Finding speed with basically just distance.

Sep 2016
19
0
ottawa
Hi,
I'm currently in the process of trying to learn how to use quadratics to find out speed/distance/time etc. My program that I'm taking is often very unclear and skims over topics regularly. The example I was given to learn the process was fairly straightforward and easy to learn, however the question I have to answer for my work is different. The first example I was given is:
A jet flies 4200km from point A to point B. On the return trip the speed was increased by 100km/h. The total trip time is 13 hours. What is the jets speed from point A to point B?
So they include distance = speed * time or time = distance / speed and that the equation should look something like this : 13 = 4200/x + 4200/(x+100)
After simplifying it should look like (13x^2)-7100x-420000=0 and factoring it to find the roots makes it look like (13x+700)(x-600)=0 The roots are: x=-53.846 and x=600
With this we know the speed is 600 because a negative number makes no sense.

Ok moving on. So the question I need help with is very similar except a very key component.
A jet flies 4800km from tokyo to bangkok. On the return trip the speed is decreased by 200km/h. If the difference in the times of flights was 2 hours, what was the jets speed from bangkok to tokyo?
So after spending quite a while just trying to piece together different combinations of numbers I've concluded that I have no idea where to start, and these people just love messing with us. I have no clue what I'm doing without the total trip time because that's just how they taught us.

If anybody would be able to help me I'd greatly appreciate it,
Thanks very much
 

romsek

MHF Helper
Nov 2013
6,837
3,079
California
well you nailed the first problem

the second one isn't that different, we know that the slower speed leg takes longer so

$\dfrac{4800}{v-200} - \dfrac{4800}{v} = 2$

and solve this identically to the way you solved the first problem
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Hi,
I'm currently in the process of trying to learn how to use quadratics to find out speed/distance/time etc. My program that I'm taking is often very unclear and skims over topics regularly. The example I was given to learn the process was fairly straightforward and easy to learn, however the question I have to answer for my work is different. The first example I was given is:
A jet flies 4200km from point A to point B. On the return trip the speed was increased by 100km/h. The total trip time is 13 hours. What is the jets speed from point A to point B?
So they include distance = speed * time or time = distance / speed and that the equation should look something like this : 13 = 4200/x + 4200/(x+100)
After simplifying it should look like (13x^2)-7100x-420000=0 and factoring it to find the roots makes it look like (13x+700)(x-600)=0 The roots are: x=-53.846 and x=600
With this we know the speed is 600 because a negative number makes no sense.

Ok moving on. So the question I need help with is very similar except a very key component.
A jet flies 4800km from tokyo to bangkok. On the return trip the speed is decreased by 200km/h. If the difference in the times of flights was 2 hours, what was the jets speed from bangkok to tokyo? T
So after spending quite a while just trying to piece together different combinations of numbers I've concluded that I have no idea where to start, and these people just love messing with us. I have no clue what I'm doing without the total trip time because that's just how they taught us.

If anybody would be able to help me I'd greatly appreciate it,
Thanks very much
This is very discouraging to read. Frankly you seem to be saying that all you have learned, and all you want to learn, is how to put numbers into memorized formulas and now you are outraged that they are "messing" with you by expecting you to think for yourself? Outrageous!

You say you know "time= distance/ speed". Here, you know that the distance from Tokyo to Bangkok is 4800 km. If call the speed, in km/h, on that leg "v", we have time= 4800/v. On the way back, so the same distance, the speed is "decreased by 200 km/h" so is v- 200. The time back is 4800/(v- 200). The "difference is times" is the difference of those numbers- subtract. Since the speed on the return trip was lower, that will take longer so 4800/(v- 200)- 4800/v= 2. Can you solve that equation?
 
Sep 2016
19
0
ottawa
This is very discouraging to read. Frankly you seem to be saying that all you have learned, and all you want to learn, is how to put numbers into memorized formulas and now you are outraged that they are "messing" with you by expecting you to think for yourself? Outrageous!

You say you know "time= distance/ speed". Here, you know that the distance from Tokyo to Bangkok is 4800 km. If call the speed, in km/h, on that leg "v", we have time= 4800/v. On the way back, so the same distance, the speed is "decreased by 200 km/h" so is v- 200. The time back is 4800/(v- 200). The "difference is times" is the difference of those numbers- subtract. Since the speed on the return trip was lower, that will take longer so 4800/(v- 200)- 4800/v= 2. Can you solve that equation?
I'm working through ILC courses on my laptop, if you were dependent on reading pdf files for your highschool education I believe you would understand my frustration. I have always preferred interaction with a teacher as opposed to basically teaching myself. I'm not only interested in learning how to put numbers into formulas, and I'm not outraged that they expect me to think for myself. Opening up a file and going through it a dozen times trying to learn every little detail is by definition "thinking for myself". As far as I can tell you are the only one outraged here, is it because not everybody has the same level of passion for mathematics as you do? I appreciate you taking the time to try to help me, but I don't appreciate the judgment.
 
Sep 2016
19
0
ottawa
well you nailed the first problem

the second one isn't that different, we know that the slower speed leg takes longer so

$\dfrac{4800}{v-200} - \dfrac{4800}{v} = 2$

and solve this identically to the way you solved the first problem
I think I'm starting to understand, so it's basically (return trip)-(first trip)=(time difference) ? Just so I know for the future.
Thank you very much
 

romsek

MHF Helper
Nov 2013
6,837
3,079
California
I think I'm starting to understand, so it's basically (return trip)-(first trip)=(time difference) ? Just so I know for the future.
Thank you very much
that's basically it but we choose (return - first) over (first - return) because the return trip is slower and thus takes longer. This way we get a positive difference.