• SpaceNoodle@lemmy.world
    link
    fedilink
    arrow-up
    67
    arrow-down
    1
    ·
    edit-2
    4 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