476

submitted 2 months ago by Track_Shovel@slrpnk.net to c/196@lemmy.blahaj.zone

45 comments fedilink hide all child comments

you are viewing a single comment's thread
view the rest of the comments

[-] ji17br@lemmy.ml 3 points 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 2 points 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.

[-] dabaldeagul@feddit.nl 2 points 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 3 points 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(b^2+g^2):

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.

[-] dabaldeagul@feddit.nl 1 points 2 months ago

Ahhh okay, thanks

[-] ji17br@lemmy.ml 1 points 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.