + ... f(u) = 0 + [0 + 1/(1 -0)^(3/2)](0) u^(2)/2 + ... f(u) = 0 + 1 u^(2)/2 + ... We've got one first-order term, so let's stop there. Dropping it back in... KE ~= mc^(2) f(v/c) KE ~= mc^(2) v^(2)/c^(2)/2 KE ~= m v^(2)/2


Chaosfox_Firemaker

My bad, entered the wrong equation, plug em both in a graphing calc and it should look right now


remarkablemayonaise

Hey, five year olds love doppler shift and emergency vehicles' sirens! If anyone can do red shift for an actual five year old that would be great!


Ghawk134

Doppler shift is really hard to explain without the basic concept of waves. But, if you can teach a 5 year old what a wave is, it's a pretty simple explanation. Imagine a wavy line. On this line, pick any section of a high hump and low hump together, then imagine that the whole line is made up of these repeating humps. That's a representation of sound. When the high and low humps in the air hit our ears, we hear sound. The closer the humps are together, the higher the sound and vice versa. Now, imagine putting a ball at the top of every high hump and imagine the wavy line coming toward you. You'll get hit by a ball every so often, let's imagine every 1 second. Now, imagine you walk forward into the balls. Now you get hit by a ball more often, right? Because the balls haven't changed speed, but you began walking toward them, this causes the balls to seem like they're moving toward you faster, and you get hit more frequently. Remember that the closer together the humps on the line are, the higher the sound we hear. So walking toward a sound causes the sound humps to be closer together for us, making the pitch higher. That's called blue shifting. Now try walking away from the wavy line. Now, the balls hit you less often. They travel faster than you, so you still get hit, but you're running away from them so it takes them longer to hit you. Because the balls are hitting you less often, the humps on the wavy line appear to be more stretched out, meaning we hear a lower sound. That's called red shifting.


Dansiman

It gets extremely easy to ELI5 with the aid of a video. Just imagine a video showing a sine wave propagating to the left (or right, take your pic) and a vertical line representing the observer. Make a little tick sound each time the top of the wave intersects the vertical line. Move the line left or right at a constant speed and the rate of the tick sounds will change. Then just speed up the wave a bunch so that the ticks come very close together (say, 30 or 40 times per second), and then when the vertical line moves left or right at a constant speed, it will actually sound like a pitch shift compared to the stationary line.


Aramec

Or you could just say 'sounds are waves and they get squeezed or stretched when the source moves'


Ghawk134

Merely stating *that* it's true doesn't help anyone understand *why* it's true. My comment was an attempt at creating an intuitive understanding of the doppler effect, not an awareness of its definition.


eyekwah2

Also, semi-unrelated but interesting tidbit that I found out recently is that even massless particles can have energy. Energy is mass times speed of light squared, but that's the rest energy formula. Moving particles have energy given by its momentum, so even massless particles can have energy. Apparently it's pretty important thing to know for particle physicists.


Ghawk134

There's a concept of "rest mass", or the mass of an object that isn't moving. Because you gain relativistic mass when moving, this rest mass is considered as a separate, intrinsic mass. However, once it starts moving any object acquires relativistic mass. A massless particle is called this because it has no rest mass. However, its movement still gives it relativistic mass. For example, photons have no rest mass, but they have relativistic mass, which means they have momentum. This momentum is the force that light sails take advantage of.


CrazyJoe29

Around 100km/h the equations that Isaac Newton (the apple/gravity guy) figures out in the 1700s describe what is happening. But close to the speed of light, which is MILLIONS of times faster, we also need the equations that Albert Einstein (the guy with the moustache and crazy hair) figured out in the early 1900s.


Eggplantosaur

About 10 million times faster, to be more accurate. Speed of light is approx 1 billion km/h


crossedstaves

Because special relativity... which is hard to explain to a five year old. This stuff with F=ma, and KE=1/2mv^2 this is all Newton stuff. But to go the speed of light you need *Einstein/Lorentz* stuff.


AtheistBibleScholar

Because the universe cheats. The speed of light is more of a different category of velocity since things either travel at the speed of light or they don't, and nothing transitions between those options. Things like photons and gravity waves can *only* move at the speed of light, while massive things like you and I can't. That's where the cheating comes in. As you blast away from me in your rocket, you'll always measure a constant acceleration and a constant speed of light inside the rocket. I would also see the same constant speed of light in your rocket because from my perspective (this is the "reference frame" if you've heard the term) you are using a shorter ruler and slower clock to measure the speed of light. As something mathematically similar, imaging a horizontal rope. If I hang a weight from the center, the rope will now be at an angle so that the tension in the rope has an upward component to balance out the weight. The question now is: How much tension do you need to apply to the rope to get it back to horizontal? And the answer is infinite tension. Increasing tension makes the rope more horizontal, but can never get it back to horizontal.


f4f4f4f4f4f4f4f4

What about those experiments in which [light is slowed down with a Bose-Einstein condensate](https://news.harvard.edu/gazette/story/1999/02/physicists-slow-speed-of-light/)?


saylevee

Please don't forget that all formulas are approximations; the natural world has plenty of tiny variance that would be ineffective to account for in each calculation. As well, these approximations are valid only within a certain range (e.g. formulas for energy are different near the speed of light versus near rest)


Villageidiot1984

You don’t need to talk about Lorenz contraction for car speeds


Jager1966

Short answer, the equation approaches division by zero as you get closer to the speed of light.


toodlesandpoodles

Thanks for this correction. Incorrectly calling a change exponential just because the graph has continually increasing slope is a pet peeve of mine.


Achack

Learned this from Donut Media not too long ago. Still haven't taken the time to really wrap my head around it but I've definitely been misusing the term for a long time.


Umbrias

Exponential just means b^x where x is the independent variable, while quadratic or more generally power means x^b, where x is the independent variable, and for quadratic b is 2.


austex3600

Quadratic is.. squared? Like with an exponent? Isn’t that exponential..?


meowtiger

quadratic growth means the number you're squaring increases, exponential growth means the exponent increases


BurnOutBrighter6

Quadratic: y = x^(2) Exponential: y = (constant)^(x) It's just a terminology thing. You're right that quadratic functions do contain exponents. But when you say "exponential function" that means a function where the variable *is in the exponent*. They grow differently and MUCH faster than quadratic functions. Energy is quadratic in velocity, aka energy scales with (velocity)^(2). So 2x the velocity means 4x the energy, 3x the velocity is 9x the energy, etc. If energy was exponential in velocity, then 2x velocity would be (energy)^(2), 3x would be (energy)^(3), etc. which isn't the case.


SerbianShitStain

An exponential function is: a(b)^x A quadratic function is: ax^2 + bx + c


PineTreesandTrailers

I thought any function with exponents were considered exponential. Is that not true?


LesbeaKF

It's when the variable is in the exponent that we call the function exponential. Otherwise, when the exponent is constant and the variable is "below", we call it polynomial.


firelizzard18

“Exponential” means the variable is in the exponent. 2^x is exponential. x^2 is not.


PoorestForm

Another (IMO simple) way to think about it is to look at the kinetic energy of the car at all 3 speeds. 100, 50, and 0 using: E=1/2mv^2 M is the mass, v is the velocity. Mass will be constant so will the 1/2, so basically you’re comparing the v^2 term only. 100^2 = 10,000 50^2 = 2,500 0^2 = 0 From this we can see that from 100 to 50 is a difference of 7500 and 50 to 0 is only a difference of 2500. Please note that these numbers are not in any actual unit like joules, since I didn’t use the mass of the vehicle in the calculation.


[deleted]

A practical demonstration of this is when you brake when driving. To come to a smooth halt you gradually release the pressure on the brake pedal, as less force on the brakes is required to maintain the same rate of deceleration. If you keep your foot still on the pedal, maintaining the same force on the brake, you will eventually just jolt to a stop. With practice, you soften your foot exponentially and get a nice smooth stop instead.


PM_ME_CODE_CALCS

I think that has more to do with the difference between static and dynamic friction. Static fiction is greater than dynamic so as you come to a stop suddenly the brakes reach a threshold where they're more static than dynamic so the brake force shoots up.


Hologram22

Another way to think about it is that for a moving object (force)=(mass)x(acceleration). But the kinetic energy of that object at any given time is (energy)=0.5x(mass)x(velocity)^2. So the amount of energy that needs to be removed from an object in motion at 100m/s to get it to 0m/s is four times as much as would be required to get that same object from 50m/s to 0m/s, even if a steady force is applied so that the acceleration is constant in both cases.


Ghawk134

One trick I like to use for basic kinetic questions is just calculate energy. What's the kinetic energy of the car at 100 kmph, 50kmph, and 0? Or what about 10, 5, and 0 m/s for easy math. .5\*m\*v^2 means that for a 1000 kg car, the kinetic energies are .5\*1000\*100=50,000 J; .5\*1000\*25=12,500 J; and 0. As you can see, 75% of the car's kinetic energy will be dissipated while dropping the first half of its velocity. This is because kinetic energy depends on the square of velocity, making the relationship between velocity and kinetic energy exponential instead of linear. That's the relationship that your intuition has wrong, that the relationship between velocity and kinetic energy is linear.


ThatOtherGuy_CA

This is also why even a difference of 10 km/h can make such a significant difference in stopping times. That's the whole reason for 30 in school zones vs 50 is residential, because you can stop in less than half the distance. https://www.ingenie.com/young-drivers-guide/stopping-distances The above website has some more good explanations and a cool visualization.


WetSound

So by symmetry, more energy is needed to accelerate a car from 50 to 100, than from 0 to 50?


jmlinden7

Assuming a perfectly efficient engine and no drag, yes. However engines aren't perfectly efficient and drag is pretty significant when accelerating.


JimmyDean82

Yes. approx 4x as much actually I believe. Edit. 3x as much. Simple math, Or requires 4x the energy to double the speed, but you’ve used 1 unit for the first half (0-50) so you can subtract that from the the 0-100 calc to get the 50-100 number, so 3x


Flo422

This is correct, 3x the energy is needed.


Drcfan

Its Energy=(1/2)*mass*velocity^2 The velocity term squared increases energy much more


PaddyLandau

This confuses me. The faster car is only faster compared to the Earth that they are driving on. As all speeds are relative, and friction (including air resistance) is being ignored, isn't reducing speed by 50Km/h identical regardless of the starting speed? After all, reducing from 50 to 0 is identical to increasing from 0 to -50 (i.e. in the opposite direction). The situation being described is identical to a rocket in space changing its speed by 50Km/h relative to whatever it's measuring itself against, and therefore direction and initial speed are irrelevant. That's why I don't understand the various comments saying that 100–50 takes three times as much energy as 50–0. Sure, if friction, air resistance, etc. were factors, but they aren't. Please explain to me :)


phiwong

Energy is also relative (similar but not identical to velocity). Two frames of reference do not have to agree on the amount of energy difference imparted or lost on a separate object. If you are stationary and see an object accelerate from 50-100 and I am moving at a constant 50 in the same direction as that object, you and I will not agree on the energy increase in that object. Energy is not a universal measure, it is a relative quantity based on the frame of reference. Since we do not share the same frame of reference, we will not necessarily agree on the energy gained or lost. And we cannot compare our energy measures either - both of us will be "correct" and have different answers. This is VERY counterintuitive but such is physics. 100 to 50 for a car of mass m means that the kinetic energy lost is (1/2)\*m\*(100\^2 - 50\^2) = 3750m. 50 to 0 for a car of mass m means that the kinetic energy lost is (1/2)\*m\*50\^2 = 1250m. Since 3750m = 3\* 1250m, therefore the energy lost going from 100 to 50 is three times that going from 50 to 0.


PaddyLandau

That is so weird! Thank you for this, u/phiwong


ChainringCalf

To continue this, it means that for the same amount of brake pressure, you will go from 100 to 50 or 50 to 0 in the same amount of time, but the brakes will be much hotter after the 100 to 50 run


NinjasOfOrca

So is the answer that it’s the same amount of force but greater energy for the faster car? I believe it, but I’m still having trouble wrapping my head around it


danegraphics

Energy in physics isn't exactly the same as how we use the word "energy" elsewhere. Intuitively, we would think that energy is "force \* time". But in physics, energy is "force \* distance". They're subtilely different concepts that are often confused.


danielv123

I mean, using time is not that wrong. It's just that you also need to add speed. Distance = time * velocity


danegraphics

True.


Christmascrae

An even simpler ELI5: It takes more energy to stop when running fast than when running slow, but you do not press your feet to the ground harder.


the-maxx

i don't think that fits the scenario super well it's intuitive that it takes more energy to stop a car going 100km/h than 50km/h it's not so intuitive to say it takes more energy to slow down your run by 1/2 than 1/2 to stopped


Christmascrae

I see your point — if precision is your concern than the analogy fails. If accuracy is your concern, the analogy holds.


the-maxx

yeah i'm not saying it wasn't a good answer, just (IMO) it's a bit harder to visualize since if i'm running, it 'feels' a lot easier to slow down at higher speed since i'm so out of shape to begin with ....


Christmascrae

I find that interesting. If you were to take 5 steps at a walking pace and stop, that feels more tiring/taxing than getting up to a running pace and stopping (we’re assuming an ideal, graceful stop)?


stdexception

You need to compare: Full running speed -> Half running speed with Half running speed -> Stopped Or 100% to 50% vs 50% to 0%. Of course 100 to 0 is more intense than 50 to 0, that's not what OP was asking.


Christmascrae

Aha. I fooled myself, and now I see.


diox8tony

Bull, its just as good an analogy. In fact it's exactly the same, so he's just repeating OPs scenario. We all intuitively feel that slowing down or turning at high speed is harder, idk how you are unique, in not feeling that when you move. Never ran fast? Never rode a bike fast?


the-maxx

the original question does not specify stopping at the higher speed, just slowing down. so that was one of the issues i had. if original question had just been 'does it take more energy to stop a car at 100kmh than 50kmh) i mean i don't think there would be a discussion here now assuming that parent poster hadn't mixed up full speed to 1/2 and 1/2 to stopped versus 1/2 to stopped, and full to stopped: i still thought it was an interestingly weaker eli5, because you are sort of disconnected physically from the work being done when breaking in a car. it barely takes more effort to depress the breaks a little or a lot, and same in a bike. if you depress the breaks the same when you're at full speed in this scenario, you also don't feel the momentum change difference, there wouldn't be one (as OP points out, don't confuse force with work) My point was also just: lets say you're running full sprint ( and especially if you're in not such good shape, haha) it 'feels' easier to slow down because you're suddenly not fighting so hard against your muscles. this versus you're going at a light jog and you come to a complete stop, feels a bit more jarring, since you have to do additional work to stop from losing your balance after. the point you brought up about changing direction though (turning). To me that makes more sense, as soon as you introduce angular momentum, then it becomes easier to 'feel' that more energy is required since i can picture the wider turn radius more easily. anyway, just sharing my visualization of the thought experiment. results may vary! edit: and just to add, the main thing that OP brought up that i think makes the ELI5 'click' was to be mindful of not just the force you're exerting but the time you spend doing it


[deleted]

[удалено]


Christmascrae

You do not need to measure it. The analogy does not need to be precise, just accurate to our experience. A child quickly learns that it requires less effort to stop walking than to stop running. Trust your gut.


Oudeis16

But... that's a different matter, surely? Yes, more "work" is being done because the car is going farther, but that's not the work of the braking. The question was specifically, does the braking itself take any more energy. Whether the car itself expends more energy or less while the braking happens isn't relevant.


upside_bluemoon

I'm a little bit confused, what am I missing here? Let's say you're in a vacuum in space and you accelerate a 1kg mass (call it A) up to 100km/h, relative to a static point X. Let's do the same with another 1kg mass but up to 50km/h (call it B), relative to a static point X. Your explanation makes sense, but what happens if we introduce a non static frame of reference Y, that follows A's path at 50km/h. At point Y's frame of reference, A is only moving at 50km/h, and it would seem, from Y's frame of reference, that A would need the same amount of energy to slow to 0km/h as B would need to slow to 0km/h from X's frame of reference. The reason I ask this is because it seems analogous to what happens on Earth. I.E. given the Earth's speed through space, or the rotation of the surface of the Earth, the two cars in OPs question are actually moving at thousands of km/h depending on your frame of reference. Hence you'd need lots and lots of energy to move anything on Earth, which isn't the case. What am I missing here?


Arquill

You're right. If you're calculating the energy of an object (either kinetic or potential), it will only make sense in that frame of reference. You can't compare the energy of an object in two different frame of references.


danegraphics

You're missing the strange but true fact that kinetic energy is relative. It depends on the reference frame. Since kinetic energy is based on velocity, and velocity is relative, energy is also relative. We don't even need acceleration to know this. If I'm still and an object passes me at 50km/h, then from my perspective, I have no kinetic energy and the object as a lot. However, from the object's perspective, *I* have all the kinetic energy and it doesn't have any.


phiwong

Energy is also relative (similar but not identical to velocity). Two frames of reference do not have to agree on the amount of energy difference imparted or lost on a separate object. If you are stationary and see an object accelerate from 50-100 and I am moving at a constant 50 in the same direction as that object, you and I will not agree on the energy increase in that object. Energy is not a universal measure, it is a relative quantity based on the frame of reference. Since we do not share the same frame of reference, we will not necessarily agree on the energy equation. This is VERY counterintuitive but such is physics.


upside_bluemoon

Interesting, thank you all for your answers, that makes sense but also raises a further question: ​ From the frame of reference onboard mass A, is the energy used to go from 100km/h to 50km/h the same as to go from 50km/h to 0km/h from the frame of reference of static point X? Because a calculable amount of energy, from let's say an on-board reverse thruster of zero mass for argument's sake, would be required to slow the mass in either case. Which frame of reference is used to calculate the energy required to slow the mass down? In this situation I could say the mass A is moving at 100km/h or equally I could say it's moving at 50km/h (depends on my frame of reference), but there is only one right answer for how much energy would actually be used in slowing the speed by 50km/h (whether that's from 100km/h to 50km/h from static point X or from 50km/h to 0km/h from point non-static point Y) My confusion is that, yes, there are two ways of looking at the relative kinetic energy/speed of mass A (point Y - 50km/h or point X - 100km/h), yet of course there is only one right answer from the point of view onboard mass A. Unless you are saying that from X's perspective you would appear to have used more fuel to slow yourself down than from Y's perspective, which I'm sure isn't the case :D


istasber

IANA physicist, so I can't say whether your answer might be right, but I'm struggling with the explanation you gave. The force doing the work isn't pushing the car in the direction of the car's existing velocity, it's pushing the car in the opposite direction. For simplicity, let's assume the cars weigh the same amount, and the acceleration of the force is 50 km/h/h. Car A would have traveled 100 km in an h without the external force, but instead has now traveled 50 + X km. Car B would have traveled 50 km in an h without the external force, but instead has now traveled Y km. Intuitively, I want to say that Y and X are the same since the time and acceleration are constant, so the cumulative residual velocity over the hour should be the same. If that's the case, both cars would have been displaced the same distance by the force, and the energy should be the same. Is my intuition wrong? Is X < Y?


aCleverGroupofAnts

I have always hated the formulation of "force x distance", it doesn't fit my intuition well at all. If I apply a force to a brick wall for an hour, it doesn't move, so no work was done, but I sure as shit expended a lot of energy!


btspike

The fibers of your muscles are being contracted within your body, so your muscles still do work and expend energy even if you do no work on the wall.


aCleverGroupofAnts

They contract and then hold in place as long as I keep pushing. I can keep pushing for a while after my muscles have contracted, continuing to expend energy. The issue is all the other forms of energy that complicate the problem. There are a lot of cases where you can't simplify it to "force applied to an object multiplied by the distance it moves". There is thermal energy and sound, and those involve countless little particles all moving small distances. It works in hypothetical physics problems where you make assumptions, but my subconscious knows that's not how it plays out in real life.


TraumaMonkey

The work you do to an object is just the force multiplied by the distance. The wasted effort is what is tripping you up. If you didn't move the object (like a wall), then you did no work to the object. Your muscles certainly used up energy, but it doesn't count as work on the wall, just wasted effort.


aCleverGroupofAnts

But work has to be done somewhere since work is equivalent to energy and energy is spent. My understanding is that the work is essentially distributed all over the place in the form of varying amounts of heat, sound, electricity, and indeed small amounts of displacement of yourself, the object, and to an incredibly small degree, the earth. All of this just adds up to me sitting in physics class hearing the teacher say "Energy is force multiplied by distance" and just thinking "well that doesn't sound quite right...".


TraumaMonkey

You're overthinking it. If you lift an object up a few feet from the ground, the potential energy the object gained is FxD, or the work that you did to it. Work usually refers to the useful energy change. All of the other things that occur are wasted energy. The energy your arm expends is higher than what the object gains, but the lessons in class aren't supposed to confuse and overload the students with all of the extra modeling you would have to do to know the entire energy change of a system (which is actually zero if you were paying attention in thermodynamics class!). tl;dr: The simple equations work and are accurate for smaller interactions, and you can add layers to increase accuracy.


aCleverGroupofAnts

I can't help overthinking, when things get too simplified (as in the teacher saying "work is energy" instead of "work is useful energy") I immediately think of real world examples where what they said doesn't fit. Don't get me wrong though, that physics teacher was great and I loved that class. That was just one of the few things that my brain didn't want to accept at the time. Even though I could apply the mathematical formulas and give the correct answers on the exams, I still didn't like how it was explained.


cbf1232

Think about it this way...if you lean a 2x4 against that brick wall it's going to apply a force against the wall but to a first approximation nothing is moving therefore no work is being done and no energy is being expended. If you push really hard against the wall, your muscle fibers are contracting and doing work, but that's just because we're inefficient. Theoretically it would be possible to make something hover in the air at a certain altitude without touching it and without expending any energy. I'd love to see us figure out how to do it.


lankymjc

Everything is elastic - just some stuff is only elastic a little bit. So when you push the wall the bricks are flexing just a little. In layman’s terms, the wall has no elasticity, but that’s just because it has no *detectable* elasticity if you don’t bring specialised equipment like electron microscopes.


aCleverGroupofAnts

That doesn't resolve the issue though. If you think of it like a spring, I can apply a force to compress the spring, but once I compress it a certain distance, I hold it there. I have to continue applying a force, so I am still expending energy, but it is no longer compressing further, just held in place. So no more distance is being travelled at that point, even though I am still expending energy.


humantarget22

Force x distance = kinetic energy. So you aren't adding any kinetic energy to the brick wall when you push on it. The elastic potential energy in a spring is also equal to the force you put in times the distance, so once you've full compressed (or stretched) a spring you won't be adding more potential energy to it. ​ But neither of those cases are to say you aren't expending energy. All the energy you are expending is being converted to thermal energy, you are heating up (and maybe also some chemical energy depending on what reactions are happening internally, I don't know much about biology or chemistry though). So you are right in that you are not adding any more energy to the wall or spring, but there is energy flowing from you to something else, in this case the environment.


aCleverGroupofAnts

Ahhh there we go, thank you! This is exactly the explanation I needed to resolve the cognitive dissonance in my mind.


soniclettuce

Put a clip/brace around the now compressed spring that holds it in place. Is the clip/brace continuously using energy to hold the spring? No. Your muscles being inefficient and wasteful isn't something that's relevant to low-level physics definitions. You are doing no work on the spring / you are not adding any energy to the spring. These definitions are ignoring waste/loss to focus on stuff like potential energy or kinetic energy, especially in the context of the object that is having stuff done to it. A brick holding up another brick loses no energy to do so. Dropping the brick that height will convert X potential energy into exactly X kinetic energy, and then into exactly X waste heat/vibration once the brick stops on the floor. Lifting the brick back to its original position will require X joules of work put in, and at the end, the brick will again have X potential energy.


aCleverGroupofAnts

Exactly, they ignore the waste, but sometimes *everything* is waste, so there are common real-life examples that don't fit the simplified explanation. Saying "energy is work and work is force multiplied by distance" gave me cognitive dissonance when I was learning physics. I didn't know anything about waste/loss because they didn't mention it, so I had no idea it was really just a simplified explanation.


soniclettuce

> work is force multiplied by distance *Work done on the object* is the part I think you're missing. If you compress a spring 1m by applying 1N of force, you've done 1J of work *on the spring* and it now has 1J more potential energy. It doesn't matter if your muscles wasted 100J or if they were perfectly efficient. The *work done on the spring* is defined by force * distance.


aCleverGroupofAnts

Yeah, that's why being clear in the wording helps. Relating back to the original post, it seems counterintuitive to calculate based on the distance the vehicle travelled in the direction opposite the force. I personally would have simply calculated the kinetic energy at 100km/h and the kinetic energy at 50km/h since we have simple formulas for those, but the other approach should also work, right? I almost feel like the "distance" variable should actually be the difference between the distance the vehicle actually travelled and the distance it would have travelled if left alone. Am I overthinking this?


TheSkiGeek

The "distance it would have traveled if left alone" is infinite unless you start dealing with friction, air resistance/drag, etc. Assuming no friction - yes, decelerating it from 100km/h to 50km/h is the same net change of kinetic energy and momentum no matter how quickly or slowly it is done. The total distance traveled during the deceleration will vary depending on quickly you decelerate it.


aCleverGroupofAnts

Ahhh right, my line of thinking there was flawed. So for the same change in kinetic energy, a weaker force would have to be applied for a longer time and therefore a longer distance. So it sounds like the math checks out, but I still feel like it's really counterintuitive to multiply a force by a distance travelled in the opposite direction.


raccoon8182

Thank you, seperate question...is G-Force an acceleration force? In other words if I feel 2g going from 0-50 will I still feel the same G's going from 50-100 in the same time? My actual question which is way more complicated (for me anyway) is....how long would it take for humans to reach the speed of light at a constant 2G force.


phiwong

There is an equivalence between gravity and acceleration. This is generally true but comes with some non-ELI5 caveats (small volumes where tidal forces can be ignored) The Newton formulations of motion, forces, gravity and acceleration are not applicable at very high masses and velocities. The next level "up" in terms of applicability will be to use Lorentz transformations (which are another set of approximations "good" enough for speeds up to perhaps 0.7 times the speed of light). Beyond that, the best currently known theory describing the effects are Einstein's equations. These are pretty much beyond ELI5. (and ELI18 for that matter) In short, there is no amount of acceleration that will ever get an object to light speed. Anything without mass must move at the speed of light and anything with mass will never travel at the speed of light regardless of acceleration. Based on our current understanding of the universe, humans will never travel at the speed of light.


pab_guy

You cannot maintain a constant 2g force. As you approach the speed of light, the energy required to impart such a force get's higher and higher as your mass increases with speed. You can never get there.


ayyeeeeeelmao

Well, you can maintain a constant acceleration. After all, your mass never changes in your own reference frame. But you’re right, it won’t lead to you going at the speed of light. https://en.m.wikipedia.org/wiki/Space_travel_using_constant_acceleration


raccoon8182

This is so interesting!! Thank you


HappinessHappening

Yes, the force felt will be the same, because the force is proportional to the change in momentum which in turn is directly related to the velocity... so 100-50 is the same as 50-0. (Kinetic) energy on the other hand goes as velocity squared, so you can clearly see 100\^2-50\^2 is larger than (and 3 times) 50\^2-0\^2. Of course, this works in non-relativistic scenarios where we assume that mass remains constant remains throughout, which is a very reasonable assumption given the values!


koolman2

>how long would it take for humans to reach the speed of light at a constant 2G force. Infinite time is the answer. You'll never reach the speed of light because the light passing you will always be going the speed of light faster than you.


raccoon8182

Mind... Blown.


koolman2

Oh it gets better. To you, the traveler, you’d surpass what you’d think of as the speed of light - you could go, say, ten light years in the span of a year - BUT to everyone else you’d be gone for over ten years.


GummyKibble

To your first question: yes. To accelerate from x to x+n in a set amount of time, you’ll feel the same force regardless of x (for values of x that aren’t near the speed of light).


CourtJester5

Physics is so interesting 🙂


hucker75

So, my brakes will heat up more from 100-50 than from 50-0, but take the same time for each? Doesn't this mean the brakes could be made to work harder from 50-0 to make me able to stop quicker?


Danne660

Most brakes are good enough to make the wheel stop completely, there is pretty much nothing to improve, working it harder does nothing if it is already 100% effective. The issue with making cars stop comes from the fact that even a non moving wheel will slide on the ground. The resistance between the wheel and the road is not infinite.


hucker75

I see. So my brakes can lock the wheel at a higher speed, but it would generate more heat, but the brakes can handle that. I've braked quite hard from 100mph to 0 without using the gears and they don't mind. I have smelt them when going down a long hill and having them on for ages though.


Danne660

Most modern cars have a system that makes sure that the brakes don't stop the wheels completely, but that is mostly to avoid the car skidding and you losing control. Must have been one hell of a long hill.


hucker75

French Alps.


NinjasOfOrca

I’m confused now. Aren’t energy and work different things?


danegraphics

Work is the amount of energy that has been made useful. So in a sense, work is energy.


Revealed_Jailor

I need some physics friend like you.


asolet

I never liked this. 😅 Speeds are supposed to be relative with no preferred reference frame. We can take 50km/h moving car as in fact stationary (with road just going the other way) and then it only takes less energy to increase the speed (another) 50km/h. I don't like that objects can have any amount of kinetic energy, just depending on the frame, and each of them is equally correct and true amount of energy. Energies are supposed to be absolute and preserved.


Astecheee

Absolutely. Kinetic energy grows in a square. A square that's 50x50 is a quarter of the size of a square that's 100x100. Going from 100x100 down to 50x50 would mean you lose three squares of 50x50, while going from 50x50 to 0x0 you would only lose one square of 50x50. So to be precise, you would have to turn exactly three times as much kinetic energy into heat energy.


PrimarilyBoobs

This is my favorite answer because of the visual analogy.


IvyM1ked

I use the same train of thought but I reduce the zeros to compare speeds when driving in adverse conditions like rain or so. Rather than using (mv^2)/2 in m/s, I simply use 90 kph, reduce to 9 and square to 81. This number can then be compared to say 60 kph (36). Is the reduction in time spent on the road worth the risk of 125% more kinetic energy?


RoryFw

This answer is so good


ArtDSellers

Bravo, good sir or madam. This is a gem of an ELI5 answer.


br094

TIL the method I use to get off expressways (slowly allow the car to decelerate without using brakes) has been saving me brakes. Well, I knew it, I just didn’t think it was THAT impactful.


Astecheee

That's certainly useful. But the biggest factor in brake wear is all the starting and stopping in city traffic.


salgherik

Even though the speed variation from 100 to 50 km/h and 50 to 0 km/h is the same, the energy involved in the first case is three times the second one. So if you apply the same force to stop the same car in the two cases you will find that the faster car will reach 50 km/h in three times the space of the second case. You need to add one more constrain to the problem, either time or space to decelerate. If you constrain by time (the car decelerate within the same time on both cases) the point of u/maximuse_ is true, but the faster car will travel more space due to the higher starting velocity. If you constrain by space (the car decelerate within the same space on both cases) you need to apply more force to the faster car due to more energy to dissipate. This example is by applying force directly to the car chassis, braking will require some equation that will only add more complexity to the problem


maximuse_

Oh cool. I got confused halfway and just deleted my answer


TheUnweeber

This is the right answer.


ChhotaKakua

I might be completely wrong. And this is just me being pedantic. But if there is no friction, there’s no braking. So infinite energy for both cases? Though I do realize you meant friction between the tyre and the road 😅


Soodeau2

I think it's still the case that if there was no friction between the tire and the road, conventional brakes would not stop the car. I would guess they meant air resistance.


XkF21WNJ

Without friction the breaks would be doing no work, so in that particular scenario no energy is lost as heat.


ChiaraStellata

One thing you're overlooking here: at high speeds, much of the braking is done by road friction and air resistance. You can get from 100 to 50 pretty quickly just by letting your foot off the accelerator. Conversely these forces are much lower at low speeds. Therefore the actual work done by the brakes to go from 50 to 0 may be similar or more, because they have to do more of the deceleration on their own.


OrganizationMoney701

I had mentioned friction should be ignored in my post you probably read my post in a hurry :-)


ChiaraStellata

Sorry about that, I did read it too quickly. To be clear, what I'm saying that your intuition for how forces work is distorted because of your experience living on Earth. The same way that we struggle to believe "an object in motion stays in motion" because that's not how things work on Earth, this is another example of how friction and gravity messes with our accurate intuitions about physics.


a_saddler

I love how everybody here is quoting ½mv² and straight up calculating the energy from that. You're all *wrong*. Kinetic energy is relative. It depends on how you're measuring it, and how fast you want acceleration to change. If you want to stop a car from 100 kmh to 50kmh over say 100m, then yes you will need 4x more energy than stopping it from 50 to 0. But if you just want to brake constantly from 100 to 0, then the amount of heat the brakes will produce is the same from 100 to 50 as from 50 to 0. You're just going to need 3 times the distance to break from 100 to 50. You've done 4 times the work, but you didn't spend 4 times the energy because you spread that energy over time.


Autoboat

Thank you for this. Reading the responses here I'm struck that something about the prevailing opinion seems intuitively and inherently wrong. I don't think we've completely nailed it yet but I appreciate seeing the dissenting opinion.


triplesdoubles

I am quite worried how far down this answer was. I think the easiest way you can see the problem with most of the other answers is to use a frame of reference of a car going in the same direction at a constant 50mph. Delta V is still 50mph either way.


Athrunz

couldn't be more wrong it doesn't matter how fast the car stop. it takes the same amount of energy to go from 100 to 0 km/h. you are thinking about power which is energy over time


slinkysuki

Also, "feels like" it needs the same force... Is not accurate. Braking from higher speed either requires more force or more time. The brakes are loosely doing Work = force * distance. If you're driving in a car, the hydraulic brake boosting system makes it very hard to appreciate how much force the car is actually applying to the rotor. If you ride a bike, its easier to notice how much longer or harder you need to squeeze the brakes to stop from higher speeds.


just_a_pyro

Braking force is the same, but braking distance increases at higher speed. If you look at the ballpark formula for braking distance you'll see something like (Speed ​​in km / h: 10) * (speed in km / h: 10) = stopping distance in meters That's where v^2 went, because work is force multiplied by distance, and also work is the change of energy.


WasabiForDinner

I know it was a really technical question, but... is that really how you'd explain it to a five year old?


[deleted]

[удалено]


OrganizationMoney701

It was a bit too technical for me But I read the answer multiple times and Now i understand it's all about the braking distance


wojtekpolska

so "Work" needed to slow car down form 100 to 50 would be higher than from 50 to 0?


souravtxt

Yes.


Syscrush

You're right that the same force applied for the same time will slow a car from 100-50 or 50-0. But when slowing from 100-50, that force is applied through a much longer distance, and work aka energy (work done equals change in energy) is force times distance. Time or velocity are not directly related to work/energy. While we're at it: as slowing from 100-50 involves doing more work in the same time as slowing from 50-0, the power associated with slowing from 100-50 is greater than the power associated with slowing from 50-0 - since power is work done divided by time.


pab_guy

While I totally get this and accept the math, I always have a hard time relating this to the relative nature of motion. Like with a rocket, we calculate delta-v. A given rocket has a total amount of delta-v capacity. Whether that rocket materializes in orbit at high velocity, or not, it has the same amount of delta-v. And then we have the Oberth effect: [https://en.wikipedia.org/wiki/Oberth\_effect](https://en.wikipedia.org/wiki/Oberth_effect) Which seems to be the opposite of what we are saying here: using the same amount of rocket fuel results in a greater change in velocity when going faster. It's weird because I can follow the math, but I can't intuit what's going on....


TheSkiGeek

>In terms of the energies involved, the Oberth effect is more effective at higher speeds because at high speed the propellant has significant kinetic energy in addition to its chemical potential energy.[2]:204 At higher speed the vehicle is able to employ the greater change (reduction) in kinetic energy of the propellant (as it is exhausted backwards and hence at reduced speed and hence reduced kinetic energy) to generate a greater increase in kinetic energy of the vehicle. This is specific to something like a rocket/jet engine, where you're generating thrust by pushing/throwing mass away from the "vehicle".


pab_guy

Ahh I forgot about conservation of momentum....


fallout_poster

Great question. I also want to know the answer. But I don't like the use of "braking" to illustrate his point, so I want to change it. Instead, mount a rocket engine on the hood of the car, and rephrase the question. Is the amount of energy used by the rocket to slow the car from 100 mph to 50 mph, the same amount of energy needed to slow the car from 50 mph to a stop? This clarifies the question to it's purest form, IMO. I don't know the answer, and I want to. I think everyone else here is wrong because they didn't ask the question this way and then answer it this way.


Iazo

No. Your rocket will consume 3 times as much energy to slow your car from 100 to 50 than to stop it from 50 to 0.


fallout_poster

Okay, great answer. Why? How? Etc... Why 3 times. Why not 4 times. What if it's from 200 mph to 150 mph. Then what's the energy difference and why?


TheSkiGeek

Kinetic energy is equal to mass times velocity squared. ``` Energy at 0 mph = 0 Energy at 50 mph = (mass) * 2500 Energy at 100mph = (mass) * 10000 ``` Therefore, accelerating from 0-50 (or 50-0) takes `m*2500` energy. Accelerating from 0-100 (or 100-0) takes `m*10000` energy, or 4x as much as 0-50. Accelerating from 50-100 (or 100-50) takes the difference of those, or `m*7500` energy, or 3x as much as 0-50.


Derangedteddy

>assume there's no friction If there's no friction then you can't brake. Did you mean air resistance?


OrganizationMoney701

i meant ignore all sorts of forces The only force acting on the car is a magical brake force


TheJeeronian

The force is the same, but the distance is different. Let's say that, for a given braking force, the rotors yield a kilojoule of braking energy. This is a simple multiple of the torque applied by the brakes by the radians in one full rotation. So, since the force is the same regardless of speed, each rotation yields the same amount of energy regardless of speed. However, at higher speeds, you get more rotations every second, and so you drain the energy of the car faster.


OrganizationMoney701

A bit more technical but still a Great answer


Foxnooku

If we assume no friction then you're definitely not braking until you hit something (but I know what you mean)


TheShroomHermit

Should ELI5 be for questions that can be answered with yes or no?


OrganizationMoney701

No there's no point in answering questions like this with a yes or no because the reason I asked it the first place is to understand exactly what the difference is assuming there's no other force except the applied force


PaxNova

Pretend you're stopping it with your hands instead of with mechanical brakes. Will the impact from stopping a 100km/h car be more than the impact from a 50 km/h car? Pretend each time you hit it, you're only slowing it 1km/h. In other words, you hit the car fifty times. Car A hits you at 100 km/h, 99 km/h, 98, etc., down to 51 km/h. Car B starts at 50 and goes to 1. Is the impact from the 51 km/h crash only twice the impact from the 1 km/h crash, or is it a lot more? This is ELI5 level calculus. Edit: From the downvotes, I assume I'm wrong. What am I doing wrong?


BA3HENOV

In real life, the force of friction on a car increases with velocity. That's probably why slowing down from 100 kmh to 50 seems as easy as slowing down from 50 to 0. According to the Law of Conservation of Energy, the energy required to speed up a car from velocity 1 to velocity 2 is equal to the energy you take away from the car when you slow from v2 to v1. You tell me: what's harder, speeding up from 0 to 50 or 50 to 100? Side note: an exponential function has the independent variable (velocity, in this case) in the exponent. The equation in question has v^2, so we'd call it *polynomial* growth, or more specifically for a 2^nd order equation: *quadratic* growth


manofoar

E=1/2mv^2. So, the energy to go from 100kph to 50kph is exponentially higher than the energy required to go from 50kph to 0kph.


the_snook

Quadratic, not exponential.


soundoftherain

Another way of framing things: Imagine you were driving on a road at 50 km/h and saw a deer, hit the brakes as hard as you could and stopped inches from it. Under those same conditions, if you were driving 100 km/h, you would hit the deer going \~86 km/h.


filipv

Yes. Kinetic energy (the one that you need to counter with braking) depends on the square of the speed.


Jokieh

Yes.If the car's kinetic energy is "x" at 100 km/h, it will be x / 2^(2) at 50 km/h because of the exponential relationship between the energy and velocity in the formula. So from 100 to 50 km/h the car loses x - x/4 = 3/4 of its kinetic energy. From 50 to 0 it loses 1/4.So the work (force applied over time by the brakes) needed for halving the speed is 3 times as much as going from half speed to full stop.


y4mat3

The energy required is bigger because kinetic energy is proprotional to velocity squared (KE = 1/2×m×v^2) so while the difference between 100 and 50 is the same as the difference between 50 and 0, the difference between 100^2 and 50^ is greater than the difference between 50^2 and 0^2.


BenCub3d

So you mentioned that kinetic energy is equal to 1/2mv^(2). Due to the conservation of energy, the energy that is dissipated by forces such as heat and sound is exactly equal to the difference between the kinetic energy before and after breaking. So you can say that the energy lost by breaking is 1/2mv2^(2) - 1/2mv1^(2), or more simply > Energy Change = 1/2m(v2^(2) - v1^(2)). You can see from this equation that the energy required to slow something down increases exponentially with velocity. 100^(2) - 50^(2) = 7500 50^(2) - 0^(2) = 2500 EDIT: Fixed equation


desolat0r

> Kinetic energy seems to be exponential according to ½mv² *squared. It would be exponential if it was ½mq^v where q would be a real number greater than 1.


uber_poutine

Let's say that a car weighs 1000 kg. Ignoring friction and air resistance, the energy is absorbed by braking. 100km/hr to 50km/hr is equal to going from 27.78m/s to 13.89m/s. You plug this into the equation Ebrake = [email protected]/hr - [email protected]/hr to get the energy difference between the two, since that's what's being absorbed by the brakes. As you mentioned, Ek = ½mv². So, plugging all of this in: 0.5\*1000kg\*(27.78m/s)\^2 - 0.5\*1000kg\*(13.89m/s)\^2 = 385864.2J - 96466.05J = 289398.15J absorbed deccelerating from 100km/hr to 50km/hr To go from 50km/hr to 0km/hr, this is the same as going from 13.89m/s to 0m/s. Plugging that into the equation, we get: 0.5\*1000kg\*(13.89m/s)\^2 - 0.5\*1000kg\*(0m/s)\^2 = 96466.05J - 0J = 96466.05J absorbed deccelerating from 50km/hr to 0Km/hr. ​ As you can see, 289398.15J is bigger than 96466.05J.


sckego

This is also why brake rotors sizing is driven more by vehicle speed than weight. I often see new riders wondering why a 400-lb superbike needs two big front rotors, while an 800-lb cruiser can get by with only one… it’s because hauling that 400-lb bike from 150mph down to 50 generates twice as much heat (kinetic energy turning into friction) as slowing the 800-lb bike from 90 to 50.


Rewiistdummlolxd

How do you break without friction?


OrganizationMoney701

I meant assume there's no friction and the brake force is just magical and can be Applied without taking friction into consideration I only have a high school understanding and most of the time we had to ignore Friction even though we knew how to calculate it