They probably would have stayed together if they had just had the sense to use SI units.
Maths/Physics with imperial units seem like a crime.
This problem would be exactly the same if they used 5 m/s
Except now the walking speed of 1 m/s (3.6 km/h) and running speed of 5 m/s (18 km/h) are realistic
Yes mph is one thing but fps!
i am running at 5 frames per second
5.099 frames per second, but allowing for rounding you could argue they are separating at 5 fps
Yeah tbh I think that’s a good next step for metrication. Currently it’s all in US customary until college with a brief lesson on metric every few years. With some high school teachers doing a bit more in metric. If you swap it these people will just instinctively use and understand metric.
I’m good with intuiting 1-3 meters, and I have a decent estimate for 1 centimeter. Beyond that, about all I got (yes I know this is all very rough estimates) is “a kilogram is a couple pounds, a kilometer is a short mile, a liter is basically a quart”. But I guess I don’t even have a good intuition for a quart because that in my head rests at the crossroads of “4 cups or ¼ gallon”.
Make metric meaningful early in life, or actively make referencing real metric measures a part of your life, if you want it to stick.
The biggest issue with swapping to metric I’m aware of isn’t the people getting used to it as much as the massive monetary and labor cost of replacing signs/notices/etc around the whole country
Nah it’s public opinion. People don’t want to change. The costs are annoying, but we can budget for them. We tried to metricate at the same time as Canada. They had less money per person than us but succeeded because they didn’t have a large mass of people unwilling to learn and change.
Well depends on what you mean by „success“
Comparatively though…
I am on my phone so sorry for terseness but maths breakup problem trapeze lady twelve feet per second south human cannonball man two hundred yards per second every direction but south how far apart after two burritos and a threesome (human cannonball not involved)
At any given time T, the coordinates form a right triangle with legs of length 5T and T. Therefore the distance D is given by D^2 = (5T)^2 + T^2 = 26T^2. This simplifies to D = T * sqrt(26). Therefore the rate of separation is sqrt(26) ft/sec regardless of time
You forgot about the curvature of the earth!
It’s flat, idiot
I didn’t know the earth was flat when using imperial, but it makes perfect sense!
Also, where’s the topographic map of the region? How can you expect us to come up with something remotely accurate without knowing this, is the third dimension a joke to you, are we all dots of ink on a paper?
Missing context: this is part of a series, all taking place in Flatland.
5ft/sec?? Jesus, that’s almost 6ft/sec!
A girl was excited for her sweet 16, and she asked her boyfriend to buy her a car. He said yes. The night of the party, he didn’t come. She was very sad. Then she found out he’d died trying to drive two cars at once to her party. Like this if you love your boyfriend.
i have the feeling that the kind of person that buys a car for his 16yo gf, is like the one that’s leading the united states
What if I love your boyfriend?
Do we need to compensate for the curvature of the earth?
That’s where the calculus comes in, otherwise it’s a basic Pythagorean theorem. I’ve compiled a list of possible latitudes (with questionable assumptions), knock yourself out
What about the local topology? What if she’s walking uphill and he’s running downhill?
No but if they’re in Mexico you gotta keep it’s sink rate in mind
thats what the unannounced bonus points are for
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It depends on the size of their feet.
Shouldn’t it be ‘after having been together’?
What is ‘at the same time’ referring to in that sentence? They wanted to break up at the same time (as in both had the idea)? They wanted to break up at the same time on the clock to continue the theme of things being same-y?
The boy is due north of what? The place? The girl? Also, the girl should be wondering about her decision, I think.
(I don’t even speak English every day anymore, so I could be wrong).
They said goodbye at a given position and are then leaving each in a different direction. They start to move at the same time from the same point.
Wait, we know their position exactly? That means we have no idea what their velocities are!
Actually, their velocities are specified precisely in the problem description.
What? Velocity too? Now we know nothing!!
(I don’t even speak English every day anymore, so I could be wrong).
You’re not wrong. I think some of it is the difference between casual speech and formal writing (people are more likely to say “after being” but write “after having been”, especially in published work)**, but some of it is also just poorly phrased. It makes enough sense to a native speaker to get what the problem is asking, though.
__
** I think the first may be correct in some cases, but idk the rule.
It only bothered me because I saw that it was a school assignment and I thought it would be to a higher standard. In casual speech, I don’t really care unless the meaning is unclear.
Sigh i miss high school maths. Even i’m lovesick now.
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Is this even calculus? Seems like simple geometry.
It’s not how far apart they are It’s how fast they are seperating. I’m not sure but the rate they are seperating might change as they get further apart due to the triangle or something? Otherwise yeah just a triangle.
Rigourously overcomplicating the problem: Let dy = distance boy travels north in time dt, and dx = distance girl travels east in time dt. We know that dy = 5dx from the question, hence:
dy | |__ dx
(This is supposed tk be formatted like a triangle but it looks janky. You get the idea.)
And the distance they separate dS in time dt is clearly the hypoteneuse. So we can write:
dS = sqrt(dx^2 + dy^2)
And divide through by dt:
dS/dt = sqrt( (dx/dt)^2 + (dy/dt)^2 )
Simply gives the rate of separation dS/dt as 5.1 feet per second.
Some confusing notation here that buries the assumption that the rate of change is constant (which is true in this case). For conceptual clarity I would explain it as:
Let y(t) be the boy’s position at time t, and x(t) the girl’s position. The distance between them is S = sqrt(x^(2) + y^(2)). The distance is changing at a rate of dS/dt = dS/dx dx/dt + dS/dy dy/dt = (xdx/dt + ydy/dt)/sqrt(x^(2) + y^(2)). We are given dy/dt = 5 and dx/dt = 1, and we can determine that at t=5 we have y = 25 and x = 5. Therefore dS/dt = 130/sqrt(650) = sqrt(26) ~= 5.1.
This is a better generalised solution, yeah. Though I do think that as far as conceptual clarity goes, doing it geometrically is a bit more transparent than using the chain rule, even if it’s sort of constrained to constant speed in this case
This is the type of question my Calc teacher would put on a test. He liked doing weird questions probably because we’re doing fucking calculus here, it doesn’t make sense in “every day word problems” (I have never once needed to know how long it will take me to fill a funnel while I let water pour out it at the same time, and where the water line will be after exactly 7 seconds) so trying to make a word problem out of it is already an exercise in nonsense.
Good thing that ain’t an english teacher
Really depends on which way the dude is going. It said that he is due North and running at 5 feet per second, is he running east? If so they would be like 4 feet apart, unless he started 4 feet further West than she did.
Really need some more info for this dumb word problem.
Instructions unclear, let’s assume he started 1,312,593,975 feet further back (see my other comment)
Reminds me that one exurb1a pinecone video.
30 volts
