document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=77322>Question 77322</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if an object is dropped off of a tower, the velocity , v, of the object after t seconds can be obtained by multiplying t by 32 and adding 10 to the results. express v as a linear fucntion of t. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #55432 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=55432\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If the velocity can be obtained by multiplying t by 32 and we add 10 we have\r<BR>\n" );
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document.write( "v=32t+10\r<BR>\n" );
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document.write( "If you want the equation in function notation then:\r<BR>\n" );
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document.write( "v(t)=32t+10\r<BR>\n" );
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document.write( "So if t=0, we have an initial velocity of v=10. If v=1, then we have a velocity of v=32+10=42. If we have v=2, then v=32*2+10=64+10=74. In other words, for each second the velocity increases by 32 (notice how 32+42=74). So 32 is our rate of increase and 10 is our initial velocity.\r<BR>\n" );
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