document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1101227>Question 1101227</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A vehicle moves at an initial speed of 20m/sec with a constant acceleration of 2m/sec square, for 5 sec before brakes are applied. if tha car comes to rest under constant deceleration in 4 sec, determine the total distance travelled during the 9 seconds? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #715805 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=715805\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Using the formula v2 = v1 + at, the speed after the 5s acceleration is:<BR>\n" );
document.write( "v2 = 20 + 2*5 = 30 m/s<BR>\n" );
document.write( "Using the formula d = v1*t + 1/2a*t^2, the distance covered during acceleration is:<BR>\n" );
document.write( "d1 = 20*5 + 1*5^2 = 125 m<BR>\n" );
document.write( "The rate of deceleration is<BR>\n" );
document.write( "a = (v2-v1)/t = (0-30)/4 = -7.5 m/s^2<BR>\n" );
document.write( "The distance during 4s deceleration is:<BR>\n" );
document.write( "d2 = 30*4 - (7.5/2)*4^2 =  60 m<BR>\n" );
document.write( "Thus the total distance traveled is 125 + 60 = 185 m </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );