David drives from his home to the airport to catch a flight.  He drives $35$ miles in the first hour, but realizes that he will be $1$ hour late if he continues at this speed.  He increases his speed by $15$ miles per hour for the rest of the way to the airport and arrives $30$ minutes early.  How many miles is the airport from his home?
$\textbf{(A) }140\qquad \textbf{(B) }175\qquad \textbf{(C) }210\qquad \textbf{(D) }245\qquad \textbf{(E) }280\qquad$
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 $\dfrac{d}{50}+1.5=\dfrac{d}{35}$, 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 $\boxed{\textbf{(C) } 210}$ miles.