document.write( "Question 1006847: Please help me solve this problem:\r
\n" ); document.write( "\n" ); document.write( "An object is moving in a straight line. The distance (in meters) from a fixed point is given by s= -2t^3 - 4t^2 + 6t + 4 where t≥0. Determine the acceleration when the velocity is -2m/s. \n" ); document.write( "
Algebra.Com's Answer #622982 by Alan3354(69443)\"\" \"About 
You can put this solution on YOUR website!
An object is moving in a straight line. The distance (in meters) from a fixed point is given by s= -2t^3 - 4t^2 + 6t + 4 where t≥0. Determine the acceleration when the velocity is -2m/s.
\n" ); document.write( "---------------
\n" ); document.write( "s= -2t^3 - 4t^2 + 6t + 4
\n" ); document.write( "Velocity is the 1st derivative.
\n" ); document.write( "s'(t) = -6t^2 - 8t + 6
\n" ); document.write( "---
\n" ); document.write( "Determine the acceleration when the velocity is -2m/s.
\n" ); document.write( "s'(t) = -6t^2 - 8t + 6 = -2
\n" ); document.write( "3t^2 + 4t - 4 = 0
\n" ); document.write( "(3t - 2)*(t + 2) = 0
\n" ); document.write( "t = 2/3 seconds (ignore t = -2)
\n" ); document.write( "------
\n" ); document.write( "Accel is the 2nd derivative
\n" ); document.write( "s\"(t) = -12t - 8
\n" ); document.write( "s\"(2/3) = -8 -8
\n" ); document.write( "Accel = -16 m/sec/sec\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );