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):
Don't forget to calculate the location where everything about them began and then include the curvature of Earth considering the latitude of said location into your speed calculation.
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.
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.
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!
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.
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.
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.
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.
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.
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.
the boy starts out running at a speed of 5ft/s then slows to 2.5ft/s over the next 5 seconds...
Then you'd need differential equations or whatever the class is, and you'd have to use more than just basic trig, but as written, it's just basic trig and the time is irrelevant.
But the speed they are moving apart from each other is increasing. Edit: IDK LOL but if you drew it as a triangle the long side would be getting longer and longer, in the first second it might grow a little bit but after some time in one second it would grow a lot. If you need anymore physics questions examined please let me know
Wouldn't that be just a right triangle? One side 5×5=25 feet long, the other 5×1=5 feet long so the hypotenuse would be the square root of (25²+5²) or around 25,5 feet.
That's basic geometry right? It's been quite a few month but I'm pretty sure I still can do highschool level maths.
53 replies
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
Don't forget to calculate the location where everything about them began and then include the curvature of Earth considering the latitude of said location into your speed calculation.
No, they're spherical children in a vacuum.
for approximation we can assume that the boy is a point mass and the girl is a lie
Oh, so we have to calculate the gravitational attraction pulling them back. Fucking hell
Augustus! Save some room for later.
https://en.m.wikipedia.org/wiki/Spherical_geometry
I couldn't find 'potatoy geometry' for a better approximation of earth.
You'll note that I already assumed that they were on a plane, not the surface of a sphere.
I'm also noting the stick up your ass. 🙄
If the potato remark and subreddit don't tip you off that I was being flippant, I don't know what will.
No, the stick would be a one-dimensional line.
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.
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.
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.
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!
You were tired so you made a spreadsheet to calculate the differential equation quiz from a meme?
Yes, compared to doing the calculations in my head lol
I work in mysterious ways
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.
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.
there is no north at the north pole so actually that's the one place it can't be
If you're at the south pole, would every direction count as north?
Sure, but there is a north say 30 ft away from the north pole.
Its pretty convenient that its raining, which means you can ignore the coefficient of friction since the surface is slippery
It doesn't matter what the actual answer is; to both the boy and the girl it feels like C.
reminds me of that one song, proof that geometric construction can solve all love affairs or something like that
what
Who hurt the math teacher?
9,14 Meter
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
Velocity is not the difference between distances.
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.
Speed is just the magnitude of velocity.
My point is that OC was completely missing the mark by not properly accounting for time.
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(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.
Ahhh okay, thanks
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.
7 Hallmarks per Homecoming.
The answer is six feet per second, right?
Can't tell if that's a genuine answer or a joke sooo...
They're moving perpendicular so unfortunately it's not that easy... I'll leave the calculations to someone who didn't fail math though
Oh that makes sense why the elapsed time would factor into the equation too.
To trick you.
Now if the question had been something like:
Then you'd need differential equations or whatever the class is, and you'd have to use more than just basic trig, but as written, it's just basic trig and the time is irrelevant.
The time matters because the angle is changing and with it the relative velocity.
No, the angle is never changing.
I just drew a picture and you're right, the triangle has the same angles, the sides just get longer.
But the speed they are moving apart from each other is increasing. Edit: IDK LOL but if you drew it as a triangle the long side would be getting longer and longer, in the first second it might grow a little bit but after some time in one second it would grow a lot. If you need anymore physics questions examined please let me know
Wouldn't that be just a right triangle? One side 5×5=25 feet long, the other 5×1=5 feet long so the hypotenuse would be the square root of (25²+5²) or around 25,5 feet.
That's basic geometry right? It's been quite a few month but I'm pretty sure I still can do highschool level maths.
That's the distance yes, but how fast are they moving away from eachother
(5^2 + 1^2)^0.5 = 5.1ft/s
Well, like I said I failed math so I'll take your word for it
Edit: autocorrect fucked me