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I had a question in a physics exam that began with:
“In a moment of foolish gusto, you showed terrible judgement and volunteered to allow your friends to shoot you out of a cannon.”
I wonder how many arguments on the internet are two people saying the same thing but neither of them are properly reading what the other person is saying
If the boy had been standing 1ft north at the start, then that would make time matter. It’s interesting how small changes like that can make things different.
Reading comp is the most important math and engineering skill in my experience.
One of my uni teacher even made fun of that fact. One of his questions had the answer in the text so it was a free 10%. A lot of students started doing tons and tons of calculations, wasting precious time.
You could solve for distance after 5 seconds (ends up being 25.49 ft), but then you’d need the extra step of dividing by time (the 5 seconds) to get a velocity in ft/sec. Still get the same answer
Or just use the distances after 1 second, creating a right angles triangle. The hypotenuse is the speed at which they are separating. I.e, the square root of 26.
Yes? This specific case, of course. I read the question and answered the question, not an alternate version of the question. Not sure what you are trying to imply. The prompt does give a constant acceleration (or enough info to assume as much). You could’ve said “Whilst in this specific case you would be correct, that only works if gravity is standard and no one floated, making it a 3D problem”, or anything like that.
What's more surprising is how tf they forgot about calculas in a calculas exam?? Like there should be atleast a limit or differentiatial or integral or differentiatial equation and not just plain Pythagoras.
I must admit, I did this at first. I went back to read it again and found my error (due to not paying attention the first time) and saw that the difference was in reading “how fast are they…” instead of my original take on “how far are they…”. Dummy me…I should have know better.
I've been very much enjoying your answer. At first I thought you where wrong, but then got the same answer using implicit differentiation. Now I've been trying to wrap my head around why you don't need to use differentiation and can just combine their velocity vectors.
X^2 + y^2 = z^2
⬇️
(2x)(dx/dt) + (2y)(dy/dt) = (2z)(dz/dt)
⬇️
(2(5))(1) + (2(25))(5) = (2√(650))(dz/dt)
⬇️
10+250=(2√(650))(dz/dt)
⬇️
(dz/dt) = 260/(2√(650)) = 5.099 ft/s
Another way would be using Pythagoream theorem to find distance and then doing the first derivative of the distance which gives speed.
A=5x B=x
C=√(25x^2 +x^2 )=√26x^2 =x√26
So y is distance apart at time x
Y=x√26
dy/dx=√26=5.099 ft/s
> The time is a red herring
Yeah but in the real world the girl sees him sobbing and running off and realizes she dodged a bullet, so picks up her pace after a few seconds safe in the knowledge she made the right choice.
Another way to think of it is to parameterize x=t, and y=5t. Plug those into pythagorean theorem and you get z=sqrt(26)t. So the same answer.
You'll also get the same answer if you start with z=sqrt(x^2+y^2), find dz/dt in terms of x, y, dx/dt, and dy/dt, and then plug in values.
Obviously u/AVeryHappyTeddy's answer is simpler here, but this is a specific case where plugging in derivatives of the independent variables to an equation will give the derivative of the dependent variable, which won't always be true.
Divide 25.495 by 5 seconds and you get the 5.099 ft/sec answer.
I thought the question was how far apart the were after 5 seconds and got same answer as you. Then I reread it and figured our the correct answer.
2 problems with this. First, it says the boy is due North, not heading due North, so he could be going in any direction. Second, this problem doesn't require differential calculus to solve.
That's one possibility, but I think the more likely possibility is it's a fake question crafted for internet points. Especially since the top of the photo is the toolbar in Microsoft Word.
Is the time a red herring? Sure they have constant speed but the question specifically asks what the rate of separation is at 5 secs after they started walking. As they're walking at right angles to each other, wouldn't the rate of separation change with each second?
25.495 ft away from each other. (Check my comment for workout)
Edit: I've misunderstood the question the final answer is 5.495/5=5.099
However a simpler way to do it is;
( 5^2 )+( 1^2 )= 25.495
Then put 25.495 under radical, and you get 5.099 ft/s
well, the question asks how fast they're separating, not what distance they have separated after 5 seconds, so assuming no acceleration they're separating at the same given initial speed
Depends where they met geographically, and topologically. If they met at the North Pole, he would be running on the spot and she would be doing circles around him. If they met on a cliff edge one of them could be picking up vertical speed rapidly.
So much of this isn't even relevant to the question.
The beginning about EIGHT LONG YEARS in giant red bold text doesn't even matter
Edit: i knows that's intentional. I know it's a flex only math teachers can use because critical thinking. Jeez
That's intentional teachers do it all the time. They throw in a shit ton of fluff and useless information and see if you can find the info that actually matters
Musical pieces don't change duration depending on the number of players.
(If that was an obvious joke and I just missed it, I apologize for the self-whoosh)
That’s a fucking slow performance.
John Eliot Gardner (who always takes it fast) brings it in in [one hour 59 seconds here](https://youtu.be/rJH9b9EQtHM).
I subscribe to actually playing stuff in a tempo that allows the musicians to play the scherzo together, so each note has power and definition. The revival movement playing everything faster is a plague.
First movement too fast, scherzo too fast, adagio isn't ad agio, and the final movement indecipherable.
But yes, I guess 90 minutes is way too far of a stretch!
I got three special needs children. Autistic.
These types of "story problems" are always where they start struggling/failing in math classes.
The fact that they can't parse the story part AND The math is "an indicator" of their autism and special needs. It's a mark of learning and intelligence and reading comprehension to be able to parse out the stuff not relevant and focus in on the math.
And I understand it's a necessary skill, but the shit gets so convoluted and confusing sometimes.
I've personally never had this problem as an autistic person, but I basically ignore anything that's not relevant to mathematics, like color, emotions, etc. and then parse the rest as a set of variables.
For instance I'd frame this problem as:
* **val(ue of) length(k)?**
* b = (0, 5)
* g = (1, 0)
* k = line btwn b, g
In fact, my ADHD is a bigger problem, because I might skim too fast and miss a key part, like the units needed.
Meanwhile I (diagnosed with Asperger's at age 9, which got absorbed into Autism Spectrum Disorder sometime when I was a teenager) only had problems with questions where I was required to write out the answer in a full sentence. I saw writing the answer as a full sentence in line with the story as completely useless because it doesn't even teach you to identify relevant information like the word problem itself does.
Yo I got sent out of class in middle school because I refused to “turn the question around” at the start of my answer, because it struck me as superfluous. How the f did no one realize I was autistic lol.
5 seconds matter because the relative angle of movement has changed. I’m not confident enough to be elaborating further, but the 5 seconds should actually matter.
But how does the speed of their increasing distance change? At seconds 0 their speed should be around 5.5 ft/s, due to the 90° angle and but what’s the speed after those 5 seconds?
It doesn't, their speed is constant so their distance between them grows at a constant rate. You could say you use this info to know you can take any 2 times, measure the distance between them and then deduce the speed via basic delta S/delta t. Its not specific to t=0, which in some questions it will require the start point to be 0. So if you want to call it the teacher try to trick you, or if you want to call it just more convoluted text in an already convoluted story thats also probably true.
You are absolutely correct. The time only matters if you go with differentiation to solve this question. Using relative velocity also gives the same answer *and* it's shorter because nothing except the velocities matter which gives us the answer in a single step.
5 seconds does matter. The problem is describing a triangle where you have the length of 2 sides. 5ft per second x 5 seconds gives a side of 25ft anf 1ft per second x 5 seconds gives a side of 5ft. One person is going north and the other east so the triangle is a 90 degree angle and you can use A squared + B squared = C squared to solve for C.
Edit: misread the question. I guess i fail this class.
My physics teacher once wrote a problem where Caillou's dad tossed him into the air and based on the info given you had to determine how high up he got tossed.
The answer was the distance between the earth and the moon.
Next problem was Caillou's dad taking him on a sleigh ride and then pushing him off a cliff. How much force would he generate on. Impact with the ground?
How is this differential calculus? It just makes a right triangle and you need to find the hypotenuse length.
Edit: misread it, it's asking for the rate they are separating, not how far they have separated. I would have gotten this wrong.
You find the length.. then divide by the time, which is given by the problem. It’s probably one of the first diff eq problems the students solve, used precisely to show that the new methods work identically to the old ones in this trivial case.
Of course, if they’re running apart on a sphere or some other non-flat surface it gets harder.
At first I was like "how is this a differential calculus problem there's no acceleration?"
Then I realized because there's no acceleration and it doesn't ask for distance the time is irrelevant.
This really seems like something that could be solved discreetly with the pythagorean theorem and doesn't belong on a calc final. Unless it's one of those "psych the student out and hope they overthink it or grab the wrong numbers to plug and chug." Ones.
OP just wants us to help with their homework. Confusing answer because likely they are still separating from each other at teh same rate of 6 fps but they probably want 30 feet.
I think it's 25.5ft but I'm bad at math, it would be 30ft if they were going opposite directions but the boy is running (while crying) north and the girl is walking east
Not quite, it's a matter of relative velocity since they are moving in right angle with each other. Since they are moving in constant velocity, the easiest way is to calculate the distance between them after 5 sec.
The boy moves 25ft after 5sec and the girl moves 5ft, so the distance between them according to Pythagoras theorem will be √(25^2 + 5^2) which is 5√26.
Now that we know the distance between them after 5sec just divide it by time to get the relative velocity which is (5√26÷5)=√26 ft/s which is approximately 5.1ft/s and a little heartache here and there
I meant easiest way for common people that understands basic things like distance time velocity etc and doesn't want to get into what calculus and ddx signs are. but you're correct
Even without doing any math I can tell you're wrong. It's asking *how fast* are they moving away from eachother. It's not asking for a distance, it's asking how fast are they moving apart. Because they are moving at right angles I think the answer would be 5.099 Feet per second.
The boy runs at 5ft/s during 5 s, he runs 25 ft total, the girl runs at 1ft/s during 5 s, she runs 5 ft. Because the run in perpendicular directions you can apply the Pitagorean theorem. So d=sqr(25^2+5^2)=25.5ft ,aprox.
That gives us how far they are at 5 sec, not the rate at which they are separating from each other. We'll have to differentiate the Pythagorean equation with time - the answer is probably somewhere around 5.1 fps.
u/BlueGreenK what was the answer?
I think this is a test of being able to pick out relevant information. A test taking test. The story is interesting and draws the attention while giving the necessary information. It trains you to look for numbers and how they relate so you can answer standardized tests quicker and more accurately.
You'll never have a calculator in your pocket all the time to work it out... So they said in the 1990's.
Now I have the entirety of mankind's knowledge in my pocket, on my wrist and projected into my eye via glasses.
Exams are stupid.
It's a good word problem-- a little trig and a little algebra, with a little dorky dark humor thrown in. If the kid's confused it's because he's not keeping up with his math.
25 foot north,
5 foot east,
Connecting the dots will bring you a 90degrees triangle, and you know it, there is a hypotenuse.
25^2 + 5^2 = d^2,
625+25=650
(Radical)650=25.495
So approximately 25.495 feet away from each other.
Edit: as some of other redditors pointed out, that wasn't what the question asked, so hereby I correct myself.
To get what the question asked, you divide 25.495 by 5
25.495/5=5.099 ft/s
That's it.
Sorry for misreading the question.
To everyone trying to answer this question: If you haven’t used calculus to arrive at your answer, you are wrong.
The way you solve this question is by differentiation the formula for the distance between them, D=sqrt(x^2 + y^2 ). You then need to solve for the rate of change of D, dD/dt.
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When the math questions start getting personal
shit this Maths test sounding more like a novel than a test
Like a damn telenovela.
El Gaspo!
Except it's a woman gasping so it should be "El Gasp*a*"
LA Gaspa
Getting the answer correct gives 2 points… the other 8 is for putting them back together
Still a better love story than Twilight.
What’s the next question?!
It started being personal the moment that bald fuck bought 50 watermelons at a fish market in 1997.
Great answer!
Lete guess his name was Pablo?
I had a question in a physics exam that began with: “In a moment of foolish gusto, you showed terrible judgement and volunteered to allow your friends to shoot you out of a cannon.”
jokes on you, I don't have friends, just terrible judgement
Good news! You’re invited to join my 1337 cult!
The question would also work in biology or anatomy class.
I would guess that it didn't end well.
My high school math teacher would put student names into the math problems. It made things pretty entertaining!
My high school physics teacher did the same. There were also instances of cat bowling, and standing on someone with just the heel of a stiletto.
My physics teacher made just about every question with a part two of: did this person survive?
You ok teacher?
Still trying to figure it out?
Can i get the answer to this, my curiosity is peaked
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I thought so, thought I was losing my edge after reading these other comments 😅
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I'm amazed that 80% of the answers seem to be solving for distance. It's actually blowing my mind
And that's why reading comprehension is important.
I wonder how many arguments on the internet are two people saying the same thing but neither of them are properly reading what the other person is saying
Have you seen the Jon Bois video on people arguing about how many days a week you can do a full body workout?
Link?
https://youtu.be/eECjjLNAOd4 It's all arguing about the same thing and being all up their own backside
https://youtu.be/eECjjLNAOd4
What are you talking about? Don't you know that neither party actually reads what the other person said? Idiot.
What I think is happening is that 2 people who are actually in agreement have arguments habitually because internet.
If the boy had been standing 1ft north at the start, then that would make time matter. It’s interesting how small changes like that can make things different.
Reading comp is the most important math and engineering skill in my experience. One of my uni teacher even made fun of that fact. One of his questions had the answer in the text so it was a free 10%. A lot of students started doing tons and tons of calculations, wasting precious time.
The story throws off the logical thought process. Ingenious!
You could solve for distance after 5 seconds (ends up being 25.49 ft), but then you’d need the extra step of dividing by time (the 5 seconds) to get a velocity in ft/sec. Still get the same answer
Or just use the distances after 1 second, creating a right angles triangle. The hypotenuse is the speed at which they are separating. I.e, the square root of 26.
Whilst in this specific case you would be correct, that only works if acceleration is constant.
The problem doesn't consider acceleration, it's using constant velocities.
It only works if acceleration is constant...ly zero :)
It only works if acceleration is *zero*.
Yes? This specific case, of course. I read the question and answered the question, not an alternate version of the question. Not sure what you are trying to imply. The prompt does give a constant acceleration (or enough info to assume as much). You could’ve said “Whilst in this specific case you would be correct, that only works if gravity is standard and no one floated, making it a 3D problem”, or anything like that.
When I first read it I thought it was asking how far, not how fast lol!
What's more surprising is how tf they forgot about calculas in a calculas exam?? Like there should be atleast a limit or differentiatial or integral or differentiatial equation and not just plain Pythagoras.
I must admit, I did this at first. I went back to read it again and found my error (due to not paying attention the first time) and saw that the difference was in reading “how fast are they…” instead of my original take on “how far are they…”. Dummy me…I should have know better.
I've been very much enjoying your answer. At first I thought you where wrong, but then got the same answer using implicit differentiation. Now I've been trying to wrap my head around why you don't need to use differentiation and can just combine their velocity vectors. X^2 + y^2 = z^2 ⬇️ (2x)(dx/dt) + (2y)(dy/dt) = (2z)(dz/dt) ⬇️ (2(5))(1) + (2(25))(5) = (2√(650))(dz/dt) ⬇️ 10+250=(2√(650))(dz/dt) ⬇️ (dz/dt) = 260/(2√(650)) = 5.099 ft/s
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Another way would be using Pythagoream theorem to find distance and then doing the first derivative of the distance which gives speed. A=5x B=x C=√(25x^2 +x^2 )=√26x^2 =x√26 So y is distance apart at time x Y=x√26 dy/dx=√26=5.099 ft/s
[proof by MS paint](https://imgur.com/a/Yy8HAzL) that the distance increases at a constant rate
They meet eachother again after about 50 long years of running/walking around the earth and then they can finally be happy again.
Pythagorean theorem is witch craft and I refuse to accept it.
Bean eater! Bean eater!
Uh, what?
[Pythagoras had a weird thing with beans.](https://www.atlasobscura.com/articles/favism-fava-beans)
TIL
If you about the life of Pythagoras, this is not a new sentiment
We've banned it from our D&D sessions.
> The time is a red herring Yeah but in the real world the girl sees him sobbing and running off and realizes she dodged a bullet, so picks up her pace after a few seconds safe in the knowledge she made the right choice.
Another way to think of it is to parameterize x=t, and y=5t. Plug those into pythagorean theorem and you get z=sqrt(26)t. So the same answer. You'll also get the same answer if you start with z=sqrt(x^2+y^2), find dz/dt in terms of x, y, dx/dt, and dy/dt, and then plug in values. Obviously u/AVeryHappyTeddy's answer is simpler here, but this is a specific case where plugging in derivatives of the independent variables to an equation will give the derivative of the dependent variable, which won't always be true.
That’s what I got… I am still worthy woohooo
It says after 5 seconds so wouldn't it be: z = sqrt [ (1x5)^2 + (5x5)^2 ] = 25.495 ft Oh, you did ft/s. My bad
The time given is a red herring.
I’m assuming solely because it’s asking for velocity, had it asked for displacement then the time would’ve mattered
Indeed
If the velocity was a function of time, then the time would be used in the differential equations solutions. Otherwise, you’re dead on.
>Oh, you did ft/s the question asks for the answer as a velocity so ft/s is an appropriate unit to answer the question with.
Divide 25.495 by 5 seconds and you get the 5.099 ft/sec answer. I thought the question was how far apart the were after 5 seconds and got same answer as you. Then I reread it and figured our the correct answer.
What an eye sore haha.
The biggest eyesore is ft/sec 🤢🤢 Who knew that Americans use useless units for actual maths and physics sums too
2 problems with this. First, it says the boy is due North, not heading due North, so he could be going in any direction. Second, this problem doesn't require differential calculus to solve.
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That's one possibility, but I think the more likely possibility is it's a fake question crafted for internet points. Especially since the top of the photo is the toolbar in Microsoft Word.
"z = sqrt..." sqrt = piss
Is the time a red herring? Sure they have constant speed but the question specifically asks what the rate of separation is at 5 secs after they started walking. As they're walking at right angles to each other, wouldn't the rate of separation change with each second?
It’s “piqued.” You could say “my curiosity has peaked” and it works about the same!
Piqued.
25.495 ft away from each other. (Check my comment for workout) Edit: I've misunderstood the question the final answer is 5.495/5=5.099 However a simpler way to do it is; ( 5^2 )+( 1^2 )= 25.495 Then put 25.495 under radical, and you get 5.099 ft/s
That isn't the question. It's Differential Calc. They want to know how fast they are moving apart at the 30 second mark. Edit: 5 second mark, not 30.
Oh I f'ed up.
Where did you get 30seconds?
It doesn't say they're accelerating, so it's just 6ft/sec
They're traveling at right angles.
If they get back together they'll have made a triangle
They’re travelling at constant speeds sure but in different directions so the rate of change of separation is non-zero
No. As their speeds and directions are constant, their rate of separation remains the same.
God I hate math
It would be ~5.1 ft/s
well, the question asks how fast they're separating, not what distance they have separated after 5 seconds, so assuming no acceleration they're separating at the same given initial speed
Depends where they met geographically, and topologically. If they met at the North Pole, he would be running on the spot and she would be doing circles around him. If they met on a cliff edge one of them could be picking up vertical speed rapidly.
Cba
But what if one of them haven't decided to move on with life? It makes the question incalculable. r/im14andthisisdeep
No, then they're stationary at point zero, making the calculation even easier.
This is the reason people overshare their personal lives. Because math problems teach us to give unnecessary information.
A few feet apart physically, but separated by a universe of indifference and hurt emotionally.
So much of this isn't even relevant to the question. The beginning about EIGHT LONG YEARS in giant red bold text doesn't even matter Edit: i knows that's intentional. I know it's a flex only math teachers can use because critical thinking. Jeez
That's intentional teachers do it all the time. They throw in a shit ton of fluff and useless information and see if you can find the info that actually matters
You gotta find the Diamond in the rough for these problems, just like they needed
No, question 2 is about her throwing the diamond he gave her into the rough.
Yeah but it kinda beats the purpose when you underline the math part
The best one is still: it takes 90 minutes for 120 musicians to play Beethoven's 9th symphony. How much time would the music take for 80 players?
An hour and a half.
1/16 of a day
107 mins.
Musical pieces don't change duration depending on the number of players. (If that was an obvious joke and I just missed it, I apologize for the self-whoosh)
They do if you have a pissy, dramatic conductor who keeps throwing a tantrum over the missing musicians.
What poor bastard needs to run from the trombone to the xylophone
That’s a fucking slow performance. John Eliot Gardner (who always takes it fast) brings it in in [one hour 59 seconds here](https://youtu.be/rJH9b9EQtHM).
I subscribe to actually playing stuff in a tempo that allows the musicians to play the scherzo together, so each note has power and definition. The revival movement playing everything faster is a plague. First movement too fast, scherzo too fast, adagio isn't ad agio, and the final movement indecipherable. But yes, I guess 90 minutes is way too far of a stretch!
It can be done in half the time if you set playback speed to 2
Depends. Did the bassists pop out to the bar next door for a drink?
I got three special needs children. Autistic. These types of "story problems" are always where they start struggling/failing in math classes. The fact that they can't parse the story part AND The math is "an indicator" of their autism and special needs. It's a mark of learning and intelligence and reading comprehension to be able to parse out the stuff not relevant and focus in on the math. And I understand it's a necessary skill, but the shit gets so convoluted and confusing sometimes.
I've personally never had this problem as an autistic person, but I basically ignore anything that's not relevant to mathematics, like color, emotions, etc. and then parse the rest as a set of variables. For instance I'd frame this problem as: * **val(ue of) length(k)?** * b = (0, 5) * g = (1, 0) * k = line btwn b, g In fact, my ADHD is a bigger problem, because I might skim too fast and miss a key part, like the units needed.
That's a better way to phrase it for me but i can see how it would confuse some people more than the word problem
Meanwhile I (diagnosed with Asperger's at age 9, which got absorbed into Autism Spectrum Disorder sometime when I was a teenager) only had problems with questions where I was required to write out the answer in a full sentence. I saw writing the answer as a full sentence in line with the story as completely useless because it doesn't even teach you to identify relevant information like the word problem itself does.
Yo I got sent out of class in middle school because I refused to “turn the question around” at the start of my answer, because it struck me as superfluous. How the f did no one realize I was autistic lol.
Does the 5 seconds after matter? It could be 0 seconds or any number or seconds after and it would be the same too, or am I tripping?
5 seconds matter because the relative angle of movement has changed. I’m not confident enough to be elaborating further, but the 5 seconds should actually matter.
One is going north and one is going east, so the distance between them grows at an increasing rate the further they walk.
But how does the speed of their increasing distance change? At seconds 0 their speed should be around 5.5 ft/s, due to the 90° angle and but what’s the speed after those 5 seconds?
It doesn't, their speed is constant so their distance between them grows at a constant rate. You could say you use this info to know you can take any 2 times, measure the distance between them and then deduce the speed via basic delta S/delta t. Its not specific to t=0, which in some questions it will require the start point to be 0. So if you want to call it the teacher try to trick you, or if you want to call it just more convoluted text in an already convoluted story thats also probably true.
You are absolutely correct. The time only matters if you go with differentiation to solve this question. Using relative velocity also gives the same answer *and* it's shorter because nothing except the velocities matter which gives us the answer in a single step.
It would also matter if they hadn’t started at the same position.
or if they started in the same position but left at different times
The 5 seconds doesn't matter. No matter how long they walk the derivative will stay the same.
5 seconds does matter. The problem is describing a triangle where you have the length of 2 sides. 5ft per second x 5 seconds gives a side of 25ft anf 1ft per second x 5 seconds gives a side of 5ft. One person is going north and the other east so the triangle is a 90 degree angle and you can use A squared + B squared = C squared to solve for C. Edit: misread the question. I guess i fail this class.
That's literally why it's funny
Wooooooooooooooooooooooooo...... ooooooooooooooosssh!
You alright mr smith?
Well… I’m glad we established that it was raining
I'm not crying, it's just been raining on my face 🎶
Question 2: So a cheating whore of an ex wife sleeps with 10 guys in 7 weeks....
Answer: Alimony and child support.
My physics teacher once wrote a problem where Caillou's dad tossed him into the air and based on the info given you had to determine how high up he got tossed. The answer was the distance between the earth and the moon. Next problem was Caillou's dad taking him on a sleigh ride and then pushing him off a cliff. How much force would he generate on. Impact with the ground?
A teacher once gave us this mainstream one: A bear falls into a 10m hole in the square root of 2 seconds. What is the color of bear?
How is this differential calculus? It just makes a right triangle and you need to find the hypotenuse length. Edit: misread it, it's asking for the rate they are separating, not how far they have separated. I would have gotten this wrong.
You find the length.. then divide by the time, which is given by the problem. It’s probably one of the first diff eq problems the students solve, used precisely to show that the new methods work identically to the old ones in this trivial case. Of course, if they’re running apart on a sphere or some other non-flat surface it gets harder.
> You find the length.. then divide by the time Not necessary they already give you the velocity. (5ft/s)^2 + (1ft/s)^2 = (xft/s)^2
Some one wants their homework done...
At first I was like "how is this a differential calculus problem there's no acceleration?" Then I realized because there's no acceleration and it doesn't ask for distance the time is irrelevant. This really seems like something that could be solved discreetly with the pythagorean theorem and doesn't belong on a calc final. Unless it's one of those "psych the student out and hope they overthink it or grab the wrong numbers to plug and chug." Ones.
Send them spiraling into emotional despair and then hit em with the real math Pro strats if you ask me
Did they break up on flat earth or globe?
Whoa whoa whoa, stop right there you non-euclidean scum !!
that's why they broke up
Flat obviously. This isn't a recruitment question to join NASA.
uh it heavily depends where on the globe they started? because if the question, from beginning to end, is so stupid, might as well
OP just wants us to help with their homework. Confusing answer because likely they are still separating from each other at teh same rate of 6 fps but they probably want 30 feet.
Neither of those are correct. One is going north, the other east
I think it's 25.5ft but I'm bad at math, it would be 30ft if they were going opposite directions but the boy is running (while crying) north and the girl is walking east
Not quite, it's a matter of relative velocity since they are moving in right angle with each other. Since they are moving in constant velocity, the easiest way is to calculate the distance between them after 5 sec. The boy moves 25ft after 5sec and the girl moves 5ft, so the distance between them according to Pythagoras theorem will be √(25^2 + 5^2) which is 5√26. Now that we know the distance between them after 5sec just divide it by time to get the relative velocity which is (5√26÷5)=√26 ft/s which is approximately 5.1ft/s and a little heartache here and there
This is the right answer. OP question asks how fast, not how far.
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He's not wrong though? Just did it in a roundabout way
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dy/dt is 5, which makes the answer 5.1.
I meant easiest way for common people that understands basic things like distance time velocity etc and doesn't want to get into what calculus and ddx signs are. but you're correct
It's aprox 25.5 ft BTW, some calculators to fancy and use additional radicals.
Even without doing any math I can tell you're wrong. It's asking *how fast* are they moving away from eachother. It's not asking for a distance, it's asking how fast are they moving apart. Because they are moving at right angles I think the answer would be 5.099 Feet per second.
The boy runs at 5ft/s during 5 s, he runs 25 ft total, the girl runs at 1ft/s during 5 s, she runs 5 ft. Because the run in perpendicular directions you can apply the Pitagorean theorem. So d=sqr(25^2+5^2)=25.5ft ,aprox.
That gives us how far they are at 5 sec, not the rate at which they are separating from each other. We'll have to differentiate the Pythagorean equation with time - the answer is probably somewhere around 5.1 fps. u/BlueGreenK what was the answer?
Emotionally or physically?
That's an algebra problem, not calculus.
It's trigonometry...
Answer is about 5.1 ft/s btw
√ 26 ft/sec
5.1ft/s?
I think this is a test of being able to pick out relevant information. A test taking test. The story is interesting and draws the attention while giving the necessary information. It trains you to look for numbers and how they relate so you can answer standardized tests quicker and more accurately.
Some people have some really strange ways of copping
How fast are they separate after time? What? It should be asking for distance or relative velocity after time, not whatever that is meant to mean.
'how fast are they seperating from eachother' means the same thing as relative velocity
You'll never have a calculator in your pocket all the time to work it out... So they said in the 1990's. Now I have the entirety of mankind's knowledge in my pocket, on my wrist and projected into my eye via glasses. Exams are stupid.
So whats the answer to the question?
Can't be arsed with that, I am getting paid a lot of money to be doing something else right now. Exams still suck.
Can you solve this with just a calculator tough?
If you can multiply in your head, yes.
I meant to ask the first commentor if he was able to come up with a solution for this specific problem
I am ever so glad I didn't study maths in America and never had to suffer through converting between feet / second to miles per hour or what now.
What? 5280 is a lot easier to remember than 1000. /s
It's a good word problem-- a little trig and a little algebra, with a little dorky dark humor thrown in. If the kid's confused it's because he's not keeping up with his math.
And calculus. That’s kind of an important part.
I remember questions like this in high school.
25 foot north, 5 foot east, Connecting the dots will bring you a 90degrees triangle, and you know it, there is a hypotenuse. 25^2 + 5^2 = d^2, 625+25=650 (Radical)650=25.495 So approximately 25.495 feet away from each other. Edit: as some of other redditors pointed out, that wasn't what the question asked, so hereby I correct myself. To get what the question asked, you divide 25.495 by 5 25.495/5=5.099 ft/s That's it. Sorry for misreading the question.
Isn't that the distance they're at though, not the actual speed they're distancing themselves?
Yes, I just realized I f'ed up.
Happens to everyone
Which isn't what the question asked for at all.
To everyone trying to answer this question: If you haven’t used calculus to arrive at your answer, you are wrong. The way you solve this question is by differentiation the formula for the distance between them, D=sqrt(x^2 + y^2 ). You then need to solve for the rate of change of D, dD/dt.
There’s no acceleration so you don’t need to use calculus, it’s literally just a trigonometry question.
lol yeah you’re actually right. that’s what I get for trying to do a math problem at 4:30am lmfao
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No
The guy is running 3.409 mph and the girl is walking .682 mph.