Mechanics Revision Thread

  • #1
Can you help me with some of these questions? I cant do the second part of Q1!!!
 

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  • #2
Hootenanny
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magicuniverse,

This is the second thread you've started where you haven't shown any work. Can I remind you that you are required to show the you have attempted any homework before posting as well as detailing any thoughts you have on your problem.

"I have no idea!" doesn't count as detailing your thoughts.
 
  • #3
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.
 
  • #4
Hootenanny
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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.
Okay, so what have you done so far in question (1)?
 
  • #5
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.
 
  • #6
Hootenanny
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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.
May i see your attempt?
 
  • #7
May i see your attempt?

Yer but you cant see my diagram as I cant 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.
 
  • #8
Hootenanny
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Yer but you cant see my diagram as I cant 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.
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?

In addition, the the boy's maximum speed relative to the water is 3 m/s.
 
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  • #9
I figured that the current was pulling him down? Il have alook at it, thanks.
 
  • #10
Hootenanny
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I figured that the current was pulling him down? Il have alook at it, thanks.
Note my second point above, the boy's total speed cannot exceed 3 m/s
 
  • #11
Thanks I have to got the answer to 200/root(5), your time is appretiated. Any help with question 2 would be appretiated now thanks.
 
  • #12
Hootenanny
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Thanks I have to got the answer to 200/root(5), your time is appretiated.
Good, and no problem :smile:
Any help with question 2 would be appretiated now thanks.
What have you attempted thus far?
 

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