document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=123084>Question 123084</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Lois and Clark are covering separate stories and have to travel in opposite directions.  Lois leaves the Daily Planet at 8:00 a.m. and travels at 35 mph.  Clark leaves at 8:15 a.m. and travels at 40 mph.  At what time will they be 140 mi apart?\r<BR>\n" );
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document.write( "I'm not sure how to write the equation for this problem.  Thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #90351 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=90351\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The basic equation that you use in solving this problem is the Distance equation. It states<BR>\n" );
document.write( "that the Distance (D) that is traveled by a moving object is equal to the Rate of travel (R)<BR>\n" );
document.write( "of the object (R) multiplied by the time the object has been traveling at that rate (T).<BR>\n" );
document.write( "In equation form, this is:<BR>\n" );
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document.write( "D = R*T<BR>\n" );
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document.write( "Since Lois & Clark are traveling in opposite directions, the sum of the distances that each <BR>\n" );
document.write( "has traveled is equal to the distance separating them.<BR>\n" );
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document.write( "Because Lois departs at 8:00, she travels for 1/4 hour (15 minutes) before Clark leaves. At<BR>\n" );
document.write( "her rate of travel (35 mph) in that 1/4 hour she has gone a distance given by the equation:<BR>\n" );
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document.write( "D = R*T = 35*(1/4) = 35/4 = 8.75 miles<BR>\n" );
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document.write( "So at the instant Clark leaves, the distance between Lois and Clark is already 8.75 miles.<BR>\n" );
document.write( "From then on the distance between them is the distance that Lois goes (35*T) plus the distance<BR>\n" );
document.write( "that Clark goes in the same time T ... (40*T). Notice that T is the common time that the <BR>\n" );
document.write( "two are on the road together, and T starts at 8:15.<BR>\n" );
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document.write( "So we can write the expression for the distance between them as the sum of:<BR>\n" );
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document.write( "8.75 + 35*T + 40*T<BR>\n" );
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document.write( "and the problem tells us that this distance is to equal 140 miles. In equation form this is:<BR>\n" );
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document.write( "8.75 + 35*T + 40*T = 140<BR>\n" );
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document.write( "On the left side, add the two terms that contain the factor T, and this addition reduces the<BR>\n" );
document.write( "equation to:<BR>\n" );
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document.write( "8.75 + 75*T = 140<BR>\n" );
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document.write( "Get rid of the 8.75 on the left side by subtracting 8.75 from both sides and the result is:<BR>\n" );
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document.write( "75*T = 131.25<BR>\n" );
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document.write( "Solve for T by dividing both sides of this equation by 75 and the result is:<BR>\n" );
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document.write( "T = 131.25/75 = 1.75<BR>\n" );
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document.write( "This tells you that starting at 8:15, Lois and Clark will be 140 miles apart in an hour and<BR>\n" );
document.write( "three-quarters.  The three-quarters of an hour (0.75) is 45 minutes.<BR>\n" );
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document.write( "So starting at 8:15, one hour later is 9:15 and 45 minutes after that is 10:00. All times are<BR>\n" );
document.write( "A.M. (in the morning) so the answer to this problem is that Lois and Clark are 140 miles<BR>\n" );
document.write( "apart at 10:00 A.M.<BR>\n" );
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document.write( "Hope this helps you to understand the problem and to see a way that you can think it through<BR>\n" );
document.write( "to a solution.<BR>\n" );
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