A race car going 150 mph went around a circular track in 5 minutes, how wide is the track in miles?

Algebra and geometry word problem with circles, rate, time and distance.
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[@joesmith-rm7px]

Happy to see many others agree that how wide is the wrong question.

[@gemini22581]

Absolutely..a math teacher using the wrong term to ask a question is unacceptable

[@davidstecchi9501]

The "width" of the race track cannot be determined from the information given, the diameter of the path that the racecar travels can be. The width of a race track is the distance between its inner boundary and its outer boundary, in this case the track would be annular. Whilst the inner boundary can be no less than the approximately, there is no evidence that the car traveled the minimum distance around the track, so it could be less.

[@joezolo9986]

150/60=2.5 miles per minute 2.5*5=12.5 times 5 minutes. The circle is 12.5 miles in circumference. 12.5/3.14=3.98 divide by Pi. The answer is 3.98 miles wide.

[@JamesParsonsIT]

Width of the track, size of the track? The problem is using the wrong term. The width of the track would be about 10ft, wide enough for a car to drive on.

[@christopherjz]

Everyone who doesn't have a degree in being pedantic knows they meant diameter when they said "wide".

[@tediouspieceofjunk]

​@christopherjz Anoyne talking about a race track knows that the 'width' of the track is the width of the surface that the cars, motorcycles, go karts, whatever, travel on. The width is a key dimension, as it determines overtaking opportunities, lines taken through corners and so on.

[@briankemp2669]

In plain English this teacher would be marked could do better ...by asking wide when should meant lap of track length...

[@andrewchristofi8386]

How wide is the track? At least as wide as the car. But the diameter of the track is 12.5 / pi = 3.98 miles.

[@stevewaters449]

Diameter of the circuit.

[@darellpiper7227]

I agree

[@throngcleaver]

4 feet, 21 inches wide. (single lane track, of course) The diameter of the track should be sought in your next video. 😊

[@richardl6751]

That is a very narrow track.

[@michaelbaughman8524]

Yes, the problem was poorly worded. Should have been, "... what is the diameter of the track, in miles?" And your joke answer should be 5 feet, 9 inches (since 21 inches is 9 inches over a foot), expressed in miles of course: 5.75/5280 = 0.001089 miles.

[@laurendoe168]

Daytona International Speedway's track is 2.5 miles long and 40 feet wide. Assuming this fictional track uses the same proportion, the track is about 64 feet wide.

[@gerryleddy5412]

Me thinks the questioner is good at mathematics but poor in English , me thinks he wants the width of the circuit not the track

[@jadebaker3629]

2 miles

[@joezolo9986]

150/60=2.5 miles per minute 2.5*5=12.5 times 5 minutes. The circle is 12.5 miles 12.5/3.14=3.98 divide by Pi.

[@11211lcb]

For a math guy you failed this one. As others have pointed out you used the wrong terms. You should have said what is the diameter of the track instead how wide is the track. The width of the track and the diameter of the track are two completely different things. I got the answer right because I discounted the width of the track which has no value to solving the problem.

[@joezolo9986]

What's the difference between the width and the diameter of a circle? I don't know. I understand the difference is in an oval. I don't know how to calculate for an oval. But I understand the difference. This is how I got the answer to this problem. 150/60=2.5 miles per minute 2.5*5=12.5 times 5 minutes. The circle is 12.5 miles 12.5/3.14=3.98 divide by Pi.

[@user-iy6wp6rl3c]

Circle is different from oval

[@phill6859]

​@@joezolo9986a track is made up of two parallel lines. A circular track is therefore two circles. The width of the track is the difference between the two circles diameters. This question asks how far the car travelled, which has no relationship to the tracks width.

[@peterallen2904]

God help us if this guy is teaching young people math (or to use the correct term, maths) . He got it all wrong.

[@RT22-pb2pp]

no should have asked how wide a circle did car travel in track could be 100 miles wide and he still went 3.98 miles LOL

[@_Aardvark_]

This problem is so badly worded. A track 3.98miles wide! Imagine a road 3.98miles wide. That's a lot of lanes.

[@thecosmologist]

When you ask a maths question you have to be both specific and precise.

[@robertstuart6645]

I agree with the comment below, in that the question could be better phrased. Instead of "wide" say "diameter." It won't make the problem any less difficult to solve.

[@NegativeAL]

If this guy is teaching our kids, god help us and them 😮

[@nelsonx5326]

Did you mean to say 'god help us and them'? You wrote 'helps', which completely changes the statement. He's not a bad teacher. I'm not a math person so I might not be able to tell a bad teach from a good.

[@jimbishop8667]

The track width could be anything... The answer you are looking for would be the sum of the diameter of the circle described by the length of the track!

[@michaelbaughman8524]

Just the diameter, not the "sum of the diameter".

[@jimbishop8667]

@@michaelbaughman8524 Thank you Sheldon.

[@nostera5142]

I'm starting to think that maths prompts are just worded to mess with the student's brain, and this is exactly what he's showing right there.

[@celsopinheiro]

The right questions are more important than the right answers...

[@mikemoore9757]

I have noticed in another video you did as well as this one , that you refer to a circles diameter as it's width. A circle has a diameter. a radius. a circumference, an area.... etc, but not a width. That's just my opinion . I enjoy your presentations. Thanks and keep them coming. You are the patron saint of the mathematically challenged!

[@lukeknowles5700]

At 8:12, where you show the two steps, it is important to realize that the meaning of the 'D' in each formula is different. The 'D' in step 1 stands for Distance. The 'D' in step 2 stands for Diameter.

[@dorischoo]

Wish you were my Math teacher when I was, in my primary one class. I was slow in absorbing and missed out all the basic u r explaining. Thank you.

[@harpleblues]

Hopefully it’s wide enough to contain all 4 wheels of the car.

[@niallhutchinson00]

Or that 5 mins won't count

[@MissJean63]

I started by getting the miles per minute by dividing 150/60 then multiplied by 5. No fractions needed here.

[@martinseal1987]

Thanks this makes so much more sense

[@dunc71]

This is the normal person's way

[@larrywiener6327]

I did it a slightly different way. I used a proportion to find the circumference after I changed hours to minutes. Then I divided by pi to get the diameter

[@stephensharma4994]

i got this correct ,,, but the problem was not stated correctly you should 've asked for the diameter of the track and not how wide it was (the track actually is about 10ft along the circumference..

[@stevewaters449]

....of the circuit!

[@johnbeck9629]

This is a trick question. I'm going to hazard a guess. It is probably at least the width of a standard two lane highway. The distance from one inside edge of a circular track to the opposite inside edge of the same track is probably a little less than 4 miles...I'm guessing again

[@topkatz58]

D = R × T Circumference of the Circle = 3.1416 × Width of the Track = D = 150 miles/hr × 5 min × 1hr/60 min = 150/12 miles = 12.5 miles Width of the Track = 12.5 miles/3.1416 ≈ 3.98 miles

[@Don-je8sm]

This guy is professing he is a math teacher, but failed English. The width of the track is at minimum the width of the car, but with the car travelling in a circular manner, there would need to be more pavement to allow for the sliding. He calculates the diameter of the teack, however, using a line, the diameter may be what he calculates, but he does not answer his own question. Even in his picture, he shows the car, see the yellow anular shown. So you have to add 2 times the width of the car to get the minimum width of the circle required. This is why some students have problems in later years, because they are taught poorly from the start .

[@johnnyfearlesszrx]

I asked my road maintenance guy for a circular track to be laid. He asked how long and how wide. I said 12.5 miles long and 3.98 miles wide. He said oh, so you don't want a track, you just need a car park in the middle of the track!

[@nicoleb3636]

3.98 miles. At 150 mph, the driver traveled 12.5 mi to get around the circle in 5 min. Circumference is 2*pi*radius or pi*diameter. This means 12.5=3.14*d so d=3.98 miles.

[@GoofballFlyer]

If the track width = diameter then 150/60 = 2.5 Miles Per Minute * 5 minutes = 12.5 miles traveled = circumference / 3.1416 = 3.979 miles. This assumes the car straddles the circumference line.

[@rogerfroud300]

Some of these questions are very poorly worded. You want the radius or diameter of the track. The width is something completely different.

[@peterbrewster6661]

The question is bogus. I guess you are asking for the diameter,which is pretty trivial stuff, calculate the circumference from the speed and time, and divide by pi to get the diameter.

[@peterdodwell5549]

Wrong category! He should be asking about the track's circumference. The width of the track relates to the track's running surface . I.e how many cars can fit side by side

[@loskylp]

I agree with most people, when reading the problem, the "track" is the path the car goes on, in this case the 'corona" around the circle. And the width of it should be enough for the car to fit in the track

[@user-py1kr3sq9t]

Who cares?.

[@loskylp]

@@user-py1kr3sq9t apparently you that are watching the video and reading the comments 🤷🏼‍♂️

[@vanderpoolfarmsl.l.c.9983]

Another example among many in which our great math teacher fails in the English part of the story problem. Nice try anyway. Thanks for the daily problems.

[@stevewaters449]

Agreed, if you can't write correct English you'll never get the answer you want/expect.😊

[@thomasperchalski6004]

What you are actually looking for is the length of the track of the car stays the same racing lane all the way around. The width is the distance from the inner diameter to the outer diameter.

[@orion7592]

I'm from SA, so I use the metric system: 150mph needs to be converted to kilometres per hour: thus: (150)(1.61) = 241.5km/h Convert to metres: (241.5)(1000)= 241 500 metres... Divide by 60 to get metres per minute: 241 500/60 = 4025 metres per minute. So, the race car completed a circuit of the track in 5 minutes, therefore; (4025m)(5) = 20 125 metres. Which is: 20.125km. The track is circular, thus it will have a diameter: So: 20.125/3.1415265...(pi) = 6.405986459km. Convert back to miles: 6.405986459/1.61 = 3.978873577... Use three significant figures... Width (diameter) of track: 3.979miles. The "width" of the track in this problem refers to the diameter, not the "width" of the actual track. It is an idealized question, as is most questions in Physics and maths. One needs to be intuitive about this. The examiner will not ask you to calculate the impossible, such as the width of the actual track, since that can be anything, in this case. You need to understand that "width" here, means diameter. Always look at context of question!

[@richardfrankham7522]

The first rule in setting a question is to be unambiguous.

[@Aereaux]

I agree with those commentors that the question is not worded correctly. It would have been much better ask what is the diameter of the track. Of course since the track has to be wide enough for a vehicle (say a race car with a width of 6 feet) then the diameter would vary from inside to middle to outside.

[@richreddell3569]

He’s looking for the length or perimeter not the width. Teachers using wrong terminology are why I had such a hard time in math

[@nyneeveanya8861]

Can’t answer this. How many lanes on the track. Is car on inside lane or outside lane or sometimes both. Are you looking for the circumference of the lane the car is in rather than the width if the track. Now I’ll go watch this video and see what you meant.

[@MrMousley]

Car travels at 150mph and takes 5 minutes to go round the track once. 5 minutes is 1/12 of an hour ... so we need to work out 1/12 of 150 1/12 of 150 is 12.5 .. the track is a circle with a circumfrence of 12.5 miles. We need the diameter ... Pi x D = 12.5 ... 3.14 x D = 12.5 ... D = 12.5/3.14 ... 3.98 The diameter of the circular track is 3.98 miles

[@Harker777]

The math is pretty easy as long as one hasn't forgotten pi

[@richardl6751]

Here's another way: 150 mph times (5280/3600) = 220 feet per second 220 fps times 60 = 13,200 feet per minute 13,200 fpm times 5 minutes = 66,000 feet 66,000 feet / 5280 = 12.5 miles 12.5 miles / Pi = 3.97887 miles QED

[@joezolo9986]

150/60=2.5 miles per minute 2.5*5=12.5 times 5 minutes. The circle is 12.5 miles 12.5/3.14=3.98 divide by Pi.

[@kevinphanson]

@@joezolo9986Why waste steps converting miles into feet, and then back to miles before the end? The whole problem can be done in miles, and be finished at least 2 steps faster than your way.

[@joezolo9986]

@@kevinphanson Exactly. That's why I did it this way. If miles was an absurd hard to calculate number, then convert to feet. But it wasn't. So I didn't.

[@thomasringa6099]

Now we know why most people hate math. The teacher should have explained the problem in 3 minutes instead of 25!!!!

[@StephanBuchin]

5 minutes is 1 hour divided by 12 so if the car can travel 150 miles in 1 hour, It will travel 150 miles divided by 12 in 5 minutes which is 12.5 miles and this is the circumference of the circle. Now the circumference is equal to 2πr so the radius is 12.5 divided by 2π equals 1.989 mile. If the car drives on the edge of that circle and the surface is paved all the way to the center, it should be 1.989 mile wide though the question is rather ambiguous. If, instead, the car drives in circles on a very wide straight road, then you must consider the diameter which is 2r or 1.989 mile times 2 = 3.979 miles and this would be the width of that road. Edit: you decided it must be driving in a circle on a wide straight road 🙂

[@stevewaters449]

Presumably you mean the diameter of the circuit. The track only needs to be wide enough to accommodate the car.

[@r.c.brousseau9655]

John, why did you chose to refer to the diameter as “width of the track”, which is a misnomer. From what I’ve seen of your questions, getting past your terminology is the first challenge to overcome before actually dealing with the math question.

[@gregdillon5352]

This is the inner circle. The outer circle wide be about an extra 6 feet so that other side of the car would be travelling slightly faster. Physics is fun when applied to the real world.

[@ellentronicmistress4969]

I feel this is a badly worded question. For me, the width of a track is the distance from the inner part of the track to the outer part, in the same way as its lengh would be the distance you would have to travel to do a complete cycle.

[@stevedietrich8936]

3.98 miles, car is traveling 2.5 miles per minute, times 5 minutes is a track circumference of 12.5 total miles. Circumference = pi times the diameter of the track, 12.5/pi = Diameter

[@martinseal1987]

Went full Mr Mackey on this one

[@kwp9138]

Trick Question the Width of the track cannot be determined by information given.

[@donfacundo2118]

3.98 mi is my answer to the diameter, meaured on the track median. The track width should be 6 to 7 times the car width so 6x2 or 7x2 is about 12 meters to 14 meters

[@stevewaters449]

Why that wide? 3or 4 metres wide should be adequate.

[@jsand8301]

Whatever it is that is a darn big race track.

[@danieldennis9831]

πD=150/12 (πD=2πr) πD=25/2 D=25/(2π) D≈3.98 or 3.9788 or 4 depending on how precise you need. I then saw other comments and agree that the wording is off. What is the track diameter is the question. How wide is the track ... most tracks are 4 or 5 lanes ... so 40-50 feet wide?

[@user-cb6nu3em9g]

He might be a maths teacher but certainly isn't an English teacher!

[@richardwilson3874]

How do we know that pi is infinitely repeating and non terminating number?

[@limogesfarmer6326]

26 minutres..... I literally figured this out in under a minute, unsiong a calculator.

[@needhelpwaquaregiadrop562]

Good looking and smart

[@texbacalian5036]

The width of the tract cannot be found using the figures. But if you are asking for the diameter of the tract at the middle, it's 12.5 miles.

[@norincobro6421]

The track is 12.5 miles in LENGTH

[@williamofravenshire]

this seems long - I did 150/60*5/3.14 came out exactly like your answer. showing my work, 150mph there are 60 minutes in 1 hour divide by 5 minutes gives you 12.5 circumference which you divide by 3.14 equals 3.98 diameter

[@jadebaker3629]

Surely the width of track at most would equal the radius

[@kennethwright870]

150mph/12incriments per hour=2rpi, 50/4 = 2rpi, r=50/8pi = approx 1.989, so, Diameter= 2*r = approx 3.98

[@markuswx1322]

I come up with about 3.98 miles 150/12 = 12.5 miles traveled since 5 minutes is 1/12 hour. We then divide by pi to get the diameter of the track. Your wording is a bit tricky because you use the word 'wide' when describing the circular track. That is one heck of a big race track.

[@tonybloomfield5635]

I did it your way also. Some other solutions here overcomplicate a simple equation.

[@dunc71]

I haven't even watched the video but the width of the track is not worked out using the numbers provided.The track could be 10 miles wide and the car could be on the inside therefore covering a much smaller distance in one lap than if it were on the outside. Also is the car starting at 150 mph or starting at 0 and accelerating to 150? Basically this is nonsense

[@michaelbaughman8524]

Race car's speed 150 mph divided by 60 minutes = 2.5 miles per minute; 2.5 mpm * 5 min = 12.5 miles traveled. So the Circumference, "C", of the track is 12.5 miles. Using the formula for the circumference of a circle, C = 2πr where r is the radius of the circle, we get 2πr = 12.5 miles. Divide each side of the equation by 2π (approx. 6.283) to get r = (12.5/6.283) = 1.99 miles. Finally, the radius X 2 gives the diameter of the circle: 2 * 1.99 = 3.98 miles.

[@markuswx1322]

All correct, of course. My only suggested improvement is to do the math without any unit conversions. This isn't always obvious from a teaching point of view. I elected to avoid minutes altogether by first converting 5 minutes back to 1/12 hour before completing the computation, and using the more direct πd rather than 2πr to derive the diameter.

[@michaelbaughman8524]

@@markuswx1322 Great comment! Yeah I did actually divide 150 by 12 to get 12.5 miles, but broke it into the minutes conversion as easier to explain, although your formulation "convert 5 minutes to 1/12 hour" is clear too. I was too fixated on the classic 2πr = C though and didn't think to simplify to πd. (Why is the formula always stated as 2πr, if πd is simpler? Presumably because radius also applies to arcs whereas diameter only applies to a circle?)

[@markuswx1322]

@@michaelbaughman8524 Funny how the simplest problems are the most important when learning or teaching. For many, the initial conversion to minutes would illustrate the problem most clearly. TC Math man's solution emphasized the Rate-Time-Distance issue in order to drive this home. For the rest of us the shortest way is often the best. In this instance since there were no arcs to measure in degrees (as you noted) it's better only to consider diameter (which is what was called for, though he worded the problem strangely) rather than radius to solve the problem.

[@jlaz920]

Is one Mille diameter divided by pi

[@devonwilson5776]

Greetings. The width of the track is the diameter of the circular track. First, we need to find the circumference, distance around the track. To do that, we use D= VT and D = (150×5)/60 miles. Now that we have the value of the circumference, and we know that the circumference of a circle is 2 PI R. We can equate 2PI R= 150×5/60, and R= (150×5)/60 ÷(2×22/7)= (150×5)/60 ×7/44=1.9886 miles. We have just determined the radius of the circular track. Now we will double the radius to find the diameter, the width of the track. That is 1.9886 ×2= 3.98 miles. Lovely.

[@roberthennebert571]

You make it so complicated there is a very simple way to calculate the diameter of the track

[@doreenleeming4630]

I think the question is wrong.The question should be how long is the track.

[@williambarnes3868]

The question is ambiguous! It should ask 'what is the diameter of the track'. The width of the track is the distance beyween the inner and outer edges of the track. Poor question setting!

[@tedsmith3061]

Another good catch up. When asked how wide track was I wasn't sure if you meant just circumference. I got that number in my head in 3 seconds. You wanted actual diameter. No racetrack in the world is 12.5 in a circle. Doesnt happen anywhere on earth. Anyways the math brush up for me was plugging in the formula for diameter of a circle. 12.5 over pi. Trick question for a mechanic. A racetrack that long would be at the very least the world's largest oval track and woo-hoo what a race spectacle that could be or, also a world breaking road course track. In Nascar a typical road course layout in the pro circuit is under 3 miles. So yeah...a 12.5 circular race track...uh no. Corny. But still you add a little bit to making the rockin world go round wit yo fun math problems! Keep em coming.

[@randallhoward3231]

That's my answer as well, after rounding up to the three most significant digits. The measurement to the right of the decimal point goes subatomic.

[@laurendoe168]

You're right, three decimal places is close enough. Since the unit of measurement is miles, three decimal places is measured in feet (5.28 ft to be exact). Six decimal places is a millimeter-ish. Nine is micrometer.... you need about a dozen decimal places to go subatomic.

[@michaelnunan1302]

Width of the car or wider.

[@bobcornwell403]

The track is approximately 3.98 miles wide.

[@xmurli]

question is not right. it should be " what is the diameter of the track". ((150/60) x 5)/3.14 (Pi)=3.988.....

[@ThomasMahoney-kn3qb]

The diameter of the track has nothing to do with this answer. What you found was the diameter of the circle that the car traveled. No wonder our kids aren't proficient in math or English.

[@olubunmiafolabi5636]

Diameter = 2r = 3.98miles

[@fayekeller7411]

Simpler 150÷5=30 then 1-5+5=9 ... Eliminate zero from the 30..3.9...close enough without having to go into too much detail that's how I figured it out on my own and it is approximate for the width of the track

[@keithschipiour4684]

I have no idea but i am not applying for the job of crew chief of a race team either .

[@raynarks]

Is this a maths or an English language question?

[@MrTalkingzero]

I agree in principle with another comment here that wording in WORD PROBLEMS is important. The width of the track is like the width of a road. That cannot be determined by the information given. Also, doubly confusing, when you ask for the width of the track and I realize that your word choice is poor, do I guess that you mean the DIAMETER OR THE CIRCUMFERENCE of the circle that forms the track? Probably not length but who refers to a diameter of a circle as the width? Math problems should not be ambiguous.

[@ratwest]

What's a silly question completely ambiguous

[@kurtmooreca]

One second guess using percentage based math (disputes with wording provided below), The width of the track is impossible to determine without further definition. Is it an off road track, a single lane track, a multi lane track? All I can determine is the length of the track, or distanced traveled by the vehicle in 5 minutes. around a circular track, and since you defined 150mph as a constant speed, then in 5 minutes we get this. 5 minutes is 1/12 of an hour or 8.3% (.083333) If the vehicle maintained a constant speed of 150mph this gives us. 150*.0833 = or 12.5 Miles. give or take some rounding privilege. Now to see if I am right, or got caught in the ambiguous wording of the conditions. * I was not correct and a victim of ambiguous definition, you were asking for the diameter of the track from an arbitrary undefined center. A 4 mile wide race track...is insane, especially if it spans 12.5 Miles in length Edit Also fwiw based on the gained knowledge and my knowledge of physical property deviations, the Lane in question is ~7.89 ft wide vs the center of the vehicle. The Diameter says the Radius says 1.99 + 1.99 based on property deviation at the objects center To clarify in a math sense, 1.99 Mi, 1.99 Mi = 3.98 Yards (first deviation) = 7.89 ft (second deviation). (this is why metric is so much better btw as a deviation is just x10 away) In order to maintain a speed of 150MPH on a congruent circle of 3.89D then the lane must be at least 7.89ft wide to allow for sustained traction and distance between outer edges of the lane. So the answer to the question as asked is, 7.89 ft is the width of the track.

[@BlackhawkPilot]

Don’t use a subject that you obviously do not know the vocabulary in a problem. Width is form side to side of the racing service. You are looking for the diameter of the circle?

[@MostlyPennyCat]

"what is the diameter of the circle described by the track"

[@janinapalmer8368]

Did it in my head .. easy 😊

[@lylemorton8641]

It’s a circular track and he wants to know the diameter of the circle not the “width of the track” assuming the whole thing isn’t black top. Math is great but as is as it was years ago, math teachers suck at English or perhaps they like being vague to create more problems. Still enjoy the videos though :)

[@elliotlambert3817]

Its not width it is diameter

[@kevinreist7718]

Okay, 5 minutes is 1/12 of an hour. so 150/12=12.5 miles long for the length of the track. Now Circumference=Pi x diameter which would give you the width. 12.5 miles/pi=3.979 miles, which is the approximate width. Now. I'll watch to see if you have some clever trickery in store for us. Alright, I've said this before, but I'll say it again, your communication skills are much better with story problems than with poorly expressed equations. Which are clearly designed to confuse. The teachers who design trick questions can really discourage students who are just trying to learn. It is also a disservice to our educational system and is not very professional or ethical. Remember, teachers are paid to educate their students, not to confuse them.

[@harrynielsen4162]

that is the diameter of the track, not the width of the track

[@50andBad]

Its a dud question....the width cant be determined even if the length of the track can

[@richard2co]

Math is the language of the Universe.

[@theobgshow]

Isn't

[@russelllomando8460]

i got 4, cool, thanks.

[@Ayelmar]

At the title card: answer is just s smidge under 4 miles -- 3.9788.... (150 mph / 5 min) / pi = 3.97887... miles diameter. BTW, "diameter" is the term you should have used, not "width" -- I was able to guess what you meant, but terminology is important.

[@terben7339]

I don't think that 'wide' means what you think it means. I think you are actually looking for the diameter of a circle. And I'm not sitting through a 27 minute video about a problem easily solved in less than 10 seconds. Badly done.

[@stephenbroadhurst7653]

150 divided by 12 then divided by 3.142 which is approximately 4 miles,and now I’m going to watch the rest of the video and see if I just showed myself up. Well after watching the rest of the video yes the answer was right but the explanation was way above my pay grade. I just thought of a clock face in that 5 minutes is one twelfth of an hour and that I had the figure of 3.142 drilled into me when I was at school all those years ago, which is funny because I can’t remember what I did yesterday.🤪

[@doreenleeming4630]

Why not ask...what is the diameter of the track?

[@victorcapetillo2070]

Very good. I concur. I support this answer 😁. I don't know anything about race tracks, but who cares. Obviously we are dealing with a circle not an actual structure.

[@charlesthomas7962]

The circular track is 12.5 miles

[@PaulSands]

Question is unanswerable as the width of the track, i.e. the bit you race on, could be infinite. Looking at the answer I'm assuming the question at worst should've read how wide, at its widest point is interior of the circuit, or couched in actual maths terms whats the diameter of the interior. Pendantic maybe bit it's important to present unambiguous questions if you're after a specific answer.

[@AmillJaime-ub3lc]

75 miles

[@Charlie-os4gk]

the track is 1.91 miles wide with a circumference of 12 miles

[@Charlie-os4gk]

that's the radius x 2 = 3.82 miles wide

[@jlaz920]

Perimeter divided by pi