Answer
$s=\dfrac{t^3}{16}$
Work Step by Step
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, $s=\dfrac{t^3}{16}$
