smc

Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives. Let $d$ be the distance still needed to travel after $1$ hour. We have that $\frac{d}{50}+1.5=\frac{d}{35}$, where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early. Simplifying gives $7d+525=10d$, or $d=175$. Now, we must add an extra $35$ miles traveled in the first hour, giving a total of $\boxed{210}$ miles.