anonymous
  • anonymous
A naval ship travels from Portsmouth, England to the west end of the Kiel Canal in northern Germany, and later returns to Portsmouth at a speed which is 10.25 km/h faster. Portsmouth is 820 km from the Kiel Canal. If the trip to the Kiel Canal takes 4 hours longer than the trip back to Portsmouth, find the speed of the ship in each direction and the total travel time.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
ok just give me 2 mins
anonymous
  • anonymous
its pretty simple tho
anonymous
  • anonymous
thank you i don't understand it :( i don't like these type of problems :(

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anonymous
  • anonymous
the problem is pretty simple you just need a pen and a paper just visualise the ship as it is moving talking note of direction it give like a triangular or rectangular figure then you can go from there.....i sued to solve these things in 10th grade now im a college senior
anonymous
  • anonymous
so just follow my pattern
anonymous
  • anonymous
ive tried trust me, i just cant put them into perspective... anything else i can do but not these word ones :( i know horrible :(
anonymous
  • anonymous
ok following....
anonymous
  • anonymous
ok ill solve it and give you the solution just give me time
anonymous
  • anonymous
thank you so much, i greatly appreciate it
anonymous
  • anonymous
when you say time? do you mean a few times or later on during the day?
anonymous
  • anonymous
let x be the speed from England to Germany, so speed from G to E is x+10.25, because the ship return at 10.25km/h faster let y be the time going from G to E, so time from E to G is y+4 (4hrs longer) England to Germany Germany to England speed x km/h x+10.25 time y+4 y distance=speed * time so x(y+4)=820 y(x+10.25)=820 solve that get y=16 or y=-20 (not logical since time cant be negative) so y=16 -> x=41 then speed from England to Germany = 41km/h speed from Ger to End = 41+10.25=51.25 km/h
anonymous
  • anonymous
awesome... but what is the total travel time?
anonymous
  • anonymous
y+y+4=16+16+4=32 hrs
anonymous
  • anonymous
oh damn, 36 i mean
anonymous
  • anonymous
Let subscript 1 denote the trip there and subscript 2 denote the trip back. Velocity is simply distance divided by time. v1=d/t1=820/(4+x) with x being the time taken for the trip back.Since it tells you the trip is 4 hours longer than the trip back, naturally it is 4+x. v2=d/t2=820/x=v1+10.25 because it tells you the second trip is 10.25 faster so you have 2 equations for v2. Isolate for v1=820x/10.25 and equate it with the first v1 equation from before, so v1=820/x-10.25=820/(4+x). Solve for x which is 16. Plug x back into all equations and you have the speeds.
anonymous
  • anonymous
Great, thank you so much for your help, i greatly appreciate it
anonymous
  • anonymous
np

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