Maths/Physics with imperial units seem like a crime.
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
It depends on the size of their feet.
5ft/sec?? Jesus, that’s almost 6ft/sec!
They probably would have stayed together if they had just had the sense to use SI units.
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.
The “5 seconds after they started moving” is relevant. If we assume this takes place on Earth (i.e. on the surface of a sphere with a set pair of north/south poles), the angle between the two vectors changes depending on their current position.
If it’s not on the equator, it’s also slightly up to interpretation if “Due East” means they’ll turn to stay on the same latitude, always adjusting to stay moving east forever or if they’ll do a great circle. In the former case, the north moving one will eventually get stuck at the north-pole too instead of continuing their circle around the globe. Most likely not within 5 seconds though, unless the place they started was within 25 feet of the north-pole.
To actually do the math we’ll need to know (or somehow deduce) where “the place where everything about them began” is though.
What a convoluted way of asking the teacher to spill the beans. I like it.
I guess the calculus portion of this is to write the separation as a function of time, s = √26*t, and then realize that the rate of separation is the same regardless of time, because the first derivative is a constant.
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!
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.
Good thing that ain’t an english teacher
Sigh i miss high school maths. Even i’m lovesick now.
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.
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** 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.
Reminds me that one exurb1a pinecone video.
a^2 + b^2 = c^2
He walks 25ft north, she walks 5ft east. The distance between them is D.
25^2 + 5^2 = D^2 625 + 25 = D^2 D = √650 = 25.5ft
Read th question again. It’s not about how far apart they are after 5 seconds. It’s about how fast they are separating from each other, so it’s about a rate, not distance.
