Question to due with constant speed, distance and time (thread 2)

[Cambridge101]

Ok so just for background, I am completely a self taught math lover, ( I am going to Cambridge university to study Economics in September), but until then I have been teaching myself higher level maths while going over old stuff I never was taught as a kid and wanted to fully understand.

I had one particularly random shower thought to do with the speed distance and time.
Ok so, say I am travelling 50km/h. And I graph this (on a distance time graph) to show that in 60 mins I travel 50km. Speed is constant and the gradient is 50/1 or 50/60 for this graph = 5/6. So for every 6 mins in time, we travel 5km. What interested me is that 5 is 10% of 50 and 6 is 10% of 60. They're equally as far along their axes as each-other.

Now people are taught simply s = d/t therefore, t = d/s (juts learn it). But I want to know intuitively why this rearrangement makes sense. But because I learn all my maths alone I have to try and make connections myself.

In my case say I travel 10km at a speed of 50km/h and I want to find the time is took.
So using the formula t = 10km / 50km/h

They way I think about this is this, if were saying 10/50 this is 10 out of 50. Thus, 10 km is 1/5th of 50km as a distance, 20%. So you can see as shown on my graph. A distance of 10km is 20% of the total distance traveled in the time of 1 hr. So because the answer to is 1/5 or 0.2hr because if it is 20% of the distance it must take 20% of the time? Because to me it doesn't seem to make intuitive sense for some reason - I just cant see the connect as to why. The answer is 12 mins when you plug it into the formula and 12 is also 20% of 60 mins. So it is like if we go 20% up the y-axis, we go 20% up the x-axis. Likewise, if we simply the ratio to 0.83 recurring / 1 I find that these are also both worth 1.6 recurring % of their axis lengths. Why is this?

Thank you guys.

 

[Deleted member 4993]

Ok so juts for background I am completely a self taught math lover, ( I am going to Cambridge university to study Economics in September), but until then I have been teaching myself higher level maths while going over old stuff I never was taught as a kid and wanted to fully understand.

I had one particularly random shower thought to do with the speed distance and time.
Ok so, say I am travelling 50km/h. And I graph this (on a distance time graph) to show that in 60 mins I travel 50km. Speed is constant and the gradient is 50/1 or 50/60 for this graph = 5/6. So for every 6 mins in time, we travel 5km. What interested me is that 5 is 10% of 50 and 6 is 10% of 60. They're equally as far along their axes as each-other.

Now people are taught simply s = d/t therefore, t = d/s (juts learn it). But I want to know intuitively why this rearrangement makes sense. But because I learn all my maths alone I have to try and make connections myself.

In my case say I travel 10km at a speed of 50km/h and I want to find the time is took.
So using the formula t = 10km / 50km/h

They way I think about this is this, if were saying 10/50 this is 10 out of 50. Thus, 10 km is 1/5th of 50km as a distance, 20%. So you can see as shown on my graph. A distance of 10km is 20% of the total distance traveled in the time of 1 hr. So because the answer to is 1/5 or 0.2hr because if it is 20% of the distance it must take 20% of the time? Because to me it doesn't seem to make intuitive sense for some reason - I just cant see the connect as to why. The answer is 12 mins when you plug it into the formula and 12 is also 20% of 60 mins. So it is like if we go 20% up the y-axis, we go 20% up the x-axis. Likewise, if we simply the ratio to 0.83 recurring / 1 I find that these are also both worth 1.6 recurring % of their axis lengths. Why is this?

Thank you guys.

Have you answered YOUR questions in thread (1)?

Question to due with constant speed, distance and time (thread 2)

Ok so just for background, I am completely a self taught math lover, ( I am going to Cambridge university to study Economics in September), but until then I have been teaching myself higher level maths while going over old stuff I never was taught as a kid and wanted to fully understand. I had...

www.freemathhelp.com

Looks like you are asking the same question again!!

 

[Cambridge101]

Have you answered YOUR questions in thread (1)?

Question to due with constant speed, distance and time (thread 2)

Ok so just for background, I am completely a self taught math lover, ( I am going to Cambridge university to study Economics in September), but until then I have been teaching myself higher level maths while going over old stuff I never was taught as a kid and wanted to fully understand. I had...

www.freemathhelp.com

Looks like you are asking the same question again!!

I am asking the same question yes. But i think in a better way.

 

[lev888]

Let's simplify a bit. Instead of 20% let's use 50%.
You travel 25km at 50km/h. And then you travel another 25km at 50km/h. Do you see intuitively that these 2 trips take the same amount of time? Do you see that the first trip's distance is 50% of the total distance? Do you see that the first trip's time is 50% of the total time?
The same "intuition" applies to 20%. If you don't see how, add 3 more trips of 25km at 50km/h. Now 1 segment's distance and time are 20% of the total distance and time.

 

[Cambridge101]

Let's simplify a bit. Instead of 20% let's use 50%.
You travel 25km at 50km/h. And then you travel another 25km at 50km/h. Do you see intuitively that these 2 trips take the same amount of time? Do you see that the first trip's distance is 50% of the total distance? Do you see that the first trip's time is 50% of the total time?
The same "intuition" applies to 20%. If you don't see how, add 3 more trips of 25km at 50km/h. Now 1 segment's distance and time are 20% of the total distance and time.

Yes, thank you.