Answer
$$645.43 \mathrm{mi}$$
Work Step by Step
\[
\begin{array}{l}
{\left[9-191,000 \int_{0}^{k} \frac{1}{(x+3963)^{2}} d x\right]^{1 / 2}=1.5} \\
{\left[9-191,000\left(-\frac{-1}{3963}+\frac{-1}{k+3963}\right)\right]^{1 / 2}=1.5}
\end{array}
\]
Evaluate the integral for $b=k, a=0,$ when the total result is equal to 1.5 (half the initial speed), using the antiderivative:
$\left[9-191,000 \cdot \frac{(k+3963)-3963}{(k+3963) 3963}\right]^{1 / 2}=1.5$
$9-191,000 \cdot \frac{k}{(k+3963) 3963}=2.25$
This simplifies as
$\frac{191,000 k}{3963 (k+3963) }=6.75$
\[
\begin{aligned}
&3963 (k+3963)6.75= 191,000 k \\
k & \approx 645.43 \mathrm{mi}
\end{aligned}
\]
