Say you took a day to get to you destination, then if you wanted the time it takes to get your next rest stop within your journey and it happen's to be: 1/24 th of total distance, constant speed all day, then you will take 1 hour, so: time(1h rest stop)=time(1d journey)/24 You can keep on going with this idea, associating times with actions... To atomic scales and beyond! ⚛️ 💥
u no d/t breaks when you get to the speed of light right? so an accurate speed would in fact be dist/angle - I think that was the formula, I'm sure someone will correct me if they're more familia with it
speed=ẋ=|x1-x0|/(T1-T0) If x1»x0, T1=t+T0, then ẋ→|x1|/t. If ẋ→c, then |x1|/t ≈ f λ'→f λ=c So, for slightly slower speeds: ẋ≈ |x1-x0|/|x1|/c≈c(1-|x0|/|x1|) |x0|/|x1|, |dx|/|x1| are like trig ratios.. 🤔 If |x1|≈|x0|, then the trig ratio |dx|/|x1| will be very small, and so the angle you could associate with the sides will also be small, around the same value as the ratio. Speed of light in terms of these effects: c≈ẋ*(trig ratio) To make use of small angle approx, would use the fact that trig functions repeat themselves over various intervals, with the absolute value of the function going from 0 to max in angle intervals of π/2 in length. So, |small angle|≈|trig(small angle)| ≈|trig(small angle+kπ/2)|. Substitute tanh function as trig ratio, then |trig(angle)|=|tanh(angle)| would probably work best because it seems that the angles can never be _exactly_ zero or kπ/2, as x1 can't be zero, etc. So, I guess that c≈|ẋ|*|tanh(angle)| and the times in the derivatives on both sides can be eliminated, giving: λ= |dx|*|tanh(angle)|. speed=ẋ=|x1-x0|/(T1-T0) If x1»x0, T1=t+T0, then ẋ→|x1|/t. If ẋ→c, then |x1|/t ≈ f λ'→f λ=c So, for slightly slower speeds: ẋ≈ |x1-x0|/|x1|/c≈c(1-|x0|/|x1|) |x0|/|x1|, d|x|/|x1| are like trig ratios.. 🤔 If |x1|≈|x0|, then the trig ratio |dx|/|x1| will be very small, and so the angle you could associate with the sides will also be small, around the same value as the ratio. Speed of light in terms of these effects: c≈ẋ*(trig ratio) To make use of small angle approx, would use the fact that trig functions repeat themselves over various intervals, with the absolute value of the function going from 0 to max in angle intervals of π/2 in length. So, |small angle|≈|trig(small angle)| ≈|trig(small angle+kπ/2)|. Substitute tanh function as trig ratio, then |trig(angle)|=|tanh(angle)| would probably work best because it seems that the angles can never be _exactly_ zero or kπ/2, as x1 can't be zero, etc. So, I guess that c≈|ẋ|*|tanh(angle)| and the times in the derivatives on both sides can be eliminated, giving: λ=|dx|*|tanh(angle)|. Next step is to differentiate both sides with respect to angle...I guess. 😕 I will leave that to you, if you want to continue this train of thought...
@@The_Green_Man_OAP there are some math symbols there that I forget the meaning of so I can't even continue the thought, btw I don't mean angle as in 0-90(360), I mean angle as in 0-100(400) at least I think that was the unit limit, been a long time since I saw the vid. I think they call this variant radians.
i still cant figure out, i ran 2 kilometers in 15/kmh, how long did it take me to run these 2 kilometers? distance = 2 km, speed = 15/kmh. how can i figure out the time?
Could you please do one of those hateful problems Like train A leaves the station at 2o'clock it's going 6o.1 , train B leaves the station at 4 O' clock 3 days ago and was going 42.5 at what time will train A over take train B during the summer solstice in 100 yrs.? Those type
God bless u I have a lazy teacher he doesn’t explain anything
Good video as always. I had no idea could you explain more of the university style equations?
This was so stupidly easy. I need you to teach me velocity acceleration negative acceleration like this pleaseeee.
Say you took a day to get to you destination,
then if you wanted the time it takes to get your next rest stop within your journey and it happen's to be: 1/24 th of total distance, constant speed all day, then you will take 1 hour, so:
time(1h rest stop)=time(1d journey)/24
You can keep on going with this idea, associating times with actions...
To atomic scales and beyond! ⚛️ 💥
u no d/t breaks when you get to the speed of light right? so an accurate speed would in fact be dist/angle - I think that was the formula, I'm sure someone will correct me if they're more familia with it
I think 🤔 they call that _Rapidity_ 💨...
speed=ẋ=|x1-x0|/(T1-T0)
If x1»x0, T1=t+T0, then ẋ→|x1|/t.
If ẋ→c, then |x1|/t ≈ f λ'→f λ=c
So, for slightly slower speeds:
ẋ≈ |x1-x0|/|x1|/c≈c(1-|x0|/|x1|)
|x0|/|x1|, |dx|/|x1| are like trig ratios.. 🤔
If |x1|≈|x0|, then the trig ratio |dx|/|x1| will be very small, and so the angle you could associate with the sides will also be small, around the same value as the ratio.
Speed of light in terms of these effects:
c≈ẋ*(trig ratio)
To make use of small angle approx,
would use the fact that trig functions repeat themselves over various intervals, with the absolute value of the function going from 0 to max in angle intervals of π/2 in length.
So, |small angle|≈|trig(small angle)|
≈|trig(small angle+kπ/2)|.
Substitute tanh function as trig ratio,
then |trig(angle)|=|tanh(angle)| would probably work best because it seems that the angles can never be _exactly_ zero or kπ/2, as x1 can't be zero, etc.
So, I guess that c≈|ẋ|*|tanh(angle)| and the times in the derivatives on both sides can be eliminated, giving: λ= |dx|*|tanh(angle)|.
speed=ẋ=|x1-x0|/(T1-T0)
If x1»x0, T1=t+T0, then ẋ→|x1|/t.
If ẋ→c, then |x1|/t ≈ f λ'→f λ=c
So, for slightly slower speeds:
ẋ≈ |x1-x0|/|x1|/c≈c(1-|x0|/|x1|)
|x0|/|x1|, d|x|/|x1| are like trig ratios.. 🤔
If |x1|≈|x0|, then the trig ratio |dx|/|x1| will be very small, and so the angle you could associate with the sides will also be small, around the same value as the ratio.
Speed of light in terms of these effects:
c≈ẋ*(trig ratio)
To make use of small angle approx,
would use the fact that trig functions repeat themselves over various intervals, with the absolute value of the function going from 0 to max in angle intervals of π/2 in length.
So, |small angle|≈|trig(small angle)|
≈|trig(small angle+kπ/2)|.
Substitute tanh function as trig ratio,
then |trig(angle)|=|tanh(angle)| would probably work best because it seems that the angles can never be _exactly_ zero or kπ/2, as x1 can't be zero, etc.
So, I guess that c≈|ẋ|*|tanh(angle)| and the times in the derivatives on both sides
can be eliminated,
giving: λ=|dx|*|tanh(angle)|.
Next step is to differentiate both sides with respect to angle...I guess. 😕
I will leave that to you, if you want to continue this train of thought...
@@The_Green_Man_OAP there are some math symbols there that I forget the meaning of so I can't even continue the thought, btw I don't mean angle as in 0-90(360), I mean angle as in 0-100(400) at least I think that was the unit limit, been a long time since I saw the vid. I think they call this variant radians.
@@zxuiji Radians go from 0 to 2π for a full rotation.
I think the unit measure you refer to is the _gradian_ .
Good job 👍
👍🏼 Liking the triangle method.
i still cant figure out, i ran 2 kilometers in 15/kmh, how long did it take me to run these 2 kilometers? distance = 2 km, speed = 15/kmh. how can i figure out the time?
Time = distance/speed
2km/15kmh = 0.1333hr
0.1333 in mins is 0.1333x 60 = 8 mins
Who's clock ⏰ do you use to measure the time?
How make out boat and stream plz make video sir
Cool
might be a bit easier on the eyes of the background was a little darker
Could you please do one of those hateful problems
Like train A leaves the station at 2o'clock it's going 6o.1 , train B leaves the station at 4 O' clock 3 days ago and was going 42.5 at what time will train A over take train B during the summer solstice in 100 yrs.? Those type
nice
❤
❤