Answer
$16ft/s^2$
Work Step by Step
1. Since $\frac{d^2s}{dt^2}=-k$, we have $\frac{ds}{dt}=\int(-k)dt=-kt+C$. At $t=0, \frac{ds}{dt}=88$, we have $-k(0)+C=88$ and $C=88$. Also $s=\int(-kt+88)dt=-\frac{k}{2}t^2+88t+D$ and $t=0, s=0$, thus $-\frac{k}{2}(0)^2+88(0)+D=0$ and $D=0$.
2. When $\frac{ds}{dt}=0$, we have $-kt+88=0$ which gives $t=\frac{88}{k}$
3. At the above time, the car stops and $s=242ft$, thus $-\frac{k}{2}(\frac{88}{k})^2+88(\frac{88}{k})=242$ which leads to $484k=88^2$ and $k=16ft/s^2$ (deceleration).
