Answer
$r=4t^{(5/2)}+4t^{(3/2)}-8t$
Work Step by Step
As we know that $\int x^{a} dx=\dfrac{x^{a+1}}{a+1}+C$
Now, we have $\dfrac{d^2 r}{dt^2}=15 \sqrt t+\dfrac{3}{\sqrt t}$
$\implies r'=15 (\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1})+3 (\dfrac{t^{\frac{-1}{2}+1}}{-1/2+1})+C=10t^{(3/2)}+6^{(1/2)}+C$
Here, $r'(1)=8$ and $C=-8$
Thus, we get $r'=10t^{(3/2)}+6^{(1/2)}-8$
Now, $r=(10)[\frac{t^{3/2+1}}{3/2+1}]+(6)[\frac{t^{1/2+1}}{1/2+1}]-8t+C$
and $r=4t^{5/2}+4t^{3/2}-8t+C$
and $r(1)=0$ and $C=0$
Hence, $r=4t^{(5/2)}+4t^{(3/2)}-8t$
