In order to pass $ B$ going $ 40$ mph on a two-lane highway, $ A$, going $ 50$ mph, must gain $ 30$ feet.  Meantime, $ C$, $ 210$ feet from $ A$, is headed toward him at $ 50$ mph.  If $ B$ and $ C$ maintain their speeds, then, in order to pass safely, $ A$ must increase his speed by:

$ \textbf{(A)}\ \text{30 mph} \qquad
\textbf{(B)}\ \text{10 mph} \qquad
\textbf{(C)}\ \text{5 mph} \qquad
\textbf{(D)}\ \text{15 mph} \qquad
\textbf{(E)}\ \text{3 mph}$
1. Let \( a \) be the new speed of \( A \). Assume that \( A \) can make an instantaneous perpendicular lane change. Let \( x \) be the time \( A \) has to make the change.

2. The distance \( A \) needs to gain relative to \( B \) is 30 feet. The relative speed of \( A \) with respect to \( B \) is \( a - 40 \) mph. Therefore, the time \( x \) it takes for \( A \) to gain 30 feet is given by:
   \[
   (a - 40)x = 30 \quad \text{(1)}
   \]

3. Meanwhile, \( C \) is 210 feet away from \( A \) and is moving towards \( A \) at 50 mph. The relative speed of \( C \) with respect to \( A \) is \( a + 50 \) mph. Therefore, the time \( x \) it takes for \( C \) to cover 210 feet is given by:
   \[
   (a + 50)x = 210 \quad \text{(2)}
   \]

4. We now have two equations:
   \[
   (a - 40)x = 30 \quad \text{(1)}
   \]
   \[
   (a + 50)x = 210 \quad \text{(2)}
   \]

5. Solving equation (1) for \( x \):
   \[
   x = \frac{30}{a - 40}
   \]

6. Substituting this value of \( x \) into equation (2):
   \[
   (a + 50) \left( \frac{30}{a - 40} \right) = 210
   \]

7. Simplifying the equation:
   \[
   \frac{30(a + 50)}{a - 40} = 210
   \]

8. Multiplying both sides by \( a - 40 \):
   \[
   30(a + 50) = 210(a - 40)
   \]

9. Expanding and simplifying:
   \[
   30a + 1500 = 210a - 8400
   \]

10. Bringing all terms involving \( a \) to one side:
    \[
    1500 + 8400 = 210a - 30a
    \]
    \[
    9900 = 180a
    \]

11. Solving for \( a \):
    \[
    a = \frac{9900}{180} = 55
    \]

12. The original speed of \( A \) was 50 mph. Therefore, the increase in speed is:
    \[
    55 - 50 = 5 \text{ mph}
    \]

The final answer is \(\boxed{5 \text{ mph}}\).