Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 241: 103

See explanations.

Step 1. Given $\frac{d^2s}{dt^2}=a$, we can get $\frac{ds}{dt}=\int(a) dt=at+C$ and $s=\int(at+C)dt=\frac{1}{2}at^2+Ct+D$, where $C, D$ are constants. Step 2. At $t=0, \frac{ds}{dt}=v_0$, we have $a(0)+C=v_0$ and $C=v_0$ Step 3. At $t=0, s=s_0$, we have $\frac{1}{2}a(0)^2+v_0(0)+D=s_0$ and $D=s_0$ Step 4. Thus $s=\frac{1}{2}at^2+v_0t+s_0$