• SpaceNoodle@lemmy.world
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    2 months ago

    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

  • De_Narm@lemmy.world
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    2 months ago

    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.

    • Da Bald Eagul@feddit.nl
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      2 months ago

      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.

      • De_Narm@lemmy.world
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        2 months ago

        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.

        • Da Bald Eagul@feddit.nl
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          2 months ago

          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!

      • booly@sh.itjust.works
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        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.

  • Bumblefumble@lemm.ee
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    2 months ago

    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.

  • Missmuffet@lemmy.world
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    2 months ago

    Its pretty convenient that its raining, which means you can ignore the coefficient of friction since the surface is slippery

    • havid_dume@lemmy.ml
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      2 months ago

      Each of their speeds is constant, but different, and they’re walking in different directions.

      • luciole (he/him)@beehaw.org
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        2 months ago

        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.

  • Saki@lemmy.blahaj.zone
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    2 months ago

    reminds me of that one song, proof that geometric construction can solve all love affairs or something like that

    • Da Bald Eagul@feddit.nl
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      2 months ago

      The question states “how fast”, not “how far”, thus you need to give the acceleration at that moment.

      At t=0, the boy and girl both haven’t moved, so their positions are 0. The distance between them is also 0, as is their acceleration.

      The boy’s distance in meters is t*1.524, the girl’s distance is t*0.3048. The distance between them is sqrt( b^2 * g^2 ). The velocity is the current distance minus the previous distance.

      At t=1, b=1.524m, g=0.305, d=sqrt( g^2 * g^2 )=0.465, v=d-d^(t-1)=0.465m/s.

      At t=5, b=7.62, g=1.524, d=11.613, and v=4.181m/s.

      Edit: fixed markdown

        • ji17br@lemmy.ml
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          2 months ago

          It’s the difference of distances apart over time. Aka how fast bf is moving away from gf, aka what the question is asking for.

          Yes, if you want to be pedantic, velocity a vector with direction, so I guess you’d have to frame the question relative to either the boyfriend or girlfriend, but I don’t think the difference between speed and velocity is part of the question.

          • SpaceNoodle@lemmy.world
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            2 months ago

            Speed is just the magnitude of velocity.

            My point is that OC was completely missing the mark by not properly accounting for time.

            • Da Bald Eagul@feddit.nl
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              2 months ago

              Hi, I made this in 5 mins because I was bored, but it’s late and I’m tired, so could you please explain what I would have to fix in my comment?

              • ji17br@lemmy.ml
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                2 months ago

                You want to figure out distance per second. One way to do this is calculate distance apart at t=0,1,2…

                The difference between each point would be the average speed over that second.

                Using sqrt(b2+g2):

                t0 = 0 t1 = 1.554m
                s1 = (1.554m-0m)/1s = 1.554m/s t2 = 3.108m
                s2=(3.108m-1.554m)= 1.554m/s

                As you continue this you will see they travel at a constant speed apart from each other. The reason this is working is because you need to divide distance by time. Dividing by 1 second won’t change the value of the number after you subtract. If you notice you can do (t2-t0)/2s and also get the same answer.

            • ji17br@lemmy.ml
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              2 months ago

              My mistake, I didn’t check his math. I thought he was saying if you take distance apart at t(n) and subtract distance apart at t(n-1) you will get distance/sec.

              • SpaceNoodle@lemmy.world
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                2 months ago

                Only if you divide by time. Including units is an essential sanity check.

                Also, the rest of the math needs to be correct.

                • ji17br@lemmy.ml
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                  2 months ago

                  Well that’s my point. The answer is correct in this specific case, because it’s already “built-in” so to speak.