Answer
a. $\frac{ds}{dt}=10t^{3/2}-6t^{1/2}$
b. $s=4t^{5/2}-4t^{3/2}$
Work Step by Step
a. Step 1. Given $a=\frac{d^2s}{dt^2}=15\sqrt t-\frac{3}{\sqrt t}$, we can get $v=\frac{ds}{dt}=\int(15\sqrt t-\frac{3}{\sqrt t})dt=10t^{3/2}-6t^{1/2}+C$ and $s=\int(10t^{3/2}-6t^{1/2}+C)dt=4t^{5/2}-4t^{3/2}+Ct+D$, where $C$ and $D$ are constants.
Step 2. At $t=1, v=4$, we have $10(1)^{3/2}-6(1)^{1/2}+C=4$ and $C=0$. Thus $v=\frac{ds}{dt}=10t^{3/2}-6t^{1/2}$
b. At $t=1, s=0$, we have $4(1)^{5/2}-4(1)^{3/2}+D$ thus $D=0$ and $s=4t^{5/2}-4t^{3/2}$
