You don't need calculus to do this. Neither one is accelerating, so "5 seconds after they started moving" is irrelevant. Just calculate the velocity of one in the reference frame of the other by subtracting the vectors: from the point of view of the boy, the girl's velocity vector has orthogonal components of -5 ft/sec north and 1 ft/sec east, so the magnitude is 26^0.5 ft/sec.
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They probably would have stayed together if they had just had the sense to use SI units.
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.
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
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?
Do we need to compensate for the curvature of the earth?
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 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.
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!!
Is he due north of the starting point, or of the girl? If the former, he is traveling due north. If the latter, his 5ft/sec velocity has a 1ft/sec eastward component, and we need to calculate the northern component.
(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.
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** I think the first may be correct in some cases, but idk the rule.
Reminds me that one exurb1a pinecone video.
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.
One of my teachers in uni always used sex potion and leather wear examples. He had been reported for sexual abuse a year prior but continued in his assignments
One of my game development lecturers always used giving handjobs to stakeholders as an example of balancing priorities and keeping everyone happy while being pulled in multiple directions. I'm assuming someone complained though because when I had him for a different class he'd stopped.

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