Question 13253
 [quote]
 Consider this fortunate incident. A girl was crossing a railroad bridge when, halfway across, she saw a train 50 meters away moving toward her. She immediately turned and ran so that the train missed her by the narrowest of margins. If she had tried to cross the bridge, the train would have struck her one meter before she could have reached the end. How long is the bridge? [/quote]
 Refer tp this diagram:

Train  50 m   B  M   A   (AB = L meters, M at the middle of AB)
*------------*……*.……*  
 When she was at point M(the middle),her distance to the train was 
 50+BM = 50 + (L/2)
 Sol: 
When she ran back to A, time spending = {{{(50+ BM)/(V[1]+V[2])}}} =
    {{{(50 + L/2)/(V[1]+V[2])}}} = {{{ (L/2)/V[2] }}} = {{{(50+L)/V[1] }}} 
If she ran forward to B, time spending = {{{(50+L/2)/(V[1]+V[2])}}} = 
{{{((L/2) - 1)/ V[2]}}} = {{{(50+1)/ V[1]}}}
 Sol: Assume {{{V[1]}}} is the speed of the train and {{{V[2]}}} is the running speed of the girl. Let the length of the bridge be L meters. Note she began to run from the middle of the bridge. As the diagram above.
When she ran back to escape at one end (point A) of the bridge, we have the time
 {{{(L/2)/V2 = (50+L)/V1 }}}. Hence, {{{V[1]/V[2] = (100+2L)/L }}}…(1).
If she ran to the other end B and hit by the train, then we have the time          {{{ 51/V[1] }}}= {{{(L/2 -1) /V[2]}}}. 
 Hence {{{V[1]/V[2] = 51/(L/2-1) }}}…(2)
By (1),(2), we get {{{(100+2L)/L = 51/(L/2-1) }}}. 
By cross multiplication and simplification, we have {{{L^2 - 3 L - 100 }}}= 0 so
 {{{L = (3 + sqrt(409))/(2) }}}. 
Thus, we obtain L, the length of the bridge is {{{( 3+ sqrt (409))/2}}} meters.   

[Also, you can use it to find {{{V[1]}}} and {{{V[2]}}}] 

 Read carefully to understand all the details.

 Kenny