document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1125555>Question 1125555</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A car goes 40 yards / sec.<BR>\n" );
document.write( "A bike goes 20 yards / sec.<BR>\n" );
document.write( "If they start at the same point, when do they meet?<BR>\n" );
document.write( "(I don't think they do, since the bike will never catch up to the car.)<BR>\n" );
document.write( "If the bike is given a 30 yard head start, at what time (seconds) will they meet?  And at what distance?  <BR>\n" );
document.write( "I also know speed = distance / time.<BR>\n" );
document.write( "Thank you!!! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #741866 by </B><A HREF=/tutors/aboutme.mpl?userid=math_helper><B>math_helper(2461)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_helper><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=math_helper><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=741866\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Part (i):  You are correct, if they are moving in the same direction and the bike and car start at the same point.<br>\r<BR>\n" );
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document.write( "Part (ii):<BR>\n" );
document.write( "speed = distance / time  —>  distance = speed*time   <BR>\n" );
document.write( "d(bike) = 30yds + 20yds/sec*(t)<BR>\n" );
document.write( "d(car) = 40yds/sec*(t)  \r<BR>\n" );
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document.write( "They meet when  d(car) = d(bike)  ==>  30+20t = 40t,  solving for t:  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28+t=1.5s+%29+ ALIGN=MIDDLE  ALT=\"highlight%28+t=1.5s+%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" > and the distance from the car's starting point is  1.5s*40yds/s = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+highlight%28+60yds+%29 ALIGN=MIDDLE  ALT=\"+highlight%28+60yds+%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "—————————————<BR>\n" );
document.write( " I chose the car's starting point as the reference point, the \"beginning.\"   Thus, the 1.5*40 in the 2nd part (\"where they meet\") is the distance from where the car started, so I used the car's speed. <br>\r<BR>\n" );
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document.write( "It would be just a valid (although a bit obscure?) to choose the bike's starting point, but remember the bike had a 30yd head start over the car.   *If* one chooses the bike's starting point, then they meet at 1.5*20 = 30yds from the BIKE's starting point (and yes, you use the bike's speed in this case), and that 30yds for the bike corresponds to a distance traveled of 30+30 = 60yds for the car. <BR>\n" );
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document.write( " </SPAN>\n" );
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