1969 AHSME Problems/Problem 27
Problem
A particle moves so that its speed for the second and subsequent miles varies inversely as the integral number of miles already traveled. For each subsequent mile the speed is constant. If the second mile is traversed in hours, then the time, in hours, needed to traverse the th mile is:
Solution
Let be the distance already traveled and be the speed for the mile. Because the speed for the second and subsequent miles varies inversely as the integral number of miles already traveled, If the second mile is traversed in hours, then the speed in the second mile () equals miles per hour. Since one mile has been traveled already, . Now solve for the speed the mile, noting that miles have already been traveled. Thus, the time it takes to travel the mile is hours.
See Also
1969 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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