487 /www/nrich/html/content/97/02/six2/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><p>Brian swims at twice the speed that the river is flowing. He swims directly downstream, from one moored boat to another and then back again, taking 12 minutes altogether. How long would it have taken him in still water?</p></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p>In this problem, consider the following:<br></br> When Brian swims with the flow, his total speed is his speed in the water plus the speed of the moving water. When he swims against the flow, his total speed is the difference between his speed in still water and the speed of the moving water.</p> <p>This leads to the following solution:</p> <p>Let the speed of the river be <em>v</em> , distance to the moored boat = <em>d</em> , and the time to complete the downstream journey = <em>t</em> .</p> <table border="0"> <tbody> <tr align="left" valign="top"> <td align="left" valign="middle"><strong>Downstream</strong></td> <td align="left" valign="middle"><strong>Upstream</strong></td> </tr> <tr align="left" valign="top"> <td align="left" valign="middle">Speed of man = 3 <em>v</em></td> <td align="left" valign="middle">Speed of man = <em>v</em></td> </tr> <tr align="left" valign="top"> <td align="left" valign="middle">Distance = <em>d</em></td> <td align="left" valign="middle">Distance = <em>d</em></td> </tr> <tr align="left" valign="top"> <td align="left" valign="middle">Time = <em>t</em></td> <td align="left" valign="middle">Time = 12 - <em>t</em></td> </tr> </tbody> </table> <p>Again, using distance = speed x time, gives the following equations:</p> <p><strong>(1)</strong> d = 3 <em>vt</em></p> <p><strong>(2)</strong> d = <em>v</em> (12- <em>t</em> )</p> <p>combining <strong>(1)</strong> and <strong>(2)</strong> gives</p> <p>3 <em>vt</em> = <em>v</em> (12 - <em>t</em> )</p> <p>which leads us to <em>t</em> = 3 minutes</p> <p>Using this information tells us that the speed of the river is d/9</p> <p>and since Brian swims at twice this speed, his speed in the still water is 2d/9.</p> <p>He swims a distance of 2 <em>d</em> at this speed.</p> <p>So it will take him 9 minutes.</p> <p><span class="editorial">Correct solutions were sent from</span> <strong class="editorial">Nicholas</strong> <span class="editorial">and</span> <strong class="editorial">Adrian</strong> <span class="editorial">(South Greenhoe Middle School).</span></p> <br></br></mdoxml> 2 5 0 0 0 1 0 Brian swims at twice the speed that a river is flowing, downstream from one moored boat to another and back again, taking 12 minutes altogether. How long would it have taken him in still water? Measures and Mensuration Speed Admin Short problems Algebra Manipulating algebraic expressions/formulae