- #1
Okay, so what have you done so far in question (1)?magicuniverse said:Well its not homework for a start. This is revision for an exam so by explaining/hinting at a method you arnt going to be put me at an unfair advantge.
May i see your attempt?magicuniverse said:I have worked out the time to be 120s with the use of a diagram and orking out resultsnt vectors. Tried doing the second part but got it wrong, my answer was massively out so I though that I must be doing it worng and wanted to know how.
Hootenanny said:May i see your attempt?
But surely if he swims at a speed of [itex]\sqrt{10}[/itex] m/s, that would mean that he is swimming at an angle of [itex]\tan^{-1}(1/3) \approx 18^o[/itex] above due west and would therefore miss the buoy?magicuniverse said:Yer but you can't see my diagram as I can't get it on here.
Basically know that the currrent will drag the bo down, but the boy can swim upwards at 1m/s so if he can swim to the west at 3 m/s his velocity will be root(10) (from pythagorus) we then divide the distance 200m by root10. This comes out with 63.25s rather than the required 89.4s.
Note my second point above, the boy's total speed cannot exceed 3 m/smagicuniverse said:I figured that the current was pulling him down? Il have alook at it, thanks.
Good, and no problemmagicuniverse said:Thanks I have to got the answer to 200/root(5), your time is appretiated.
What have you attempted thus far?magicuniverse said:Any help with question 2 would be appretiated now thanks.
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