1955 AHSME Problems/Problem 50

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:

$\textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}\ \text{10 mph}\qquad\textbf{(C)}\ \text{5 mph}\qquad\textbf{(D)}\ \text{15 mph}\qquad\textbf{(E)}\ \text{3 mph}$


Let $V_A, V_B, V_C$ be $A, B, C$'s velocity, respectively. We want to pass $B$ before we collide with $C$. Since $A$ and $B$ are going in the same direction and $V_A>V_B$, $A$ will pass $B$ in $\frac{30\mathrm{ft}}{V_A-V_B}$ time. Since $A$ and $C$ are going in opposite directions, their relative velocity is $V_A+V_C$, so the amount of time before $A$ will collide with $C$ is given by $\frac{210\mathrm{ft}}{V_A+V_C}$. We want to pass $B$ before we collide with $C$, so $V_A$ must satisfy the inequality $\frac{30\mathrm{ft}}{V_A-V_B}<\frac{210\mathrm{ft}}{V_A+V_C}$. We can eliminate all the units, simplifying the inequality to $\frac{1}{V_A-V_B} < \frac{7}{V_A+V_C}$. Solving this and substituting our known values of $V_B$ and $V_C$ yields $330<6V_A$, so $A$ must increase his speed by $\boxed{\textbf{(C) \ } 5 }$ miles per hour.