Solving a Jet Speed Problem: Tokyo to Bangkok Distance and Time

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A jet traveling from Tokyo to Bangkok covers 4800 km, with the return speed reduced by 200 km/h, resulting in a 2-hour longer flight time. The correct equation to solve for the jet's speed is derived from the difference in travel times, leading to 4800/(x - 2) - 4800/x = 200 km/h. Participants discussed the importance of accurately setting up the equations and the implications of time versus speed in the calculations. There were also inquiries about learning LaTeX for better formatting of mathematical expressions. The discussion emphasized solving the quadratic equation to find the jet's speed on the return trip.
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Homework Statement


A jet flew from Tokyo to Bangkok, a distance of 4800 km. On the return trip, the speed was decreased by 200km/h. If the difference in the times of the flights was 2 hours, what was the jet's speed from Bangkok to Tokyo?


Homework Equations



Distance=(speed)(time)
Time=Distance/Speed
Speed=Distance/Time

The Attempt at a Solution


On sample questions they showed us how to solve a problem like this, i attempted to solve it in a similar manner however as you will see i was unable to

Let x represent speed

D=4800 Speed=? Time =?
----------------------------------------------------------------------------------------
Distance Speed Time
---------------------------------------------------------------------------------------
A to B 4800 x 4800/x (since time = d/s)
Tokyo
to Bangkok
----------------------------------------------------------------------------------------
B to A 4800 (x-200) 4800/(x-200)
Bangkok Since on return
to Tokyo trip speed was
reduced by 200
----------------------------------------------------------------------------------------
Time -2
----------------------------------------------------------------------------------------

4800/x +4800/(x-200)=-2

Lowest common denominator = x(x-200)
Here we multiply everything by x(x-200)

4800/x*x(x-200)+4800/(x-200)*x(x-200)=-2x(x-200)
Here some of the variable cancel out and the calculations are made and we are left
with the end result which is

4800x-960000+4800x=-2x(x-200)
4800x-960000+4800x=-2x^2+400x (On next step the 2x^2 and the right side is moved to
2x^2-400x+4800x-960000+4800x=0 the left side, hence the change in signs)
2x^2-400x+4800x+4800x-960000=0

Basically at this point my calculations got sketchy because there were several ways of going about this problem at this point, also I believe i had made some errors early in my calculations.

Sorry if it isn't very coherent i did my best to make it so, but I just got LaTex(Tex) and i didnt have the time to learn it just yet, also ms word won't let me paste the equations as they are here without altering them.

Thanks in advance and for bearing with me @_@
 
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supernova1203 said:

Homework Statement


A jet flew from Tokyo to Bangkok, a distance of 4800 km. On the return trip, the speed was decreased by 200km/h. If the difference in the times of the flights was 2 hours, what was the jet's speed from Bangkok to Tokyo?


Homework Equations



Distance=(speed)(time)
Time=Distance/Speed
Speed=Distance/Time

The Attempt at a Solution


On sample questions they showed us how to solve a problem like this, i attempted to solve it in a similar manner however as you will see i was unable to

Let x represent speed

D=4800 Speed=? Time =?
----------------------------------------------------------------------------------------
Distance Speed Time
---------------------------------------------------------------------------------------
A to B 4800 x 4800/x (since time = d/s)
Tokyo
to Bangkok
----------------------------------------------------------------------------------------
B to A 4800 (x-200) 4800/(x-200)
Bangkok Since on return
to Tokyo trip speed was
reduced by 200
----------------------------------------------------------------------------------------
Time -2
----------------------------------------------------------------------------------------

4800/x +4800/(x-200)=-2

I think this line is wrong. What you're after is 4800/x-200 - 4800/x = 2, as the slower speed takes 2 hours longer, where x here is the time. Can you solve this equation?
 
Or if it helps you understand it better in terms of writing it out in the difference from the base case, you could arbitrarily say that 4800=v*t. Then 4800=(v-200)*(t+2) to take into account both the slower speed and longer duration. You can then substitute, reduce, and use the ratio t=4800/v to solve for v (quadratic would have two answers, only one makes sense since you're dealing with positive distance).
 
Apphysicist said:
Or if it helps you understand it better in terms of writing it out in the difference from the base case, you could arbitrarily say that 4800=v*t. Then 4800=(v-200)*(t+2) to take into account both the slower speed and longer duration. You can then substitute, reduce, and use the ratio t=4800/v to solve for v (quadratic would have two answers, only one makes sense since you're dealing with positive distance).

Hm... i believe that makes sense, i recall them solving a similar problem previously this way, Thanks!
 
sjb-2812 said:
I think this line is wrong. What you're after is 4800/x-200 - 4800/x = 2, as the slower speed takes 2 hours longer, where x here is the time. Can you solve this equation?

Almost...

4800 represents distance in km.
So, if x is time (in hours) then 4800/x would represent km/hr.

Therefore, you are finding the difference between 2 velocities, which we know is 200 km/hr.


The correct equation is:

\frac{4800}{x - 2} - \frac{4800}{x} = 200~km/hr
 
zgozvrm said:
Almost...

4800 represents distance in km.
So, if x is time (in hours) then 4800/x would represent km/hr.

Therefore, you are finding the difference between 2 velocities, which we know is 200 km/hr.


The correct equation is:

\frac{4800}{x - 2} - \frac{4800}{x} = 200~km/hr

You are, of course, right. Or change the word time for speed in my answer ;)
 
zgozvrm said:
Almost...

4800 represents distance in km.
So, if x is time (in hours) then 4800/x would represent km/hr.

Therefore, you are finding the difference between 2 velocities, which we know is 200 km/hr.


The correct equation is:

\frac{4800}{x - 2} - \frac{4800}{x} = 200~km/hr

Is there a resource from where i can learn LaTex? I got it i just have no clue as to how to start and where to begin? I use ms word for most of the work but if i copy paste, the format changes... @_@
 
I thought there was a tutorial here, but I can't seem to find it.
Otherwise, try Google!

(Also, you can click the "Quote" button on a post that has LaTex to see how it was entered).
 
supernova1203 said:
Is there a resource from where i can learn LaTex? I got it i just have no clue as to how to start and where to begin? I use ms word for most of the work but if i copy paste, the format changes... @_@

For many examples, see the PF sticky here:
https://www.physicsforums.com/showthread.php?t=8997
 
  • #10
Thanks a bunch! I have a feeling ill end up needing it eventually :)
 
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