Hi V=IR, (This is tongue in cheek by the way) Is it 'cause you is from Africa that you confused things back on page one? Your first language might be French but I think the original question is quite clear.
Hi V=IR, (This is tongue in cheek by the way) Is it 'cause you is from Africa that you confused things back on page one? Your first language might be French but I think the original question is quite clear.
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Well, you now have several 'answers' from which to choose, viz:
Spark123 : 1000m
holmslaw : 1000m
Softus : 1000m
tim west : It will take off
Adrian80 : 1300m
mikeyd : 1500m
mikeyd : 1750m
mikeyd : 1600m
pipme : 1600 m
S
Shutpa
Bad news guys, my daughter tells me that ten and eleven year olds were given the same problems.
S
Shutpa
You are right Softus, it doesn't change a thing. I am simply pointing out that there are some really bright youngsters out there.
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bolo is older than nine surely?
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1st journey: in 1 unit of 'my' (travelling) time the two ferries pass each other, distances traversed amount to one width of the Hudson.
The slower ferry has covered 650m in that single time unit from it's starting dock.
The ferry boats do what they do, apart from speed - the same thing - Then on the second leg they pass each other just 350m from the opposite river bank... Another time period has passed, again the two ferries have completed a total combined distance equal to a width of the Hudson.
The slower ferry has been sailing for 2 time units at 650m per unit, hence it has covered 1300m since the first passing point.
350m of that distance was on the return journey to second passing point, hence 1300 - 350 = 950m from 1st passing point to passenger dock at end of journey.
Therefore 650m + 950m = river width = 1600m.
The slower ferry has covered 650m in that single time unit from it's starting dock.
The ferry boats do what they do, apart from speed - the same thing - Then on the second leg they pass each other just 350m from the opposite river bank... Another time period has passed, again the two ferries have completed a total combined distance equal to a width of the Hudson.
The slower ferry has been sailing for 2 time units at 650m per unit, hence it has covered 1300m since the first passing point.
350m of that distance was on the return journey to second passing point, hence 1300 - 350 = 950m from 1st passing point to passenger dock at end of journey.
Therefore 650m + 950m = river width = 1600m.
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I've searched the topic, especially the first post. I've looked down the back of the sofa. Heck, I've even looked in the least used compartment in the fridge, but I can't find anything that supports your assertion that the slower boat covers 650m from the port to the first meeting point.
No, it has taken 2 of your time units plus 10 minutes to cover the supposed 1300 metres.
As your time unit is unknown, how do you account for these 10 minutes.
Using your method of calculation, you can assume any river width and your maths still works out.
Surely the answer must be as simple as adding the two distances together?
Time and speed doesn't come into it - does it?
If they both meet at a point of 650m from the left bank in one direction, then 350m from the right bank in the other direction, do you just not add the two together?
And don't shoot me down - I did say I was crap at maths!
I know a couple of teachers - I'm gonna ask them cos this is bugging me too now.
Time and speed doesn't come into it - does it?
If they both meet at a point of 650m from the left bank in one direction, then 350m from the right bank in the other direction, do you just not add the two together?
And don't shoot me down - I did say I was crap at maths!
I know a couple of teachers - I'm gonna ask them cos this is bugging me too now.
1000 meters lol
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