Percent grade from stopping distance difference

Approach

The extra stopping distance down the grade comes only from the braking distance, because the lag distance is the same:

$$SSD_{down}-SSD_{up}=\frac{V^2}{254}\left[\frac{1}{f-n}-\frac{1}{f+n}\right]=12\ \text{m}$$

Here $V=65$ km/h, $f=0.38$ and $n$ = grade as a ratio. $\dfrac{V^2}{254}=\dfrac{65^2}{254}=16.634$.

Solving for the grade

$$16.634\left[\frac{2n}{f^2-n^2}\right]=12\ \Rightarrow\ \frac{2n}{0.1444-n^2}=0.7214$$ $$0.7214\,n^2+2n-0.10417=0\ \Rightarrow\ n=0.0511$$

Grade $\approx5.11\%$.

Stopping sight distance up the grade ($t=2.5$ s)

$$SSD_{up}=0.278Vt+\frac{V^2}{254(f+n)}=0.278(65)(2.5)+\frac{65^2}{254(0.4311)}=45.18+38.58=83.8\ \text{m}$$

Check: $SSD_{down}=45.18+\dfrac{65^2}{254(0.3289)}=95.8$ m; difference $=12.00$ m, which matches 12 m.

Answer: grade = 5.1%; stopping sight distance up the grade = 83.8 m (down the grade 95.8 m).