this post was submitted on 28 Aug 2024
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[–] [email protected] 4 points 2 months ago* (last edited 2 months ago) (2 children)

~~They each move at a constant speed, but the distance between them doesn't increase at a constant pace. See my other comment.~~

Edit: I am dumb, and looked at the wrong number.

[–] [email protected] 12 points 2 months ago* (last edited 2 months ago) (1 children)

I'm trying to apply the most simple math possible and it seems to add up.

After one second, their distance is √(5² + 1²) = ~5.1 ft

After two seconds, their distance is √(10² + 2²) = ~10.2 ft

After three seconds, it's √(15² + 3²) = ~15.3 ft

As speed is the rate of change of distance over time, you can see it's a constant 5.1 ft/s. You're free to point out any error, but I don't think you need anything more than Pythagoras' theorem.

The question specifically asks for their seperation speed at 5s to ignore any initial change in their speed as they first need to accelerate, I'd assume.

[–] [email protected] 7 points 2 months ago (1 children)

Ah sorry, I'm tired and made a mistake. I quickly made a spreadsheet (because keeping track of numbers is hard), and I was looking at the wrong column in the sheet. My bad!

[–] [email protected] 8 points 2 months ago (1 children)

You were tired so you made a spreadsheet to calculate the differential equation quiz from a meme?

[–] [email protected] 3 points 2 months ago

Yes, compared to doing the calculations in my head lol

I work in mysterious ways

[–] booly 7 points 2 months ago

I don't see why the distance between them isn't growing at a constant speed.

At any given time t seconds after separation, the boy is 5t north, and the girl is 1t east. The distance between them is defined by the square root of ((5t)^2 + (t)^2 ), or about 5.099t.

In other words, the distance between them is simply a function defined as 5.099t, whose first derivative with respect to time is just 5.099.