## hermommy1101 4 years ago 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.

1. 3lroy

ok just give me 2 mins

2. 3lroy

its pretty simple tho

3. hermommy1101

thank you i don't understand it :( i don't like these type of problems :(

4. 3lroy

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

5. 3lroy

6. hermommy1101

ive tried trust me, i just cant put them into perspective... anything else i can do but not these word ones :( i know horrible :(

7. hermommy1101

ok following....

8. 3lroy

ok ill solve it and give you the solution just give me time

9. hermommy1101

thank you so much, i greatly appreciate it

10. hermommy1101

when you say time? do you mean a few times or later on during the day?

11. ttt

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

12. hermommy1101

awesome... but what is the total travel time?

13. ttt

y+y+4=16+16+4=32 hrs

14. ttt

oh damn, 36 i mean

15. spaceknight

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.

16. hermommy1101

Great, thank you so much for your help, i greatly appreciate it

17. ttt

np

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