document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=30584>Question 30584</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A train, an hour after starting, meets an accident which detains it one hour, after which it proceed at 3/5 of its former rate and arrives 3 hours late.  But, had accident happened 80km farther, it would have been 1.5 hours late only.  Find the original rate of the train and the length of the journey. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #17291 by </B><A HREF=/tutors/aboutme.mpl?userid=Fermat><B>Fermat(136)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Fermat><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Fermat><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=17291\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let v1 be the speed before the breakdown<BR>\n" );
document.write( "Let v2 be the speed after the breakdown<BR>\n" );
document.write( "v2 = (3/5)v1<BR>\n" );
document.write( "============<BR>\n" );
document.write( "The two journeys are identical except for the 80km part<BR>\n" );
document.write( "let s1 be the distance up to the 80 km part<BR>\n" );
document.write( "The difference in time taken between travelling the 80 km at v1 and at v2 is equal to the difference in time of the two late journeys - which difference is 1.5 hrs.<BR>\n" );
document.write( "In other words, it takes 1.5 hrs longer to travel 80 km when trvelling at v2 than it does when travelling at v1.\r<BR>\n" );
document.write( "\n" );
document.write( "time to travel 80 km at v1 = 80/v1 = t1<BR>\n" );
document.write( "time to travel 80 km at v2 = 80/v2 = t2<BR>\n" );
document.write( "t2 - t1 = 1.5<BR>\n" );
document.write( "80/v2 - 80/v1 = 3/2<BR>\n" );
document.write( "80(1/v2 - 1/v1) = 3/2<BR>\n" );
document.write( "(80/v1)(5/3 - 1) = 3/2<BR>\n" );
document.write( "(80/v1)(2/3) = 3/2<BR>\n" );
document.write( "80/v1 = 9/4<BR>\n" );
document.write( "v1 = 320/9<BR>\n" );
document.write( "v1 = 35.5556 km/hr<BR>\n" );
document.write( "==================<BR>\n" );
document.write( "Let s2 be the distance travelled after the breakdown (in the first joiuurney), such that,<BR>\n" );
document.write( "s1 + s2 = L, where L is the overall journey distance travelled from the start to the final destination.<BR>\n" );
document.write( "Let T be the journey time without breakdowns.<BR>\n" );
document.write( "L = v1T<BR>\n" );
document.write( "=======<BR>\n" );
document.write( "In the first journey, excluding the 1hrs wait time, it takes 2hrs longer to travel the distance s2, at the speed v2, than at the speed v1 (normal/standard speed)<BR>\n" );
document.write( "t = s2/v1<BR>\n" );
document.write( "t' = s2/v2<BR>\n" );
document.write( "t' - t = 2<BR>\n" );
document.write( "s2(1/v2 - 1/v1) = 2<BR>\n" );
document.write( "(s2/v1)(5/3 - 1) = 2<BR>\n" );
document.write( "s2/v1 = 2*3/2 = 3<BR>\n" );
document.write( "s2 = 3v1<BR>\n" );
document.write( "s2 = 106.6667 km<BR>\n" );
document.write( "================<BR>\n" );
document.write( "The distance s1 is travelled in 1 hr (given) at speed v1<BR>\n" );
document.write( "s1 = 35.5556*1<BR>\n" );
document.write( "s1 = 35.5556 km<BR>\n" );
document.write( "===============<BR>\n" );
document.write( "L = s1 + s2<BR>\n" );
document.write( "L = 35.5556 + 106.6667<BR>\n" );
document.write( "L = 142.2222 km<BR>\n" );
document.write( "=============== </SPAN>\n" );
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