Difference between revisions of "1955 AHSME Problems/Problem 50"
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+ | In order to pass <math>B</math> going <math>40</math> mph on a two-lane highway, <math>A</math>, going <math>50</math> mph, must gain <math>30</math> feet. | ||
+ | Meantime, <math>C, 210</math> feet from <math>A</math>, is headed toward him at <math>50</math> mph. If <math>B</math> and <math>C</math> maintain their speeds, | ||
+ | then, in order to pass safely, <math>A</math> must increase his speed by: | ||
+ | |||
+ | <math> \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} </math> | ||
+ | |||
== Solution== | == Solution== | ||
Revision as of 20:26, 31 December 2023
In order to pass going mph on a two-lane highway, , going mph, must gain feet. Meantime, feet from , is headed toward him at mph. If and maintain their speeds, then, in order to pass safely, must increase his speed by:
Solution
Let be 's velocity, respectively. We want to pass before we collide with . Since and are going in the same direction and , will pass in time. Since and are going in opposite directions, their velocity is given by , so the amount of time before will collide with is given by . We want to pass before we collide with , so must satisfy the inequality . We can eliminate all the units, simplifying the inequality to . Solving this and substituting our known values of and yields , so must increase his speed by miles per hour.
~anduran