# Problem of the Day

Carl and Alvin stand at points and , respectively, as represented in the figure above (not drawn to scale), with distances labeled in feet. Carl starts running toward point at a rate of 400 ft/min, and Alvin starts running toward point at a rate of 500 ft/min. When Alvin reaches point , he turns around and runs in the opposite direction until he reaches point , whereupon he reverses his direction again and runs toward point . If Alvin repeats this pattern indefinitely, how far will the two runners be from point when they meet? Assume that Alvin is able to reverse his direction instantaneously.

A. feet
B. feet
C. feet
D. feet
E. feet

[latexpage]

D. $47\frac{7}{9}$ feet

See the Solution

### Solution

[latexpage]

The only things you need to solve this problem are a familiarity with the formula $d=rt$ (distance=rate*time) and a tolerance for tedious long division. However, the solution is somewhat long, so it is best to break the problem up into parts.

First, let’s determine how long it takes Carl to reach point $A$. Using $d=rt$, we can set up the equation $2200=400t$. Solving for $t$ tells us that it takes him $t=5.5$ minutes to reach point $A$. In that time, Alvin has been running back and forth between points $A$ and $B$. We need to determine exactly where he is between points $A$ and $B$. Running at 500 ft/min for 5.5 minutes, Alvin would be able to cover $d=500\cdot5.5$ feet or 2,750 feet. 2,750 divided by 70 is $39\frac{2}{7}$. This means that he has run the 70 feet between $A$ and $B$ 39 times, and he is $\frac{2}{7}$ of the way to doing it again. By inspection, we can determine that every time he has just finished running the 70 feet an odd number of times, he is at point $B$. 39 is odd, so he must have just reached point $B$ and is on his way back to point $A$. $\frac{2}{7}$ of 70 is 20, so he is 20 feet from point $B$ when Carl is at point $A$.

Now, Carl and Alvin are 50 feet apart and running in opposite directions. We need to find out how long it will take them to cover the 50 feet between them. In $t$ minutes, Carl will run $400t$ feet and Alvin will run $500t$ feet. The total distance they run will be $400t+500t=900t$ feet. Solving the equation $900t=50$ for $t$ tells us that they will meet in $t=\frac{1}{18}$ minutes. In that time, Alvin will have run $d=500\cdot\frac{1}{18}$ or $d=27\frac{7}{9}$ feet. Adding this to the 20 feet that he was from point $B$ initially, tells us that they meet when they are $47\frac{7}{9}$ feet from point $B$.