In Germany, this (the quadratic behavior of braking distance in relation to speed) is required knowledge to get a driving license. The test questions even include how far you travel in the second you need to react.
In Finland, it’s not really required; as the theoretic questions are all about: ”Here’s the set-up. What’s the right thing to do?”; but they drive it home (OK; bad pun is bad), pretty well. They give you a chance to ride a car-crash-simulator (basically, it’s a chair that slides down a slope, to a sudden halt) that simulates a car-crash, at 7 km/h. I took it, and let me tell you, if the seatbelt wasn’t there, I would have gone to the orbit. Then, they give you something to think about: In Finland, the speed limit, for most city-busses, is 70 km/h, 10 times faster, than the simulator; meaning the kinetic energy is 10² = 100 times bigger; and yet, most busses don’t have seatbelts.
This wasn't really a number problem. He's basically teaching that energy is proportionally to the squared velocity. Which, you've said it, one should have learned at least in driving school if not in school. OU double the speed, you quadruple the energy.
After 30+ years as an engineer, in two days I'm interviewing to become a high school STEM teacher. I'm going to use this simple (but not obvious) observation every chance I get.
I honestly hope you get the opportunity to share this information with young drivers. This is exactly the kind of thing that, if hammered into one's mind, could save lives.
Anyone asking for km/h or other speeds, there is a general rule for any unit. For any speed x, traveling at sqrt(2)*x will double your kinetic energy. The breaking distance for x to 0 is equal to the breaking distance from sqrt(2)*x to x. Put in whatever numbers and units you like.
Starrgate I don't know how you decoded that!! There's a lot of geniuses on this channel!! Btw f-in love Stargate. I'm rewatching SG1 and will follow that with Atlantis :)
For those of you that are curious about the "hard" way to solve this: Let v1 be the initial speed of the blue car and v2 the initial speed of the red car. The distance to the tree is D. Furthermore, the blue car comes to a halt after time t1, the red car will crash after time t2. The acceleration is a for both cars (see the footnote for why this is a reasonable assumption). Then we have for the blue car: (1) D=v1*t1+1/2*a*t1^2 (2) v1+a*t1=0 a= -v1/t1 from (1) we replace a according to (2) and get (3) D=v1*t1 - 1/2*v1*t1=1/2*v1t1 t1 = 2*D/v1 replace t1 in (2) and get (4) a = -v1^2/(2*D) For the equations of the red car we have similarily: (5) D=1/2*a*t2^2+v2*t2 1/2*a*t2^2+v2*t2 - D=0 This is a quadratic equation with the positive solution (we neglect the negative solution for obvious reasons): (6) t2 = (-v2+SQRT(v2^2+2a*D))/a The speed at time t2 with a constant acceleration a is given by (7) v=v2+a*t2 we replace t2 in (7) and get (8) v=v2+ (-v2+SQRT(v2^2+2a*D)) = SQRT(v2^2+2a*D) we use the expression for a in (4) and get our final result: v=SQRT(v2^2-v1^2) As promised, i will try to reason why we can assume that a=constant for both cars. For a moving object, the force of friction F_R is very nearly constant within a reasonably small range of velocities. Newton tells us, that (9) F_R = m*a where m is the mass of the object (in our case, the mass of the cars). Since F_R is constant and the mass doesnt change either it follows, that the acceleration a is constant aswell. Again, this isnt entirely true as the force of friction is dependant on the velocity. If anyone could provide some real world values on this relation, i'd love to know about it. (Btw, assuming that F_R is constant we also get the conservation of energy as it was used in the video: The energy loss through friction is: W = (Integrate from starting point to end point) dx*F_R which is the same for both cars (since the path is the same and F_R is the same for either car))
i don't. the low speeds are mushed too closely together which makes it difficult to tell how fast you are going in lower speed limit areas, which would defeat the whole purpose.
Here's the SUVAT solution: Equations to use are: v = u + at [1] s = ut + 0.5at^2 [2] s = 0.5(u + v)t [3] v^2 = u^2 + 2as [4] s = vt - 0.5at^2 [5] Blue car: u = 70, v = 0, a?, t?, s? Using s = 0.5(u + v)t: s = 0.5(70+0)t = 35t Using v = u + at: 0 = 70 + at, therefore a = - 70 / t Given that a and s are the same for the red car: Red car: u = 100, a = -70 / t, s = 35t, v? v^2 = u^2 + 2as v^2 = 100^2 + 2(-70/t)(35t) v^2 = 5100 v = 71.4mph (3s.f) So the red car hits the tree at 71.4mph.
This is a rather inefficient way of solving under SUVAT. v^2=u^2+2as on its own is sufficient. Rearrange to v^2-u^2=2as. We have assumed as part of the parameters that a and s are equal for both cars, and that the first car reaches a final velocity of 0, therefore: (0)^2-(u_1)^2=(v_2)^2-(u_2)^2=2as -(u_1)^2=(v_2)^2-(u_2)^2 (v_2)^2=(u_2)^2-(u_1)^2 v_2=sqrt[(u_2)^2-(u_1)^2] v_2=sqrt[100^2-70^2] v_2=sqrt[5100] v_2=71 mph The fact that it's entirely solvable using v^2=u^2+2as shows its connection to a problem that can be solved entirely using energy rules, since the derivation of this rule is most simply performed from work and kinetic energy: K_i=(1/2)mu^2 K_f=(1/2)mv^2 W=K_f-K_i=Fs F=ma :. K_f-K_i=mas :. (1/2)mv^2-(1/2)mu^2=mas v^2-u^2=2as (cancelling through m from all the terms and multiplying all terms by a factor of 2 v^2=u^2+2as The nice thing about the v^2-u^2=2as where 2as is simply taken to be a constant is that we don't have to worry about units: 2as is in whatever unit we need it to be in, it's simply a constant, and the rest of the problem can be calculated in mph without any dodgy unit work at all.
Don't even need that. Assuming braking force is constant, the work done per distance traveled is equal, and we can assume both cars braking systems to the same amount of work in acceleration. The kinetic energy of an object is proportional to the square of the velocity. If the blue car dissipates some constant * 70^2, then red car dispates the same. By the proportion rule, we know the red car started with the same constant * 100^2. Thus K*V(rf)^2 = k*10000 - k*4900 The K's simply cancel. V(rf)^2 =5100 V(rf) = +- 71.4
Oh, I mean, the work/energy method is definitely the superior one for solving the problem as posed (and basically the one used in the video itself), no argument there. However, given that the original poster explicitly stated that SUVAT was the method being used, it seemed appropriate to show the most efficient SUVAT path, not the most efficient path overall (especially when the video itself had already pretty much covered that).
only in the event that coffee breaks lower the reaction time, therefore making the crash survivable, shall I declare coffee breaks legal, otherwise i would like to replace them with tea breaks... since i personally don't like coffee.
If this was in sixty symbols we wouldnt see any equations or numbers. Communicating physics to a wide audience is incompatible with squaring such big numbers as 70 or 100. And 1/2 is a terrifying monster
First of all, this is an awesome video. You two took what is a rather complicated subject/equation or set of equations and essential compressed it to an easily digestible ~4 minutes of actual math and 4 further minutes of banter and setup (two very important things). As I am just now learning to drive, seeing something like this is-as you said-very sobering. As you started the problem, I attempted some math in my head and estimated around 40mph for the red car's impact, still immensely injurious and possibly fatal. However, seeing the massive disparity in impact speed was totally shocking to me. Even the best air bagging and padding can do little to stop neck snapping at that speed of impact. There's something very powerful about cold, hard, numbers; especially in situations or hypotheticals as serious as this one. When you are shown hard math on the actual danger of your actions (rather than a purely hypothetical and often annoying warning) you are now faced with proven fact to discourage them. It is one thing to disregard a hackneyed warning, but it is an entirely different story when you are going directly against a proven fatality. When you consider the numbers in context, the extra speed does not make up for the added risk. I for one, would much rather get somewhere 30mph slower than hit an object 71mph faster (or even hit it at all). These are important facts and truths that people need to know, and often, math is the bets way to show them. But that gets me thinking, could we then calculate the speed of impact of the head on the airbag? What speeds are injurious? Which ones are fatal? By showing and proving these things with math, we give solid and undeniable reasons to not do something dangerous to oneself, AND others.
I'm a game designer, and I have a saying that every part of making a game is somewhat tricky, except the User Interface, which is incredibly difficult. Ben's idea about speedometers showing speed in a non linear way is absolutely genius.
Another easy way to justify that the red car crashes at a speed greater than 30 mph is using constant acceleration braking. Suppose, after time t, the blue guy reaches 0 mph, and is barely touching the tree. Assuming constant deceleration, the red guy is going 30 mph at time t. But since the red guy is going faster at all times, he would be *past* the tree at time t, meaning the impact speed must be > 30.
Actually I thought that was missing from the video, as it was just barely mentioned for the case of the reaction time... But it's ANYWAY quite a huge factor... Adding up to the quadratic effect (because it "shifts" the slope of the square function to where difference is even higher...).
I like your idea regarding the speedometer. Actually some car manufacturers (like my VW golf) do the oposite! All the gaps in the lower speed are further than higher speed - the gaps between 10-20-30 are the same as the gaps between 40-60-80 (KMH).
yall ever seen a marine speedometer? They work with a pitot tube, and dynamic pressure. The 'sqaure' relationship is actually shown on the guage!!! A quick google search will show it.
I did my driving license in Germany and we actually had a very similar comparison in the driving school. A car which is traveling with 50km/h and is able to stop at a given distance, would still have about 50km/h if it was driving with 70km/h. I am wondering why this example is not teached in UK? I asked some friends here in Germany and this was nothing new to them ;-) Anyways, thanks for putting this on.
tamasgal - 40 years ago, when I took my UK driving test, we had to memorize the numbers, but we were never taught where they come from. I don't know if things are different today.
in the US at least, it's hard enough making sure everyone knows when to turn on their headlights and which color on the traffic light means stop. We don't have time for additional facts
For UK driving tests we are (or were when I did them) required to know the stopping distance from various speeds... which is imo less useful than how fast you'd be going when you hit something that you could have avoided hitting.
I remember here in Australia they ran a bunch of speed awareness campaigns in the early 2000's on TV. One such campaign, which I believe you can find by searching "Slow-Mo TAC Anti-Speed TV Ad", explores a very similar scenario. Two identical cars, two identical reaction times, and an obstacle appearing in the road ahead. Just thought that might be interesting to share for anyone looking for another visual comparison. There's a direct quote in the ad which stuck with me right through to today. "In the last five meters of braking, you wipe off half your speed."
I felt so sill for thinking thinking "The red guy loses about half his KE so he's about mph." Really it's "he hits with a little more KE than it took the blue guy to go 70mph."
Its ok , the rest of the world will drive on the correct side of the road one day.... On the left ! Think about it right hand drive it makes sense , most people are right handed so you have your dominant hand on the wheel at all times even when you change gear, and as most people are right footed you always have your dominant foot on the accelerator and brake and leave your left foot for the clutch.
Andrew Joy interesting point! Could be argued that changing gears requires more coordination, and the dominant hand would help. But I think either is fine really, just wish it were standard everywhere hahah. Also, left hand drive has the same pedal setup! Right foot gas/brake, left foot clutch.
Andrew Joy The fact that the majority are right-eyed also favors driving on the left. The research supports all this, but it's only marginal. Perhaps we should have a system where right-handers drive on the left and left-handers drive on the right.
"Perhaps we should have a system where right-handers drive on the left and left-handers drive on the right." That sounds absolutely ingenious! You should go to the responsible polit.... WAIT! DON'T! They might actually do it...
Guitar players use their dominant hand to just pick the string away whilst their non-dominant hand does all the widdly bits. Seems odd but I think the reason is that your dominant hand is better at automating it's actions whilst the non requires concentration. I agree with your outcome. I don't think about my right hand to steer I just think about where I want to go. In the same way I press a note with my left hand, looking at the fretboard and my right hand picks the string without me thinking about it.
Solving using the velocity equations and distance to stop ect... shows that the distance from the start of the skid to the tree for blue vehicle to stop was 223.74 feet making the tree roughly 224 feet away from when both cars start skidding. Assuming a drag factor of .73 which is average on pavement. Making the acceleration factor 23.50. Meaning the red vehicles speed after skidding on the surface for 224ft (impact with tree) 71.41MPH. I also did it the way you did with Ke and it’s also right. Just was a lot easier! Thanks for this, helped me brush up on my Ke!
1/2mv^2 is also why high winds are so dangerous to sailors. If you leave port with full sail and then the wind doubles, taking down one sail will not get the amount of power going into the boat back down to a manageable amount. You'll need to dramatically reduce your sails. In San Francisco bay, I've seen sails reefed down to less than 25% of their original size in order to keep sailing through high winds.
Cars should have an additional gague that shows the total stopping distance for the current speed, assuming the standard reaction time and deceleration used to calculate the braking distances in the highway code (about 1.5 seconds and 2/3rds of a G).
The Highway Code shows a stopping distance of 96m for a car travelling at 31.3m/s (70mph). Of that distance, 21m is the distance travelled before deceleration takes place (the "thinking distance"). Therefore the reaction time is 21m ÷ 31.3m/s = 0.671s. For the car travelling at 44.7m/s (100mph) the thinking distance is thus 0.671s x 44.7m/s = 30m. The question now is at what velocity will it be travelling when it hits the tree 96m - 30m = 66m away if it decelerates at the same rate. What is the deceleration? The blue car, initially travelling at 31.3m/s, comes to rest in 96m - 21m = 75m. We know v² = u² + 2as, from which a = -31.3² ÷ (2 x 75) = -6.53m/s². Finally, the velocity of the red car at the moment of impact can be calculated from the same equation, v² = u² + 2as, where u = 44.7m/s, a = -6.53m/s² and s = 66m. Thus v² = 44.7² - 2 x 6.53 x 66 = 1998 - 862 = 1136, from which v = 33.7m/s or 75.4mph. Sobering, indeed.
I know nobody will believe me, but I paused the video for around 2 minutes when he said to try and work it out. I got the answer right. By thinning about the equations of motion, I worked out the answer would be (10/7)^2*70/2mph and I was so pleased I got it right. This is what makes maths so brilliant. A simple problem and a little though can be really rewarding! Go on with your day and think of all the people who found a little problem and put the effort in to solve it, and in turn, make your life easier
I like when they include some algebra in these videos. It makes it much more interesting to see something one could actively do to understand the problem or concept.
Thanks to some conversions from some other people, we can watch this video sane. I have added my own unit conversion as well, just to make sure we still have a balance of insanity :3 70 mph ≈ 113 km/h ≈ 31,3889 m/s ≈ 6.26e+07 beard-second/jiffy (bs/j) 100 mph ≈ 161 km/h ≈ 44,7222 m/s ≈ 8.94e+07 bs/j 30 mph ≈ 48 km/h ≈ 13,3333 m/s ≈ 2.68e+07 bs/j 71 mph ≈ 114 km/h ≈ 31,6667 m/s ≈ 6.35e+07 bs/j
This system precisely parallels several common systems studied in physics. The ones that come first to my mind as an electronics enthusiast are the complementary concepts of inductive/capacitive-resistive time constants. The kinetic energy stored in a moving mass is equal to one half its mass times its speed squared (E_k=mv^2/2). In an inductor, which is typically a coil of wire designed to store energy in a magnetic field, the actual energy stored will ideally be equal to one half the inductance (which depends on the size of wire and how it's wound) times the current squared (E=LI^2/2), and in a capacitor which is typically a pair of conductive plates (usually foil) separated by an electrical insulator designed to store energy in an electric field, the energy will be equal to one half the capacitance (which depends on the area of the plates and how closely they are spaced) times the voltage squared (E=CV^2/2). In a sense, inductance is very much like electrical momentum; when a voltage is first applied to an inductor, a magnetic field begins forming as current begins to flow. To make a long explanation short, the rate of deceleration of a car with a given mass under a given level of braking will be neither linear, as the video suggests we might assume, nor quadratic, as he assumes. The kinetic energy will be bled away exponentially for as long as dynamic friction in the brakes dominates. That is, there will be a characteristic time constant defined by the mass of the car divided by the resistance to motion (t=m/r) where that resistance is the amount of angular force felt by the brake pad at a given speed (It's only angular because cars have wheels that roll.) If the brakes simply dragged against the ground the calculation would be much simpler.). Within one time constant 68% of the energy will have been dissipated and within two 95%. Within 5 time constants the type of friction involved will have almost certainly dropped well below the threshold for the brakes to lock (static friction) unless the vehicle had been traveling at several times the speed of sound.
Two assumptions were made in the model : 1st: constant (de)acceleration. 2nd: same amount of energy reduction when braking. Those are not necessary the case. In fact, one can devise other models when both cars, having a common acceleration function, successfully stop just before colliding.
Hey Mr sparks you were my maths tutor and in house tutor at school, frictional force is proportion to the square of speed. Therefore applying the same coefficient of friction to the brake discs means the amount of force applied to decelatrate is greate, therefore the rate of declaration is not constant (suvat equations can not be used).
Yes, air resistance is proportional to the square of the velocity, but we assumed that the whole process only takes a very short amount of time, so we can neglect air resistance. The frictional force of the brakes is independent of the velocity, so that's why these calculations are used
@@williamhauser4290 No, it's just not. Under normal conditions sliding friction force only depends on: -The coefficient of friction, which in the case of the brakes only changes over long time periods because of heavy usage or if you overheat them by braking repeatedly in quick succession. -The normal force, which is directly proportional to how hard you brake and there's no reason to assume that driving faster will also allow you to stand on the brakes harder.
I think the guy in the blue car may still be in trouble, because if the tree doesn't remain stationary, but moves forward on the end that the red car hits, it's going to pop back on the end where the blue car stopped. I have no idea how to mathematically figure that out. This is interesting, and you have a new subscriber here.
Because velocity is assumed to decrease linearly, kinetic energy must decrease squared. Hence, delta_E1 is not equal to delta_E2 (in respect to a certain distance). In summary, this calculation cannot (I think?) be correct.
Why would velocity decrease linearly though? The force generated by the breaks is (approximately) constant, so the amount of Energy lost is linear (with respect to the distance). Thus the velocity decreases with the square root of the distance (as claimed in the video)
Never heard of the "suvat" equations, but the only equation you need is the v_f^2 = v_0^2 + 2ad. (Yes, this is the conservation of energy equation or the change in kinetic energy equals the work expended.) Here v_0 = initial velocity, v_f = final velocity. Blue car: 0^2 = 70^2 + 2ad From this you find, 2ad = -4900. Red car: v_f^2 = 100^2 + 2ad = 100^2 - 4900 or v_f is the square root of 10,000 - 4,900 or 71.4 mph
velocity =initial velocity(u) plus acceleration(a) times time(t) (s)distance= initial velocity(u) times time(t) +1/2a(acceleration)t(Time) squared s = 1/2(u + v)t v squared = u squared + 2as s = vt - 1/2a(t) squared
I'm in the UK and I agree. I think we drove on the WRONG side so it costs other countries more money to make cars for us. And so protecting car companies in the UK. Didn't really work out. :)
The units when you solving this problem do not matter. The result is valid if we consider that both cars have the same mass and acceleration. I think it was necessary to clarify it. In this way it can be stated that the work to travel the distance to the tree is the same and arrive at the proposal of the subtraction of the kinetic energies that is seen in the video.
I posted this on a car forum back in 2002, none of the petrol heads believed me, even with the maths. All believed they (and their modified cars) could '...break the laws of physics, Jim'. Wonder how many of them drove red cars and are now wrapped around a tree...
70 mph = 112.654 kph = 31.2928 mps, 100 mph = 160.934 kph = 44.704 mps; do note those are miles per hour, kilometers per hour and meters per second respectively (not miles per second, or meters per hour)
It has to do with the ratio, not the unit of measurement. I get why they used mph, but in doing so they should made made clear that it does not matter.
Another approach to understand why this happens is that the breaks of the red car have less time to reduce the car's speed, as it is approaching faster to the tree. It is actually the same reason, just a different way to explain it.
Well considering the roads and cars match...then its the correct side. Here in America we drive on the other side...but our steering wheel is also on the other side. So ultimately its the same thing.
The friction force is only proportional to the normal force, and not the speed. Which means that the friction force is actually not greater the faster you go :)
That's the _opposite_ of what he proposed though. He wants something more like 0-15-30-40-50-60-70-75-80-85-90-95-100 as we move between roughly the same angle each time. That way it's 6 steps between 0 and 70 and 12 steps between 0 and 100 which accurately reflects that the latter is approximately twice the kinetic energy.
codebeard I think we are talking we are talking about different implementations of the same thing. In his case the values of the speedometer remain the same but the gap between them changes. In VW's case the gaps remain the same but the values grow in somewhat nonlinear fashion. Also in modern cars the acceleration between 90 and a 100 for example is trivial so the needle will have to move more rapidly the higher the speed. In vw case it is the opposite: the faster you go the slower the needle moves. I guess what I'm saying is that what he proposes is a static representation of the energy of the car and how it grows and on the other hand we have a dynamic representation with the movement of the needle.
Codebeard What Sparks proposed, or the alternative you describe makes sense; giving the driver a clue as to the potential damage in a crash. I wonder what VW's reasoning was for doing the opposite. Let's lull people going over 70 into going even faster so they're more likely to crash and be removed from the gene pool?
Дичо Дичев If the gaps remain the same but the values increase faster at GREATER values, then this is the OPPOSITE of what was proposed. Which is what codebeard said.
volkswagen are made in germany and are intended primarily for the german market, same with audi (which are basically teh same company) the way their tachometres work they are designed for key speeds to be easily identified. for example thirty and fifty km/h (most common speed limits in populated areas) where highlited red in the audi i owned, 100km/h (the speed limit outside of populated areas) was right at 12 O'clock and another highlight was put at 130km/h wich is the generally recomended speed for driving on the autobahn (you can go faster if you want but if say someone else causes an accident which involves you while you where going over 130, that guys insurance can argue that you could have avoided the accident by going slower so they might not pay all of the damage because you are partially to blame.)
The intuitive explanation: You have to remember that when braking, not only is there the 30 mph speed difference that the brake has to make up for, but you are also travelling more distance in the same amount of time, so you have less time to apply the brake before the crash.
"SUVAT" encompasses energy, not dodgey at all: Vf^2=V0^2 + 2aS. That is work-energy, just with the mass divided out. In this example, modeled with constant acceleration, 2aS = -70^2 for the blue car (since Vf is 0). "2aS" is the same for the red car because it has the same acceleration and distance traveled, you get 100^2-70^2=Vf^2 when you look for the speed of the red car the instant before it touches the tree.
First comment on a Numberphile video? Worth my dollar on patreon I reckon! P.S. Haha at the end, "Don't just think "what formula do I use?"" Formula, and they're formula 1 slot cars!
whats that formular? I used 100/70=1,43 ; 1,43²=2 -> red car has twice the energy; so on impact it has the energy of the blue car when it started which is ~70mph
My way of thinking was the higher velocity meant more resistance against breaking. It still all comes down to the fact of non-linearity, that so few people fail to understand.
The bigger problem is, that the faster person travels further in their reaction time, so they start braking while being closer to the tree, than the slower person.
As the masses of the two cars are the same (and the equation is simplified to not take into account other variables dependant upon speed) then on the kinetic energy equation 1/2*m becomes a constant multiplier (and as such can be consired to be just one) and pythagoras theorem is only thing left. Area's defined by the triangles sides becomes the kinetic energy and the difference between the two areas defined and proportional to speed gets the red cars speed.
Since he is going faster the red guy would actually have less time to brake, so he would get rid of less energy. I think the actual speed he would hit at is faster than the 71.
Tayler Robinson valid point . But won't the rate of deceleration be more because more contact with the brake pads ? So won't the speed be lesser than 71?
K.R. Koushik it would be the same amount of contact because the wheels are going faster, but for less time. So maybe it would still be 71? I dunno. I'm not great with physics.
If you're talking about the reaction time it's said in the video that they start braking at the same time. If you're talking about the eq. it does include " 1/2 a.t^2 " but it still means he hits the tree at 71 mph. He just calculated it differently
This would be the assumption I guess, alternatively you could also assume that given the higher speed the red person may be paying more attention to the road given the greater implied risk of crashing and thus actually break before the blue car who we could assume is not paying as much attention to the road obstacles.
yenioyuncu255 he's not talking about reaction time. He's talking about the time he has for deceleration . And my point is that the rate of deceleration would be more because they have the same brakes and one is traveling faster.
Weil what do you mean? Infinity isn't a real number, but it is a valid result of an infinite sum (since it is part of the extended real line). I don't see why they would've needed to mention nonstandard models *or* philosophy.
you can also solve it without energy equations using kinematics: vf^2 = vi^2 +2a∆d so we have for the blue car: 0^2 = 70^2 + 2a∆d so a = ((-4900)/(2∆d)) now let's call the final velocity of the red car "f" f^2 = 100^2 + 2a∆d = 10000 + 2(((-4900)/(2∆d))∆d so f^2 = 10000- 4900 = 5100 so f is about 71.4 mph
Is Brady trying to start Physicsphile? I had to dig back to high school science to remember about kinetic energy. But I do remember that we talked about the square of speed in driver’s ed when discussing why speed kills.
We talk about kinetic energy and braking distances increasing with the square of the speed in Swedish driving schools too. But for some reason, the writers of the main driver's education material have decided to call it (kinetic energy) "the living force", which (as a science guy) drives me mad...
This should be mandatory material for every learner driver. In fact, I coincidentally came up with the same sort of question when discussing braking distances with a friend of mine. I initially thought I had done something wrong (I used 50 km/h and 100 km/h). After multiple checks I saw how dangerous driving 100 km/h actually was in comparison to 50 km/h.
Nah, think about it. The distance to slow from 100mph -> Xmph is equal to the distance it is to slow from 70mph -> 0mph, however the car travelling 100mph will pass that distance at a much quicker rate, significantly reducing the time to stop/slow.
I did the same thing. I think it has to do with metagaming the answer. In the back of your mind, you know they have control over exactly how to ask the question, so you assume they're asking it in the way that produces the most interesting answer. He even says in the video that you probably excluded 30mph because you know he wouldn't have asked if that had been the answer. Arriving at 70 intuitively is the a similar process, just taken a step further.
If you got it right, good(I'm going to guess you're one of the few), but I think much of the point of this video is that the answer is completely counter to most people's intuition, which is to say that most people would likely never guess 71 mph.
Another way to think of it is that when you brake, you're still moving forward until your energy reaches zero, which means that higher speed translates to a further distance travelled in a shorter amount of time. In the same first second the two cars move different distances, even if their brakes are of the exact same strength, the cars the same weight, the brakes were applied simultaneously, etc. etc. and each following measure of time the blue car is moving at lower and lower speeds and so gets more and more increments of time to slow down, before reaching that barrier in the road, while the red car simply moves further at each time increment, so cannot slow down as much. This is why, logically, esp. from a physics standpoint which mirrors the results of the mathematics that Ben showed, it maintains a speed higher than the blue car began with, at the point of the barrier across the road, and thus crashes.
Here's another way to think about this: If the speed at which you stop just in time (70 mph in the example) is one side of a right triangle, and the initial speed for the crash case (100 mph) is the hypotenuse, then the impact speed (71 mph) is the other side. So, if you can stop in time at 30 mph, then if you're going 50 mph, you're going to crash at 40 mph.
[Edit:] Solved. See replies. I don't understand what you mean by "brake at the same rate". Because if that meant that acceleration (a) was the same, then the answer would be 30 mph: u1 = 70, v1 = 0 u2 = 100, v2 = ? { v1 = u1 + at { v2 = u2 + at Subtract equation 1 from equation 2 and you get: v2 - v1 = u2 - u1 v2 - 0 = 100 - 70 v2 = 30
The problem here is the confusion between decelerating for some time and decelerating for some distance. It isn't wrong to say that when both cars have "lost" 70 mph with the same deceleration, the same time has passed (and Mr. Sparks probably wouldn't deny this either). However, when the red car decelerated to 30 mph, it travelled a much further distance than the blue car has decelerated to 0 mph. To explain this, I use the formular for potential energy which you can use it for generally any kind of acceleration. The advantage is that it's E = m*a*s which means, s is a variable here and time doesn't matter for this kind of formular. If I put this equal with kinetic energy and solve for s, you'll get this: s = v²/(2a). a is the same in both cars meaning that only v is a variable here. As you can see, v is squared meaning that the bigger v becomes, s becomes *much* larger, thus increasing the breaking distance by quite a bit.
![BOWKYLION - วิงวอน (ex-change) [Official MV]](http://i.ytimg.com/vi/cLdnZWDaO-w/mqdefault.jpg)
Catch a more in-depth interview with Ben on our Numberphile Podcast: th-cam.com/video/-tGni9ObJWk/w-d-xo.html
In Germany, this (the quadratic behavior of braking distance in relation to speed) is required knowledge to get a driving license. The test questions even include how far you travel in the second you need to react.
In Finland, it’s not really required; as the theoretic questions are all about: ”Here’s the set-up. What’s the right thing to do?”; but they drive it home (OK; bad pun is bad), pretty well. They give you a chance to ride a car-crash-simulator (basically, it’s a chair that slides down a slope, to a sudden halt) that simulates a car-crash, at 7 km/h. I took it, and let me tell you, if the seatbelt wasn’t there, I would have gone to the orbit. Then, they give you something to think about: In Finland, the speed limit, for most city-busses, is 70 km/h, 10 times faster, than the simulator; meaning the kinetic energy is 10² = 100 times bigger; and yet, most busses don’t have seatbelts.
This wasn't really a number problem. He's basically teaching that energy is proportionally to the squared velocity. Which, you've said it, one should have learned at least in driving school if not in school. OU double the speed, you quadruple the energy.
70 mph ≈ 113 km/h
100 mph ≈ 161 km/h
30 mph ≈ 48 km/h
71 mph ≈ 114 km/h
No sane person would drive 160km/h
Henrix98 Never been to Germany, huh?
I've done it and seen cars pass me at a lot higher speeds.
The real mvp
70 mph ≈ 113 km/h ≈ 31,3889 m/s
100 mph ≈ 161 km/h ≈ 44,7222 m/s
30 mph ≈ 48 km/h ≈ 13,3333 m/s
71 mph ≈ 114 km/h ≈ 31,6667 m/s
After 30+ years as an engineer, in two days I'm interviewing to become a high school STEM teacher. I'm going to use this simple (but not obvious) observation every chance I get.
It's in more than one introductory physics book that I've seen.
I honestly hope you get the opportunity to share this information with young drivers. This is exactly the kind of thing that, if hammered into one's mind, could save lives.
Maybe just once per class :P
Anyone asking for km/h or other speeds, there is a general rule for any unit.
For any speed x, traveling at sqrt(2)*x will double your kinetic energy.
The breaking distance for x to 0 is equal to the breaking distance from sqrt(2)*x to x.
Put in whatever numbers and units you like.
Wait so do I have this right? If I'm going 41% faster, my stopping distance will be doubled?
@@nahog99 Assuming no reaction time, yes. In practice if you also consider reaction time it's even worse.
@@danielrose1392 right I’m assuming from the moment braking is applied.
I was in a car crash, and if you can pick up from subtle clues, it didn't work out too well for me
Little Cripple “little cripple”
Starrgate I don't know how you decoded that!! There's a lot of geniuses on this channel!!
Btw f-in love Stargate. I'm rewatching SG1 and will follow that with Atlantis :)
LOL
Are you a rapper?
Little Cripple Giving subtle clues is lame.
For those of you that are curious about the "hard" way to solve this:
Let v1 be the initial speed of the blue car and v2 the initial speed of the red car. The distance to the tree is D. Furthermore, the blue car comes to a halt after time t1, the red car will crash after time t2. The acceleration is a for both cars (see the footnote for why this is a reasonable assumption).
Then we have for the blue car:
(1) D=v1*t1+1/2*a*t1^2
(2) v1+a*t1=0 a= -v1/t1
from (1) we replace a according to (2) and get
(3) D=v1*t1 - 1/2*v1*t1=1/2*v1t1 t1 = 2*D/v1
replace t1 in (2) and get
(4) a = -v1^2/(2*D)
For the equations of the red car we have similarily:
(5) D=1/2*a*t2^2+v2*t2 1/2*a*t2^2+v2*t2 - D=0
This is a quadratic equation with the positive solution (we neglect the negative solution for obvious reasons):
(6) t2 = (-v2+SQRT(v2^2+2a*D))/a
The speed at time t2 with a constant acceleration a is given by
(7) v=v2+a*t2
we replace t2 in (7) and get
(8) v=v2+ (-v2+SQRT(v2^2+2a*D)) = SQRT(v2^2+2a*D)
we use the expression for a in (4) and get our final result:
v=SQRT(v2^2-v1^2)
As promised, i will try to reason why we can assume that a=constant for both cars.
For a moving object, the force of friction F_R is very nearly constant within a reasonably small range of velocities.
Newton tells us, that
(9) F_R = m*a
where m is the mass of the object (in our case, the mass of the cars). Since F_R is constant and the mass doesnt change either it follows, that the acceleration a is constant aswell.
Again, this isnt entirely true as the force of friction is dependant on the velocity. If anyone could provide some real world values on this relation, i'd love to know about it.
(Btw, assuming that F_R is constant we also get the conservation of energy as it was used in the video:
The energy loss through friction is:
W = (Integrate from starting point to end point) dx*F_R
which is the same for both cars (since the path is the same and F_R is the same for either car))
That was an extremely elegant way of calculating that. And the idea about the speedometer is pretty nice as well.
I love the idea of the v-squared speedometer.
i don't.
the low speeds are mushed too closely together which makes it difficult to tell how fast you are going in lower speed limit areas, which would defeat the whole purpose.
I would prefer a kinetic energy display just alonside the traditional speedometer
Now this is smart (math-informed) public policy! Brilliant. I love it. And I bet Edward Tufte would love it, too.
@@windhelmguard5295 Its useful in cars with a dual analog and digital speedometer.
@@windhelmguard5295 we have the technology these days to have an automated system that switches from normal to squared depending on ur speed
Here's the SUVAT solution:
Equations to use are:
v = u + at [1]
s = ut + 0.5at^2 [2]
s = 0.5(u + v)t [3]
v^2 = u^2 + 2as [4]
s = vt - 0.5at^2 [5]
Blue car: u = 70, v = 0, a?, t?, s?
Using s = 0.5(u + v)t:
s = 0.5(70+0)t = 35t
Using v = u + at:
0 = 70 + at, therefore a = - 70 / t
Given that a and s are the same for the red car:
Red car: u = 100, a = -70 / t, s = 35t, v?
v^2 = u^2 + 2as
v^2 = 100^2 + 2(-70/t)(35t)
v^2 = 5100
v = 71.4mph (3s.f)
So the red car hits the tree at 71.4mph.
This is a rather inefficient way of solving under SUVAT. v^2=u^2+2as on its own is sufficient.
Rearrange to v^2-u^2=2as. We have assumed as part of the parameters that a and s are equal for both cars, and that the first car reaches a final velocity of 0, therefore:
(0)^2-(u_1)^2=(v_2)^2-(u_2)^2=2as
-(u_1)^2=(v_2)^2-(u_2)^2
(v_2)^2=(u_2)^2-(u_1)^2
v_2=sqrt[(u_2)^2-(u_1)^2]
v_2=sqrt[100^2-70^2]
v_2=sqrt[5100]
v_2=71 mph
The fact that it's entirely solvable using v^2=u^2+2as shows its connection to a problem that can be solved entirely using energy rules, since the derivation of this rule is most simply performed from work and kinetic energy:
K_i=(1/2)mu^2
K_f=(1/2)mv^2
W=K_f-K_i=Fs
F=ma
:. K_f-K_i=mas
:. (1/2)mv^2-(1/2)mu^2=mas
v^2-u^2=2as (cancelling through m from all the terms and multiplying all terms by a factor of 2
v^2=u^2+2as
The nice thing about the v^2-u^2=2as where 2as is simply taken to be a constant is that we don't have to worry about units: 2as is in whatever unit we need it to be in, it's simply a constant, and the rest of the problem can be calculated in mph without any dodgy unit work at all.
you are 4 hours late
Don't even need that. Assuming braking force is constant, the work done per distance traveled is equal, and we can assume both cars braking systems to the same amount of work in acceleration. The kinetic energy of an object is proportional to the square of the velocity. If the blue car dissipates some constant * 70^2, then red car dispates the same. By the proportion rule, we know the red car started with the same constant * 100^2. Thus
K*V(rf)^2 = k*10000 - k*4900
The K's simply cancel.
V(rf)^2 =5100
V(rf) = +- 71.4
remember Work = F * d, time frame doesn't matter.
Oh, I mean, the work/energy method is definitely the superior one for solving the problem as posed (and basically the one used in the video itself), no argument there. However, given that the original poster explicitly stated that SUVAT was the method being used, it seemed appropriate to show the most efficient SUVAT path, not the most efficient path overall (especially when the video itself had already pretty much covered that).
Numberphile but no kmh;(
They're British aha
you can just exchange the units then the result is 71km/h
Bah, if you want proper SI units it ought to be in meters per second anyway.
The units don't really matter here...
bah kmh is so silly when you can use m/s:)
It is not speed that causes injury, it is the sudden deceleration. For this reason I propose to make breaks illegal
Would that include coffee breaks?
brakes*
woops,
only in the event that coffee breaks lower the reaction time, therefore making the crash survivable, shall I declare coffee breaks legal, otherwise i would like to replace them with tea breaks... since i personally don't like coffee.
outlaw all acceleration
traffic math, my favourite kind
Yes!
This comment is hiding in plain sight.
How about describing the speed in barleycorns per quarter-decade? That'll make things easier.
Brilliant! +10^3 likes!
The first one equals to 291,392,640,000 bc/¼d and the second one is equal to 416,275,200,000 bc/¼d, correct me if I'm wrong.
false.
That's a nice sixty symbols episode you have there
if it was sixty symbols, they would have included relativistic effects
Would be a great shame if, say, something were to happen to it, wouldn't it?
the approximations made it seem like an engineering episode
If this was in sixty symbols we wouldnt see any equations or numbers.
Communicating physics to a wide audience is incompatible with squaring such big numbers as 70 or 100. And 1/2 is a terrifying monster
??
This dude's hairstyle straight out of 1700's
Bernoullis descendant
So are his units of measurement
Jeff Ahn Damn I was gonna write that too. xD
And also the terms he uses - carriageway seem so odd
Abhishek Verma guy is gollum
First of all, this is an awesome video. You two took what is a rather complicated subject/equation or set of equations and essential compressed it to an easily digestible ~4 minutes of actual math and 4 further minutes of banter and setup (two very important things). As I am just now learning to drive, seeing something like this is-as you said-very sobering. As you started the problem, I attempted some math in my head and estimated around 40mph for the red car's impact, still immensely injurious and possibly fatal. However, seeing the massive disparity in impact speed was totally shocking to me. Even the best air bagging and padding can do little to stop neck snapping at that speed of impact. There's something very powerful about cold, hard, numbers; especially in situations or hypotheticals as serious as this one. When you are shown hard math on the actual danger of your actions (rather than a purely hypothetical and often annoying warning) you are now faced with proven fact to discourage them. It is one thing to disregard a hackneyed warning, but it is an entirely different story when you are going directly against a proven fatality.
When you consider the numbers in context, the extra speed does not make up for the added risk. I for one, would much rather get somewhere 30mph slower than hit an object 71mph faster (or even hit it at all). These are important facts and truths that people need to know, and often, math is the bets way to show them. But that gets me thinking, could we then calculate the speed of impact of the head on the airbag? What speeds are injurious? Which ones are fatal? By showing and proving these things with math, we give solid and undeniable reasons to not do something dangerous to oneself, AND others.
Every new driver should watch this video.
I'm a game designer, and I have a saying that every part of making a game is somewhat tricky, except the User Interface, which is incredibly difficult. Ben's idea about speedometers showing speed in a non linear way is absolutely genius.
This is why I drive a green car.
The knight rider theme at the end was a nice touch!
Another easy way to justify that the red car crashes at a speed greater than 30 mph is using constant acceleration braking.
Suppose, after time t, the blue guy reaches 0 mph, and is barely touching the tree. Assuming constant deceleration, the red guy is going 30 mph at time t. But since the red guy is going faster at all times, he would be *past* the tree at time t, meaning the impact speed must be > 30.
Yeah that's so obvious, I'm really wondering how anyone could think that it's 30 mph
because we assumed they would reach the tree at the same time ... oops
Actually I thought that was missing from the video, as it was just barely mentioned for the case of the reaction time... But it's ANYWAY quite a huge factor... Adding up to the quadratic effect (because it "shifts" the slope of the square function to where difference is even higher...).
Thanks, that was the piece I was missing.
My reasoning was all "if a is constant than v grows linearally so why isn't it 30?"
I think that because people have genuinely given me that answer. I didn't say it was sensible. :)
I like your idea regarding the speedometer. Actually some car manufacturers (like my VW golf) do the oposite!
All the gaps in the lower speed are further than higher speed - the gaps between 10-20-30 are the same as the gaps between 40-60-80 (KMH).
That Speedo he proposed is amazing.
Ssshhhh
Imagine Nation ikr?! That should be patented properly!
It would make it seem like your car has less acceleration at higher speeds. I think car makers would think it would make their cars look bad.
PleaseDontWatchThese Well it would actually look faster at higher speeds, because the needle would be moving faster across the gauge.
yall ever seen a marine speedometer? They work with a pitot tube, and dynamic pressure. The 'sqaure' relationship is actually shown on the guage!!! A quick google search will show it.
So satisfying to get the correct answer before they revealed the solution :)
And yes, I used the speed/velocity/acceleration equations.
I did my driving license in Germany and we actually had a very similar comparison in the driving school. A car which is traveling with 50km/h and is able to stop at a given distance, would still have about 50km/h if it was driving with 70km/h. I am wondering why this example is not teached in UK? I asked some friends here in Germany and this was nothing new to them ;-) Anyways, thanks for putting this on.
As they worry about things in the UK driving test like reversing around a corner and not the important stuff.
Andrew Joy 😂
tamasgal - 40 years ago, when I took my UK driving test, we had to memorize the numbers, but we were never taught where they come from. I don't know if things are different today.
in the US at least, it's hard enough making sure everyone knows when to turn on their headlights and which color on the traffic light means stop. We don't have time for additional facts
For UK driving tests we are (or were when I did them) required to know the stopping distance from various speeds... which is imo less useful than how fast you'd be going when you hit something that you could have avoided hitting.
I remember here in Australia they ran a bunch of speed awareness campaigns in the early 2000's on TV. One such campaign, which I believe you can find by searching "Slow-Mo TAC Anti-Speed TV Ad", explores a very similar scenario. Two identical cars, two identical reaction times, and an obstacle appearing in the road ahead. Just thought that might be interesting to share for anyone looking for another visual comparison.
There's a direct quote in the ad which stuck with me right through to today. "In the last five meters of braking, you wipe off half your speed."
That music at the end! Took me back to my childhood! Knight rider...
I felt so sill for thinking thinking "The red guy loses about half his KE so he's about mph." Really it's "he hits with a little more KE than it took the blue guy to go 70mph."
This needs to be explained on speed awareness courses, would probably change a lot of people's behaviours
Brilliantly simple explanation. Thanks; I'll be using this one.
This was so weird expecting an overtake on the left all the time
Its ok , the rest of the world will drive on the correct side of the road one day.... On the left ! Think about it right hand drive it makes sense , most people are right handed so you have your dominant hand on the wheel at all times even when you change gear, and as most people are right footed you always have your dominant foot on the accelerator and brake and leave your left foot for the clutch.
Andrew Joy interesting point! Could be argued that changing gears requires more coordination, and the dominant hand would help. But I think either is fine really, just wish it were standard everywhere hahah.
Also, left hand drive has the same pedal setup! Right foot gas/brake, left foot clutch.
Andrew Joy
The fact that the majority are right-eyed also favors driving on the left. The research supports all this, but it's only marginal.
Perhaps we should have a system where right-handers drive on the left and left-handers drive on the right.
"Perhaps we should have a system where right-handers drive on the left and left-handers drive on the right." That sounds absolutely ingenious! You should go to the responsible polit.... WAIT! DON'T! They might actually do it...
Guitar players use their dominant hand to just pick the string away whilst their non-dominant hand does all the widdly bits. Seems odd but I think the reason is that your dominant hand is better at automating it's actions whilst the non requires concentration. I agree with your outcome. I don't think about my right hand to steer I just think about where I want to go. In the same way I press a note with my left hand, looking at the fretboard and my right hand picks the string without me thinking about it.
I've seen this math before, but your comments make it more interesting, profound, scary, sobering, the list goes on. Thank you.
My cars speedo does have larger increments after 80kmh
I too get a hard on when driving fast... but 80kmh is pretty damn slow... I prefer 150kmh...
Solving using the velocity equations and distance to stop ect... shows that the distance from the start of the skid to the tree for blue vehicle to stop was 223.74 feet making the tree roughly 224 feet away from when both cars start skidding. Assuming a drag factor of .73 which is average on pavement. Making the acceleration factor 23.50. Meaning the red vehicles speed after skidding on the surface for 224ft (impact with tree) 71.41MPH. I also did it the way you did with Ke and it’s also right. Just was a lot easier! Thanks for this, helped me brush up on my Ke!
Well I have an engineering fundamentals exam in 26 hours. This counts as revision right?
Yep. You’re sorted.
How did your exam go?
Or have you completely forgotten as I'm asking you 6 years later?
1/2mv^2 is also why high winds are so dangerous to sailors. If you leave port with full sail and then the wind doubles, taking down one sail will not get the amount of power going into the boat back down to a manageable amount. You'll need to dramatically reduce your sails. In San Francisco bay, I've seen sails reefed down to less than 25% of their original size in order to keep sailing through high winds.
nth
Cars should have an additional gague that shows the total stopping distance for the current speed, assuming the standard reaction time and deceleration used to calculate the braking distances in the highway code (about 1.5 seconds and 2/3rds of a G).
The Highway Code shows a stopping distance of 96m for a car travelling at 31.3m/s (70mph). Of that distance, 21m is the distance travelled before deceleration takes place (the "thinking distance"). Therefore the reaction time is 21m ÷ 31.3m/s = 0.671s.
For the car travelling at 44.7m/s (100mph) the thinking distance is thus 0.671s x 44.7m/s = 30m. The question now is at what velocity will it be travelling when it hits the tree 96m - 30m = 66m away if it decelerates at the same rate.
What is the deceleration? The blue car, initially travelling at 31.3m/s, comes to rest in 96m - 21m = 75m. We know v² = u² + 2as, from which a = -31.3² ÷ (2 x 75) = -6.53m/s².
Finally, the velocity of the red car at the moment of impact can be calculated from the same equation, v² = u² + 2as, where u = 44.7m/s, a = -6.53m/s² and s = 66m. Thus v² = 44.7² - 2 x 6.53 x 66 = 1998 - 862 = 1136, from which v = 33.7m/s or 75.4mph. Sobering, indeed.
I know nobody will believe me, but I paused the video for around 2 minutes when he said to try and work it out. I got the answer right. By thinning about the equations of motion, I worked out the answer would be (10/7)^2*70/2mph and I was so pleased I got it right. This is what makes maths so brilliant. A simple problem and a little though can be really rewarding!
Go on with your day and think of all the people who found a little problem and put the effort in to solve it, and in turn, make your life easier
Can you do a reaction Video where you watch the newest mathologer video with all the profs from the -1/12 Video?
I like when they include some algebra in these videos. It makes it much more interesting to see something one could actively do to understand the problem or concept.
That speedometer is lit. I want one. Hmm just have to get a car first.
or a boat. Marine speedometers look exactly like that.
Wow, for real? That's so awesome.
bon15914, there goes his patent...
Thanks to some conversions from some other people, we can watch this video sane. I have added my own unit conversion as well, just to make sure we still have a balance of insanity :3
70 mph ≈ 113 km/h ≈ 31,3889 m/s ≈ 6.26e+07 beard-second/jiffy (bs/j)
100 mph ≈ 161 km/h ≈ 44,7222 m/s ≈ 8.94e+07 bs/j
30 mph ≈ 48 km/h ≈ 13,3333 m/s ≈ 2.68e+07 bs/j
71 mph ≈ 114 km/h ≈ 31,6667 m/s ≈ 6.35e+07 bs/j
Man smiling in the thumbnail of a video about car crashes. smooth~
There be no smiling in the land of hypothetical car crashes! By decree of some guy on a computer with an (appropriately upset looking) anime avatar.
Ah yes! Hence forth there shall be no more smiling in my hypothetical crashes!
This system precisely parallels several common systems studied in physics. The ones that come first to my mind as an electronics enthusiast are the complementary concepts of inductive/capacitive-resistive time constants. The kinetic energy stored in a moving mass is equal to one half its mass times its speed squared (E_k=mv^2/2).
In an inductor, which is typically a coil of wire designed to store energy in a magnetic field, the actual energy stored will ideally be equal to one half the inductance (which depends on the size of wire and how it's wound) times the current squared (E=LI^2/2), and in a capacitor which is typically a pair of conductive plates (usually foil) separated by an electrical insulator designed to store energy in an electric field, the energy will be equal to one half the capacitance (which depends on the area of the plates and how closely they are spaced) times the voltage squared (E=CV^2/2). In a sense, inductance is very much like electrical momentum; when a voltage is first applied to an inductor, a magnetic field begins forming as current begins to flow.
To make a long explanation short, the rate of deceleration of a car with a given mass under a given level of braking will be neither linear, as the video suggests we might assume, nor quadratic, as he assumes. The kinetic energy will be bled away exponentially for as long as dynamic friction in the brakes dominates. That is, there will be a characteristic time constant defined by the mass of the car divided by the resistance to motion (t=m/r) where that resistance is the amount of angular force felt by the brake pad at a given speed (It's only angular because cars have wheels that roll.) If the brakes simply dragged against the ground the calculation would be much simpler.). Within one time constant 68% of the energy will have been dissipated and within two 95%. Within 5 time constants the type of friction involved will have almost certainly dropped well below the threshold for the brakes to lock (static friction) unless the vehicle had been traveling at several times the speed of sound.
Really interesting outcome.. :/ also, love how you put in the knight rider theme there at the end- ;)
Two assumptions were made in the model :
1st: constant (de)acceleration.
2nd: same amount of energy reduction when braking.
Those are not necessary the case. In fact, one can devise other models when both cars, having a common acceleration function, successfully stop just before colliding.
Miles per hour AND driving on the left side?!?! :/
England
+panulli4 - You understand now how a tiny savage tribe in the middle of a rainy foggy island, first dominated its neighbors tribes, then, the world...
Garry Iglesias Wow, I'm impressed.
And then reporting efficiency in miles per gallon... [facepalm]
The 2 aren't related so I don't see why that matters.
Hey Mr sparks you were my maths tutor and in house tutor at school, frictional force is proportion to the square of speed. Therefore applying the same coefficient of friction to the brake discs means the amount of force applied to decelatrate is greate, therefore the rate of declaration is not constant (suvat equations can not be used).
Yes, air resistance is proportional to the square of the velocity, but we assumed that the whole process only takes a very short amount of time, so we can neglect air resistance.
The frictional force of the brakes is independent of the velocity, so that's why these calculations are used
@@Mmmm1ch43l friction force on the break is completely dependent on velocity,
@@williamhauser4290 No, it's just not.
Under normal conditions sliding friction force only depends on:
-The coefficient of friction, which in the case of the brakes only changes over long time periods because of heavy usage or if you overheat them by braking repeatedly in quick succession.
-The normal force, which is directly proportional to how hard you brake and there's no reason to assume that driving faster will also allow you to stand on the brakes harder.
Ben has such a nice voice!
The most entertaining guy on numberphile. Wish there were more videos with him
That speedometer looks neat. Might be hard to make sure you're under 20 in that pesky residential street, tho.
I think the guy in the blue car may still be in trouble, because if the tree doesn't remain stationary, but moves forward on the end that the red car hits, it's going to pop back on the end where the blue car stopped. I have no idea how to mathematically figure that out. This is interesting, and you have a new subscriber here.
Because velocity is assumed to decrease linearly, kinetic energy must decrease squared. Hence, delta_E1 is not equal to delta_E2 (in respect to a certain distance). In summary, this calculation cannot (I think?) be correct.
Why would velocity decrease linearly though?
The force generated by the breaks is (approximately) constant, so the amount of Energy lost is linear (with respect to the distance). Thus the velocity decreases with the square root of the distance (as claimed in the video)
Never heard of the "suvat" equations, but the only equation you need is the v_f^2 = v_0^2 + 2ad. (Yes, this is the conservation of energy equation or the change in kinetic energy equals the work expended.) Here v_0 = initial velocity, v_f = final velocity.
Blue car:
0^2 = 70^2 + 2ad
From this you find, 2ad = -4900.
Red car:
v_f^2 = 100^2 + 2ad = 100^2 - 4900 or v_f is the square root of 10,000 - 4,900 or 71.4 mph
People need to know this
Do they not teach this when you get your drivers license in GB?
velocity =initial velocity(u) plus acceleration(a) times time(t)
(s)distance= initial velocity(u) times time(t) +1/2a(acceleration)t(Time) squared
s = 1/2(u + v)t
v squared = u squared + 2as
s = vt - 1/2a(t) squared
This is all caused by simple fact that in UK poeple drive on the WRONG SIDE OF THE ROAD!
Nice try mathematical troll. The tree was across both lanes 😏
Norest you mean the correct side
Alistair Shaw nah bud, here in the states we literally drive on the Right side!
Left side best side.
I'm in the UK and I agree. I think we drove on the WRONG side so it costs other countries more money to make cars for us. And so protecting car companies in the UK. Didn't really work out. :)
The units when you solving this problem do not matter. The result is valid if we consider that both cars have the same mass and acceleration. I think it was necessary to clarify it. In this way it can be stated that the work to travel the distance to the tree is the same and arrive at the proposal of the subtraction of the kinetic energies that is seen in the video.
“Calculating a Car Crash”, you mean my marriage....?
Sounds like you were going 100mph and came off worse 😂
The answer makes since once you hear it and think about it (before matching it out) but it is still surprizing.
I posted this on a car forum back in 2002, none of the petrol heads believed me, even with the maths. All believed they (and their modified cars) could '...break the laws of physics, Jim'. Wonder how many of them drove red cars and are now wrapped around a tree...
And if it was just that they wrapped around a tree the problem wouldn't be so big. But often enough they wrap around someones family car...
often they do not crash at all and make someone else crash...
yes, the best way for the red car to lose speed is to scrape the blue one
The best way for the red car to lose speed is to go autonomous. . . .
Rew Rose ( ͡° ͜ʖ ͡°) autonomous
1:15 Of course! It's travelling at a slower rate, so it has less mass and got smaller.
Can you add the km/h units?
Junker just assume one guy is going 70 km/h and the other 100, still works the same in this model
70 mph = 112 km/h, 100 mph = 160 km/h. Those speeds are really high
tommihommi1 Oh, thanks.
I figured that the speed they put had more of a purpose.
70 mph = 112.654 kph = 31.2928 mps, 100 mph = 160.934 kph = 44.704 mps; do note those are miles per hour, kilometers per hour and meters per second respectively (not miles per second, or meters per hour)
It has to do with the ratio, not the unit of measurement. I get why they used mph, but in doing so they should made made clear that it does not matter.
Another approach to understand why this happens is that the breaks of the red car have less time to reduce the car's speed, as it is approaching faster to the tree. It is actually the same reason, just a different way to explain it.
Maybe the cars should consider to drive on the correct side of the road
If left isn’t right, right’s all that’s left!
Well considering the roads and cars match...then its the correct side.
Here in America we drive on the other side...but our steering wheel is also on the other side.
So ultimately its the same thing.
They are on the right side. Only uncivilised heathens drive on the right. And Kiwis...
The friction force is only proportional to the normal force, and not the speed. Which means that the friction force is actually not greater the faster you go :)
That Girl from Ipanema 😂❤🇧🇷
Nailed it!
Substracting the kynetic energy 100^2 - 70^2 = v^2. Since 7^2 ~50, the red car crashes at v = ~70mph.
Actually VW did this with their speedometers. They have ones that go 10-20-30-40-50-60-80-100-120.... KPH that is.
That's the _opposite_ of what he proposed though. He wants something more like 0-15-30-40-50-60-70-75-80-85-90-95-100 as we move between roughly the same angle each time. That way it's 6 steps between 0 and 70 and 12 steps between 0 and 100 which accurately reflects that the latter is approximately twice the kinetic energy.
codebeard I think we are talking we are talking about different implementations of the same thing. In his case the values of the speedometer remain the same but the gap between them changes. In VW's case the gaps remain the same but the values grow in somewhat nonlinear fashion.
Also in modern cars the acceleration between 90 and a 100 for example is trivial so the needle will have to move more rapidly the higher the speed. In vw case it is the opposite: the faster you go the slower the needle moves.
I guess what I'm saying is that what he proposes is a static representation of the energy of the car and how it grows and on the other hand we have a dynamic representation with the movement of the needle.
Codebeard What Sparks proposed, or the alternative you describe makes sense; giving the driver a clue as to the potential damage in a crash. I wonder what VW's reasoning was for doing the opposite. Let's lull people going over 70 into going even faster so they're more likely to crash and be removed from the gene pool?
Дичо Дичев
If the gaps remain the same but the values increase faster at GREATER values, then this is the OPPOSITE of what was proposed. Which is what codebeard said.
volkswagen are made in germany and are intended primarily for the german market, same with audi (which are basically teh same company) the way their tachometres work they are designed for key speeds to be easily identified.
for example thirty and fifty km/h (most common speed limits in populated areas) where highlited red in the audi i owned, 100km/h (the speed limit outside of populated areas) was right at 12 O'clock and another highlight was put at 130km/h wich is the generally recomended speed for driving on the autobahn (you can go faster if you want but if say someone else causes an accident which involves you while you where going over 130, that guys insurance can argue that you could have avoided the accident by going slower so they might not pay all of the damage because you are partially to blame.)
This speedometer idea is just brilliant! They should really make these.
Why didn‘t you just use yards/minute instead of miles/hour?
because people in metric coutries have no clue how much yard is ;)
Yards per minute sounds really annoying.
Wtf
Exactly the point panulli4 was making and yet they were happy doing a video in miles per hour.
Probably happy doing it because that's where they live no?
This is EXACTLY what got me at the speed awareness course!
Wait what? This guy taught me GCSE maths!!
Did I? Sorry about that...
Haha Don’t be!! It was because of you I got an A* then later went off to get a degree in Maths!
Haha don’t be! It was due to your tutorage I got an A* and then went on to get a degree in Maths! So Thank you!
The intuitive explanation:
You have to remember that when braking, not only is there the 30 mph speed difference that the brake has to make up for, but you are also travelling more distance in the same amount of time, so you have less time to apply the brake before the crash.
That’s what I was thinking!
More videos like this!!!
"SUVAT" encompasses energy, not dodgey at all: Vf^2=V0^2 + 2aS. That is work-energy, just with the mass divided out. In this example, modeled with constant acceleration, 2aS = -70^2 for the blue car (since Vf is 0). "2aS" is the same for the red car because it has the same acceleration and distance traveled, you get 100^2-70^2=Vf^2 when you look for the speed of the red car the instant before it touches the tree.
First comment on a Numberphile video? Worth my dollar on patreon I reckon!
P.S. Haha at the end, "Don't just think "what formula do I use?"" Formula, and they're formula 1 slot cars!
It's "dad joke inception" in that final segment!!!
Maths in the hood
Or under it...
Brilliant video - I think all drivers should watch this!
10*sqrt(51) mph = ~71.4 mph?
Yea, I was corrrect
whats that formular? I used 100/70=1,43 ; 1,43²=2 -> red car has twice the energy; so on impact it has the energy of the blue car when it started which is ~70mph
If v is the final velocity of red car, I got, 100^2-v^2=70^2-0^2. Since, v^2-u^2=2aS and acceleration and distplacement of both the cars is the same.
same
Always love your videos!! Each new video opens my mind to whole new set of possibilities!! TY!
My way of thinking was the higher velocity meant more resistance against breaking. It still all comes down to the fact of non-linearity, that so few people fail to understand.
The bigger problem is, that the faster person travels further in their reaction time, so they start braking while being closer to the tree, than the slower person.
YunaQQQ reaction time is not taken into account. red travelling further WHILE breaking is the point
In the example the resistance against braking is precisely assumed to be independent of speed...
I took a very complicated path to solve that. Ended up solving 9 simultaneous equations, covered 6 A5s, but got the right answer at the end! :D
"I want car manufacturers to do a thing and one day I will patent it so they can't do that thing"
Vyngorn, he'll patient it so they have to pay him to do that thing.
As the masses of the two cars are the same (and the equation is simplified to not take into account other variables dependant upon speed) then on the kinetic energy equation 1/2*m becomes a constant multiplier (and as such can be consired to be just one) and pythagoras theorem is only thing left. Area's defined by the triangles sides becomes the kinetic energy and the difference between the two areas defined and proportional to speed gets the red cars speed.
Since he is going faster the red guy would actually have less time to brake, so he would get rid of less energy. I think the actual speed he would hit at is faster than the 71.
Tayler Robinson valid point . But won't the rate of deceleration be more because more contact with the brake pads ? So won't the speed be lesser than 71?
K.R. Koushik it would be the same amount of contact because the wheels are going faster, but for less time. So maybe it would still be 71? I dunno. I'm not great with physics.
If you're talking about the reaction time it's said in the video that they start braking at the same time. If you're talking about the eq. it does include " 1/2 a.t^2 " but it still means he hits the tree at 71 mph. He just calculated it differently
This would be the assumption I guess, alternatively you could also assume that given the higher speed the red person may be paying more attention to the road given the greater implied risk of crashing and thus actually break before the blue car who we could assume is not paying as much attention to the road obstacles.
yenioyuncu255 he's not talking about reaction time. He's talking about the time he has for deceleration .
And my point is that the rate of deceleration would be more because they have the same brakes and one is traveling faster.
Torricelli's formula got me to that result just as quickly as the kinectic energy approach.
Mathologer roasted you
It’s so cold in England right now I could totally do with some warmth.
Lol was looking for this comment
cnjdev No, because once you reach 2147483647 mph, you will start going -2147483648 mph.
Weil what do you mean? Infinity isn't a real number, but it is a valid result of an infinite sum (since it is part of the extended real line). I don't see why they would've needed to mention nonstandard models *or* philosophy.
best pr disaster ive ever seen :O)
you can also solve it without energy equations using kinematics:
vf^2 = vi^2 +2a∆d
so we have for the blue car:
0^2 = 70^2 + 2a∆d
so a = ((-4900)/(2∆d))
now let's call the final velocity of the red car "f"
f^2 = 100^2 + 2a∆d = 10000 + 2(((-4900)/(2∆d))∆d
so f^2 = 10000- 4900 = 5100
so f is about 71.4 mph
Is Brady trying to start Physicsphile? I had to dig back to high school science to remember about kinetic energy. But I do remember that we talked about the square of speed in driver’s ed when discussing why speed kills.
I have a physics channel called Sixty Symbols. th-cam.com/users/sixtysymbols
This is "applied mathematics"
Yeah, it's mechanics, which is a maths module. It was for me at least.
We talk about kinetic energy and braking distances increasing with the square of the speed in Swedish driving schools too. But for some reason, the writers of the main driver's education material have decided to call it (kinetic energy) "the living force", which (as a science guy) drives me mad...
So applied mathematics is just physics, then?
This should be mandatory material for every learner driver. In fact, I coincidentally came up with the same sort of question when discussing braking distances with a friend of mine. I initially thought I had done something wrong (I used 50 km/h and 100 km/h). After multiple checks I saw how dangerous driving 100 km/h actually was in comparison to 50 km/h.
Damn, my intuition told me 70mph.
u must be kidding
Nah, think about it. The distance to slow from 100mph -> Xmph is equal to the distance it is to slow from 70mph -> 0mph, however the car travelling 100mph will pass that distance at a much quicker rate, significantly reducing the time to stop/slow.
I did the same thing. I think it has to do with metagaming the answer. In the back of your mind, you know they have control over exactly how to ask the question, so you assume they're asking it in the way that produces the most interesting answer. He even says in the video that you probably excluded 30mph because you know he wouldn't have asked if that had been the answer. Arriving at 70 intuitively is the a similar process, just taken a step further.
If you got it right, good(I'm going to guess you're one of the few), but I think much of the point of this video is that the answer is completely counter to most people's intuition, which is to say that most people would likely never guess 71 mph.
It helps if you know that 100/sqrt(2)=70.71, so you can guess where he's going...
Another way to think of it is that when you brake, you're still moving forward until your energy reaches zero, which means that higher speed translates to a further distance travelled in a shorter amount of time. In the same first second the two cars move different distances, even if their brakes are of the exact same strength, the cars the same weight, the brakes were applied simultaneously, etc. etc. and each following measure of time the blue car is moving at lower and lower speeds and so gets more and more increments of time to slow down, before reaching that barrier in the road, while the red car simply moves further at each time increment, so cannot slow down as much. This is why, logically, esp. from a physics standpoint which mirrors the results of the mathematics that Ben showed, it maintains a speed higher than the blue car began with, at the point of the barrier across the road, and thus crashes.
Why did you use miles per hour? Doesn't most of the world go by kilometers?
There's a relation between using kilometres, and not losing wars to Vietnamese farmers, I guess.
British guy, they use miles often/mostly. Ever watched Too Gear?
Nah, when does Too Gear come on? Is that the one hosted by Jereny Clarksom?
You mean two gear? or the fifth gear?
The two go together, over there cars drive on the left side and use miles per hour, coincidentally.
This is the oberth effect applied to cars! Really interesting way of demonstrating it, given the lack of rockets...
Is that the host of SciShow Space wearing a wig? O.O
You're welcome. I'm surprised other people haven't noticed the resemblance. Or maybe it's just me?
Here's another way to think about this:
If the speed at which you stop just in time (70 mph in the example) is one side of a right triangle, and the initial speed for the crash case (100 mph) is the hypotenuse, then the impact speed (71 mph) is the other side.
So, if you can stop in time at 30 mph, then if you're going 50 mph, you're going to crash at 40 mph.
[Edit:] Solved. See replies.
I don't understand what you mean by "brake at the same rate". Because if that meant that acceleration (a) was the same, then the answer would be 30 mph:
u1 = 70, v1 = 0
u2 = 100, v2 = ?
{ v1 = u1 + at
{ v2 = u2 + at
Subtract equation 1 from equation 2 and you get:
v2 - v1 = u2 - u1
v2 - 0 = 100 - 70
v2 = 30
Unless you consider that acceleration changes over time, but then stating that they're decelarating at the same rate doesn't make much sense.
You are correct, the answer is 30 based on the same constant deceleration for both cars.
But they don't have a constant deceleration.
By "decelerating at the same rate" is meant that the breaks have the same power
The problem here is the confusion between decelerating for some time and decelerating for some distance. It isn't wrong to say that when both cars have "lost" 70 mph with the same deceleration, the same time has passed (and Mr. Sparks probably wouldn't deny this either). However, when the red car decelerated to 30 mph, it travelled a much further distance than the blue car has decelerated to 0 mph.
To explain this, I use the formular for potential energy which you can use it for generally any kind of acceleration. The advantage is that it's E = m*a*s which means, s is a variable here and time doesn't matter for this kind of formular.
If I put this equal with kinetic energy and solve for s, you'll get this: s = v²/(2a). a is the same in both cars meaning that only v is a variable here. As you can see, v is squared meaning that the bigger v becomes, s becomes *much* larger, thus increasing the breaking distance by quite a bit.
Yes, if it's the same power then 70 mph is the correct answer.