Assuming you mean the diameter of the circular racing-track circuit through the centre point (i.e. distance through the centre point of the circumference of the circuit? ). Speed x time = distance 60mph x (15÷60)= 60 x 1/4 = 15 miles. Circumference of circle or distance travelled in 15 minutes = 15 miles. Diametre whose circumference is 15 miles: If Circumference = Pi x diametre, then Circumference ÷ Pi = diametre 15 ÷ 3.142 = 4.774 Answer is b) 4.77 miles 🙏
John: Having raced for many years, the width of the track is the width of the roadway not the distance side to side. Language is important. How about how many acres are needed to build the track?
Well it would be the diameter of the track, since acre is area. But I do agree the width of a track is the paved or graded bit you drive on, edge to edge.
Seems your reply is clearer. This math question is controversial, better being asked the diameter of the track circuit than width of then track which I as assumed as the width of the road
The only thing you can say about the width of the track is that it is probably somewhat wider than the car, given that the car is stable at sixty miles per hour! On the other hand, the diameter of the circular track would be a different question altogether! It was always a real irritation concerning how many mathematics questions are so badly framed in English. Best wishes from George PS: The answer would be that the diameter of the circle in 4.77 miles to two decimal places. [15 over Pi].
At the title card, my answer is b) 4.77 mi. At 60 mph, the car is going a mile per minute. Thus, the track is 15 miles around, since it can be circled in 15 minutes. 15 miles / pi = ~4.77 miles across.
surely it depends on theminimum width of the car and grip of the car's tires and surface material and the coefficent of friction between the two. I assume you mean what is the diameter of the track?. or how far is the centre point of the curciut from the track? or even how much area does the track take up.
Mr. Mathman reminds me of my college psychics prof who would say, "no one learns math by guessing at the wrong answers". He would give us stacks of problems already worked out .
Don't forget the fact that if the diameter of the track is 4.774, that diameter must be increased at least enough such that the car fits on the pavement. Then add in the fact that most tracks are designed to allow two cars to be side by side, and sometimes even three Then add a "safety factor" of pavement.
I know the circumference and diameter of the track. But without knowing the dimensions of the car, I can't give a minimum width the track must be for the car to stay on it. I'm gonna err on the safe side and say the track is 8 feet wide. 15 Miles circumference and 4.77 miles diameter.
Anyone with track racing experience knows, track width ALWAYS means how much room the car has to drive, left to right, avoiding other cars, if any. How far it is across the entire racetrack is ALWAYS the diameter of the racetrack. But you didn't want to give away the solution with the keyword "diameter," implying π, so you said, "width." You could have done better saying, "...total width of the entire course," in lieu of track width.
Greetings. That should be easy. From the details given, we can find the perimeter of the circular track. Thereafter, we will determine the diameter of the the track. To accomplish this, we need to know how to calculate distance using the velocity formula in addition to knowing how to calculate the diameter of a circle when the circumference is given or known. I will forgo the calculations.
Perhaps you are asking the Wrong question. Aren't you actually looking for the diameter of the Track path(Assuming of course it's a perfect circle)? Just asking.... Given these assumptions, I will proceed and try to find his diameter!🙂
VERY badly worded problem. The question should have been, "What is center diameter of the track?" because that's what you're asking for. A circular track that an automobile can drive on has a width that is defined by the inner and outer radius of the circle, neither of which are given. There is not enough information given to calculate the "width" of the track, only the center diameter.
After first reading I thought 'how wide is the track' meant what you said - inner to outer edge of track. That's why John says to read each problem 3 times. You need 3 readings to sort out the ambiguities. He could have asked, "what is the diameter of the circular path on a track taken by a car completing one revolution in 15 min. at 60mph".
You can only calculate an average path the car traveled within the boundary of the track. The track can be infinitely wide. Oddly worded question and is actually irrelevant to the cars speed and velocity.
No, it's the diameter of the circle the car makes...I got it right. 4.77 Using the circumference formula...you need to first calculate the circumference of the circle the car made.....which is 15 miles . Then plug that into the formula... Nothing tricky about this . The width of the track IS the diameter of the circle the car made, WITHIN the track.....IF the track Was very wide and the car travelled on the outside of the track, it would have traveled much further than on the inside of the track.. The question is about how wide the track would be for the given values . Which is ALWAYS different from the inside to the outside, as all race car drivers know 😉
Old timer here, just subscribed, I have lots of time, would like to watch your videos from day one. How do I get to your first video on TH-cam? Thanks 👍
60 MPH is one mile per minute. This means that the distance traveled is 15 miles. We now need to define exactly what the "width of a track" means. It generally means the distance from one edge of the track to the other edge, perpendicular to the direction of travel. However, from the context of the multiple choice, it seems to be asking for the diameter of the track. This is unknowable because it is not mentioned how many laps the car went. From the phrasing, it's reasonable to assume that it completed at least one lap. However, a single lap is ridiculous because no permanent track would ever have a 15 mile long lap. One reason is because the land required would never pay for itself from the revenue. However, the phrasing does not exclude the possibility of multiple laps. If the track was the Daytona Speedway, it did 6 laps; if it was Martinsville, it did about 28.5 laps.
@@thewitt2890 The question absolutely does not say one lap, You don't get a 22 min explanation when you take a test. You just have to try to figure out what it means from the words on the page. The way this is worded is a hot mess. Width and diameter are not interchangeable Saying a car "went around a track" is not the same as "a car made 1 complete lap of the track." There's even more flaws too. Is it the inside of the track? the outside? How wide, the real meaning as in how much road does the car have, is the track? Are there other cars the driver is maneuvering around? From the context you can decipher the writer of the question isn't expecting you solve with those factors but why make students waste time trying to decipher a poorly worded question?
I've been watching your excellent videos. For this one, be sensitive to the fact WIDTH of the track is NOT the DIAMETER of the circle described by the track. While the English language is not as exacting and objective in meaning as mathematics, what little there is should not be minimized because we don't choose the right words. Otherwise, well done!
DAMN. I was all prepared to say something about no way to know how wide the track is but the diameter of the track is about 4.775 miles and that would be the center line of where the car actually traversed, but everybody already beat me to it.
Car going 60 mph = 1 mpm (miles per minute) One circumference = 2r . pi = 2 . ½d . pi = d . pi Speed = distance / time = d.pi / 15 = 1 mpm So the correct answer is d = 15/pi miles is just less than 5 so answer B (stupid multiple choice)
Wait so you are substituting diameter with how wide? I would have NEVER known that without your explanation, The width of a track is the usable track around an unused circle in the middle. I hope you don't give this question to students. You are making them spend way too much time just trying to decipher what you are even asking.
John does not get an A+++ or any stars or smiley faces for incorrectly stating the intended problem. Nor can he brag to his family and friends about his expertise in math or automotive racing. Because the DIAMETER of the track is 4.77 miles, the width of the track could be from 2.38 miles - approx. the width of the car.
Its a pizza with infinite slices that rearranged under integration gives an exact area and as such this number cannot be algebraic since it is not a polynomial root therefore must be transcendental. Driving a mile a minute for 15 minutes is 15 miles. We can check using math. 60miles/1hour = xmiles/1/4hours. I got 60miles/1hour = 4xmile/1hour Multiply both sides by 1 hour to cancel ours or you get a funky unit like milehours. I got 60miles = 4xmiles. Divide both sides by 4 to isolate x. I got 15 miles = x miles I think this is reflexive so the circle is 15 miles circumference. 15miles ÷pi = 4 miles 4,090 feet 1.71 inches. The fractional part is irrational and cannot be written as a ratio of two integers because pi is irrational and any mantissa of a result from it will be irrational as well. This is annoying because I really wanted it to be 1.75 inches. If I could take pi to high enough powers and prove it was and integer then maybe I could write the answer as a rational using pi number of roots. Or not. Or some other method. Or not. Best bet for exact values would be 15/pi. If an engineer needed a decimal approximation he could easy find it to be 4.77 or as many digits of precision he needed.
How long does it take to explain miles per hour? Answer: 15 minutes. 20 minutes to remind me of a simple equation that I should have remembered anyway and if not I could Google it in less than 10 seconds. Simple, 15÷π.
Did you do this on purpose for some reason? Or did you actually screw up this bad? Kinda reinforces the notion that mathematicians don't live in the real world.
Easier to convert from hours to minutes, since we need to know how far we went in 15 minutes. 1 hour=60 minutes..so the car travels 1 mile per minute..therefore in 15 minutes you travelled 15 miles.... Much easier solution, IMHO Then plug it into the formua and you're done. 15 divide by pie...easy peasy 😜
There is no data present to allow the calculation of the actual width of the track. Instantly we know the track is 15 miles long, just use Pi to calculate the diameter. There's far too much talking waffle. This is simple maths that an 11 year old would be taught in the UK
Assuming you mean the diameter of the circular racing-track circuit through the centre point (i.e. distance through the centre point of the circumference of the circuit? ).
Speed x time = distance
60mph x (15÷60)=
60 x 1/4 = 15 miles.
Circumference of circle or distance travelled in 15 minutes = 15 miles.
Diametre whose circumference is 15 miles:
If Circumference = Pi x diametre, then
Circumference ÷ Pi = diametre
15 ÷ 3.142 = 4.774
Answer is b) 4.77 miles
🙏
The track is wide enough for a car to drive on, presumably
😂😂😂 Always a comedian in the crowd.
Hey there Mr TH-cam Math Man, thanks for being there and raising our collective IQs one video lesson at a time 😸👍
John: Having raced for many years, the width of the track is the width of the roadway not the distance side to side. Language is important. How about how many acres are needed to build the track?
Well it would be the diameter of the track, since acre is area. But I do agree the width of a track is the paved or graded bit you drive on, edge to edge.
Seems your reply is clearer. This math question is controversial, better being asked the diameter of the track circuit than width of then track which I as assumed as the width of the road
About 11,459 acres, assuming you're including the interior of the circle.
The width of the track is not the diameter of the track.
The only thing you can say about the width of the track is that it is probably somewhat wider than the car, given that the car is stable at sixty miles per hour!
On the other hand, the diameter of the circular track would be a different question altogether!
It was always a real irritation concerning how many mathematics questions are so badly framed in English.
Best wishes from George
PS: The answer would be that the diameter of the circle in 4.77 miles to two decimal places. [15 over Pi].
Please don't confuse poor Jon...😱 😱 😂
assume the car is a point - centre of the car.
At the title card, my answer is b) 4.77 mi.
At 60 mph, the car is going a mile per minute.
Thus, the track is 15 miles around, since it can be circled in 15 minutes.
15 miles / pi = ~4.77 miles across.
got it 4.77. 15/(3.14) thanks for the fun. width is the diameter. pure translation of the problem.
If the question is what is the diameter of the track, why not ask what is the diameter, width implies the width of the road. Total nonsense
surely it depends on theminimum width of the car and grip of the car's tires and surface material and the coefficent of friction between the two.
I assume you mean what is the diameter of the track?. or how far is the centre point of the curciut from the track? or even how much area does the track take up.
Nun of the above.
Probably, somewhere in between 6 and 25 meters wide.
The diameter = circumference = 15 miles : pi = about 4,77 miles
Mr. Mathman reminds me of my college psychics prof who would say, "no one learns math by guessing at the wrong answers". He would give us stacks of problems already worked out .
The Diameter is 4.774 miles. But that doesnt tell you how wide the track is.😅.
My work, 2piR=C
C=15
15/2pi=r
R=2.387 x2=D
D=4.774
Don't forget the fact that if the diameter of the track is 4.774, that diameter must be increased at least enough such that the car fits on the pavement.
Then add in the fact that most tracks are designed to allow two cars to be side by side, and sometimes even three
Then add a "safety factor" of pavement.
I know the circumference and diameter of the track. But without knowing the dimensions of the car, I can't give a minimum width the track must be for the car to stay on it. I'm gonna err on the safe side and say the track is 8 feet wide. 15 Miles circumference and 4.77 miles diameter.
60 mph = 1 mile per minute, so the circumference = 15 miles, the width = the diameter, C = pi(D), therefore C/pi = the width = 4.77 miles
Anyone with track racing experience knows, track width ALWAYS means how much room the car has to drive, left to right, avoiding other cars, if any. How far it is across the entire racetrack is ALWAYS the diameter of the racetrack. But you didn't want to give away the solution with the keyword "diameter," implying π, so you said, "width." You could have done better saying, "...total width of the entire course," in lieu of track width.
Greetings. That should be easy. From the details given, we can find the perimeter of the circular track. Thereafter, we will determine the diameter of the the track. To accomplish this, we need to know how to calculate distance using the velocity formula in addition to knowing how to calculate the diameter of a circle when the circumference is given or known. I will forgo the calculations.
Perhaps you are asking the Wrong question. Aren't you actually looking for the diameter of the Track path(Assuming of course it's a perfect circle)? Just asking.... Given these assumptions, I will proceed and try to find his diameter!🙂
Guessing that dividing 15 by 3.172 is about a bit under 5, so b. By the process of elimination.
VERY badly worded problem. The question should have been, "What is center diameter of the track?" because that's what you're asking for. A circular track that an automobile can drive on has a width that is defined by the inner and outer radius of the circle, neither of which are given. There is not enough information given to calculate the "width" of the track, only the center diameter.
After first reading I thought 'how wide is the track' meant what you said - inner to outer edge of track. That's why John says to read each problem 3 times. You need 3 readings to sort out the ambiguities. He could have asked, "what is the diameter of the circular path on a track taken by a car completing one revolution in 15 min. at 60mph".
Correct Sir 🤠😁🎉😂
I understood that width meant diameter so I knew what to do. Using diameter instead of width would be clearer
You overthinking it doesn't make it badly worded.
@@Vipre- As opposed to not thinking at all...?
Width isn't diameter. Math good, logic bad. 😂
Aha... High School Math... Shame it wasn't High school English or Critical thinking. 😂
You can only calculate an average path the car traveled within the boundary of the track. The track can be infinitely wide. Oddly worded question and is actually irrelevant to the cars speed and velocity.
No, it's the diameter of the circle the car makes...I got it right. 4.77
Using the circumference formula...you need to first calculate the circumference of the circle the car made.....which is 15 miles . Then plug that into the formula...
Nothing tricky about this . The width of the track IS the diameter of the circle the car made, WITHIN the track.....IF the track Was very wide and the car travelled on the outside of the track, it would have traveled much further than on the inside of the track..
The question is about how wide the track would be for the given values . Which is ALWAYS different from the inside to the outside, as all race car drivers know 😉
Old timer here, just subscribed, I have lots of time, would like to watch your videos from day one. How do I get to your first video on TH-cam?
Thanks 👍
I got it right 🙌
Nice one 👌
The question makes no sense. There is no track or road that is miles wide.
That is true. In my opinion is how long the car travelled.
Fermilab but that's a track for atoms.
There is if you're crossing the road.
60 MPH is one mile per minute. This means that the distance traveled is 15 miles.
We now need to define exactly what the "width of a track" means. It generally means the distance from one edge of the track to the other edge, perpendicular to the direction of travel. However, from the context of the multiple choice, it seems to be asking for the diameter of the track.
This is unknowable because it is not mentioned how many laps the car went. From the phrasing, it's reasonable to assume that it completed at least one lap. However, a single lap is ridiculous because no permanent track would ever have a 15 mile long lap. One reason is because the land required would never pay for itself from the revenue.
However, the phrasing does not exclude the possibility of multiple laps. If the track was the Daytona Speedway, it did 6 laps; if it was Martinsville, it did about 28.5 laps.
Your wrong. He said one lap. Stop trying to be clever, you blew it. He did not say it was reality A H.
The question, as asked, does NOT say one lap. Read it yourself, or are you illiterate?
@@thewitt2890 The question absolutely does not say one lap, You don't get a 22 min explanation when you take a test. You just have to try to figure out what it means from the words on the page. The way this is worded is a hot mess. Width and diameter are not interchangeable Saying a car "went around a track" is not the same as "a car made 1 complete lap of the track." There's even more flaws too. Is it the inside of the track? the outside? How wide, the real meaning as in how much road does the car have, is the track? Are there other cars the driver is maneuvering around? From the context you can decipher the writer of the question isn't expecting you solve with those factors but why make students waste time trying to decipher a poorly worded question?
@@thewitt2890You're* and also definitely didn't say once around, just "around" which could be any number of laps but at least one.
If you had listened to John @ 1 minute and 34 seconds in his video he clearly says 1 time around.
@@Spudz76
Sorry, but in a word problem , words do matter.
I've been watching your excellent videos. For this one, be sensitive to the fact WIDTH of the track is NOT the DIAMETER of the circle described by the track. While the English language is not as exacting and objective in meaning as mathematics, what little there is should not be minimized because we don't choose the right words. Otherwise, well done!
DAMN. I was all prepared to say something about no way to know how wide the track is but the diameter of the track is about 4.775 miles and that would be the center line of where the car actually traversed, but everybody already beat me to it.
Everytime I click on Mr. Mathman's vid I hear that familiar sound ...Okayyy ... thks Mr. Mathman.
60 mph is a mile a minute, therefore the circuit is 15 miles in circumference. Divide 15 by pi = 4.77
How wide is the car😂😂😂😂
Its not how wide.
Car going 60 mph = 1 mpm (miles per minute)
One circumference = 2r . pi = 2 . ½d . pi = d . pi
Speed = distance / time = d.pi / 15 = 1 mpm
So the correct answer is d = 15/pi miles is just less than 5 so answer B (stupid multiple choice)
Wait so you are substituting diameter with how wide? I would have NEVER known that without your explanation, The width of a track is the usable track around an unused circle in the middle. I hope you don't give this question to students. You are making them spend way too much time just trying to decipher what you are even asking.
John does not get an A+++ or any stars or smiley faces for incorrectly stating the intended problem. Nor can he brag to his family and friends about his expertise in math or automotive racing.
Because the DIAMETER of the track is 4.77 miles, the width of the track could be from 2.38 miles - approx. the width of the car.
Its a pizza with infinite slices that rearranged under integration gives an exact area and as such this number cannot be algebraic since it is not a polynomial root therefore must be transcendental.
Driving a mile a minute for 15 minutes is 15 miles. We can check using math.
60miles/1hour = xmiles/1/4hours.
I got 60miles/1hour = 4xmile/1hour
Multiply both sides by 1 hour to cancel ours or you get a funky unit like milehours.
I got 60miles = 4xmiles.
Divide both sides by 4 to isolate x.
I got 15 miles = x miles
I think this is reflexive so the circle is 15 miles circumference.
15miles ÷pi = 4 miles 4,090 feet 1.71 inches. The fractional part is irrational and cannot be written as a ratio of two integers because pi is irrational and any mantissa of a result from it will be irrational as well. This is annoying because I really wanted it to be 1.75 inches. If I could take pi to high enough powers and prove it was and integer then maybe I could write the answer as a rational using pi number of roots. Or not. Or some other method. Or not.
Best bet for exact values would be 15/pi. If an engineer needed a decimal approximation he could easy find it to be 4.77 or as many digits of precision he needed.
Thank you
How long does it take to explain miles per hour? Answer: 15 minutes. 20 minutes to remind me of a simple equation that I should have remembered anyway and if not I could Google it in less than 10 seconds. Simple, 15÷π.
add the time for the adverts.
Why string things out by reading threee times what is already clearly shown in the video
Yet another poorly worded problem. I assume you mean the diameter of the track.
D=15÷3.14
=4.77 miles.
Did you do this on purpose for some reason? Or did you actually screw up this bad? Kinda reinforces the notion that mathematicians don't live in the real world.
60mph = 1 mile a minute x 15 min, 15 miles, didn't need to use fractions!
How wide is the track? The track is at least as wide as the car is. Oh, maybe you wanted to know the diameter of the circle? Or perhaps the radius?
I hated math because I had horrible teachers. Later I learned to appreciate math.
Do you want the width of the track or how long is the track?
It was asking for the diameter of the center line of the track, not its width.
simple question, had the answer in about 10 seconds.
Easier to convert from hours to minutes, since we need to know how far we went in 15 minutes.
1 hour=60 minutes..so the car travels 1 mile per minute..therefore in 15 minutes you travelled 15 miles....
Much easier solution, IMHO
Then plug it into the formua and you're done.
15 divide by pie...easy peasy 😜
Just needs to be about 10 feet 15 would be more comfortable.
b) 4.77 miles is correct
it is b 4.77 miles
I don't need to add anything--you guys all saw the problem, but the only CORRECT answer is "insufficient information to answer the question"
Car goes 1 mile in a minute = 15 miles circumference divided by pie= 4.77 😝
Love pie !
b) 4.77
no idea.
4.77 miles
15/pi=4.77
B) 4.77mi
4.77
4milies
B 4.77
There is no data present to allow the calculation of the actual width of the track. Instantly we know the track is 15 miles long, just use Pi to calculate the diameter. There's far too much talking waffle. This is simple maths that an 11 year old would be taught in the UK
Why is this 22 min long!!! 😱😱😱
Because you forgot to switch to 1.5x playback speed
3.98 mi
15
You are by far the Worst Teacher of Mathmatics I've ever had the misfortune to listen to...😱 😱 😱
Oh I beg to disagree. My HS algebra II teacher was awful.
B
too long explaining simple principles.
b
What … Wide doesn’t called Diameter…. Simple things explain in very difficult way…. Difficult Teaching…?
4.77
b
4.77
b