Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 240: 86

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

The anti-derivative for $\dfrac{d^2s}{dt^2}=\dfrac{3t}{8}$ is $\dfrac{ds}{dt}=\dfrac{3}{16}t^2+c$; $s=\dfrac{1}{16}t^3+ct+c'$ Now, apply Initial conditions $s'(4)=3 ; s(4)=4$, we have This implies $3+c=3$ or, $ c=0 $ and $4+0+c'=4$ and $c'=0$ Hence, $s=\dfrac{t^3}{16}$