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.
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.
Speed is just the magnitude of velocity.
My point is that OC was completely missing the mark by not properly accounting for time.
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.
Only if you divide by time. Including units is an essential sanity check.
Also, the rest of the math needs to be correct.
Well that’s my point. The answer is correct in this specific case, because it’s already “built-in” so to speak.
No, their answer is wrong.
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?
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.