## Calculus: Early Transcendentals 8th Edition

(a) $v(t)=4t^{3}-6t^{2}+2t-1$ $a(t)=12t^{2}-12t+2$ (b) a(1) = 2 s (c) See graph.
$s(t)=t^{4}-2t^{3}+t^{2}-t$ (a) Velocity is the derivative of position, so to get the velocity function in terms of t, we just need to take the derivative of s(t): $s'(t) = v(t)=(4)t^{3}-2(3)t^{2}+2t-1$ $v(t)=4t^{3}-6t^{2}+2t-1.$ Acceleration is the derivative of velocity: $v'(t) =a(t)=4(3)t^{2}-6(2)t+2-0$ $a(t)=12t^{2}-12t+2.$ (b) To find the acceleration at t = 1, just plug it into a(t): $a(1)=12(1)^{2}-12(1)+2 = 2 s$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}