Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 273: 108

$s=\dfrac{t^3}{16}$

We need to find the anti-derivative for $\dfrac{d^2s}{dt^2}=\dfrac{3t}{8}$ We have: $\dfrac{ds}{dt}=\dfrac{3}{16}t^2+c$; $s=\dfrac{1}{16}t^3+ct+c'$ Apply the initial conditions $s'(4)=3 ; s(4)=4$ Thus, $3+c=3 \implies c=0 $ and $4+0+c'=4 \implies c'=0$ Hence, $s=\dfrac{t^3}{16}$