document.write( "Question 1147472: The position of a particle moving along the 𝑥-axis at any time 𝑡 is given by
\n" ); document.write( "𝑠(𝑡)=𝑡^3−11𝑡^2+24𝑡−17.
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Algebra.Com's Answer #768846 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "(a) The velocity is the first derivative of the position function over time\r\n" );
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document.write( "        v(t) =  = 3t^2 - 22t + 24.     (1)\r\n" );
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document.write( "     To find v(5),  substitute t= 5 into the formula (1)\r\n" );
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document.write( "    The acceleration is the second derivative of the position function over time\r\n" );
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document.write( "        a(t) =  = 6t - 22.       (2)\r\n" );
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document.write( "     To find a(5),  substitute t= 5 into the formula (2).\r\n" );
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document.write( "(b)  At t= 5,  v(5) = -11,   a(5) = 8.\r\n" );
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document.write( "     So, the particle moves in negative direction and has positive acceleration.\r\n" );
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document.write( "     In the normal human language, it means that the particle DECELERATES (slowing down, moving in negative direction).\r\n" );
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