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EXAMPLE 6 A particle moves in a straight line and has acceleration given by a(t) = 12t+4. Its initial velocity is v(0) = -4 cm/s and its initial displacement is s(0) = 8 cm. Find its position function, s(t). SOLUTION Since v'(t) = a(t) = 12t+4, antidifferentiation gives v(t) = Note that v(0) = C. But we are given that v(0) = -4, so C = v(t) = Since v(t) = s'(t), s is the antiderivative of v: |) + 4( [ s(t) = = + 4t + C = ( s(t) = + D. This gives s(0) = D. We are given that s(0) = 8, so D = is + C. - 4t+ D and and the required position function

EXAMPLE 6 A particle moves in a straight line and has acceleration given by $a(t)=12t+4$. Its initial velocity is $v(0)=?4cm/s$ and its initial displacement is $s(0)=8cm$. Find its position function, $s(t)$. SOLUTION Since $v_{?}(t)=a(t)=12t+4$, antidifferentiation gives $v(t)=+4t+c=+C.$ Note that $v(0)=C$. But we are given that $v(0)=?4$, so $C=$ and $v(t)=$ Since $v(t)=s_{?}(t)$, $s$ is the antiderivative of $v$ : $s(t)?=6()+4()?4t+D=+D.?$ This gives $s(0)=D$. We are given that $s(0)=8$, so $D=$ and the required position function is $s(t)=$

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