They’re traveling away from their origin at constant velocities, so they’re traveling relative to each other at constant velocities as well.
The magnitude of the resulting vector (i.e., speed) can be calculated trivially since their movement is perpendicular on a plane, as the root of sum of squares, which many could recognize as the Pythagorean theorem:
√((5 ft/s)² + (1 ft/s)²) = √26 ft/s ≈ 5.1 ft/s
You can verify this by finding that their average speed apart is the same at all times (for all t > 0):
Vavg = √((t * 5 ft/s)² + (t * 1 ft/s)²) / t = √(t² * ((5 ft/s)² + (1 ft/s)²)) / t = √26 ft/s
Don’t forget to calculate the location where everything about them began and then include the curvature of Earth considering the latitude of said location into your speed calculation.
No, they’re spherical children in a vacuum.
for approximation we can assume that the boy is a point mass and the girl is a lie
Oh, so we have to calculate the gravitational attraction pulling them back. Fucking hell
Augustus! Save some room for later.
https://en.m.wikipedia.org/wiki/Spherical_geometry
I couldn’t find ‘potatoy geometry’ for a better approximation of earth.
You’ll note that I already assumed that they were on a plane, not the surface of a sphere.
I’m also noting the stick up your ass. 🙄
If the potato remark and subreddit don’t tip you off that I was being flippant, I don’t know what will.
No, the stick would be a one-dimensional line.
It’s been a while, but I think it’s quite trivial.
After one second, they span a right angled triangle, therefore (using a² + b² = c²) their distance is √(5²+1²) = ~5.1 ft
They move at constant speed, therefore they seperate at 5.1 ft/s. That means at 5s it’s just 5.1 × 5 = 25.5 ft for the distance and their speed is still the same.
Depends on where they met each other. If they for example fell in love during the main event of a trip to the north pole, that would change things a lot.
there is no north at the north pole so actually that’s the one place it can’t be
If you’re at the south pole, would every direction count as north?
Sure, but there is a north say 30 ft away from the north pole.
Its pretty convenient that its raining, which means you can ignore the coefficient of friction since the surface is slippery
It’s* pretty convenient that it’s* raining
Differential calculus? That looks more like algebra. Their speed is constant.
I agree, it is not calculus, it’s trigonometry.
Each of their speeds is constant, but different, and they’re walking in different directions.
Their distance is the hypotenuse of a triangle with sides 5t and t which will be root((5t)2 + t2). So the distance at time t of the ex lovers will be root(26) × t. You can basically grasp intuitively that the speed is indeed constant and equals to the root(26)=5.1 ft/sec. Technically you’d use the derivative power rule to drop the t and get the speed.
Look. Teachers have some unresolved shit as well.
It doesn’t matter what the actual answer is; to both the boy and the girl it feels like C.
reminds me of that one song, proof that geometric construction can solve all love affairs or something like that
what
Who hurt the math teacher?
9,14 Meter
