I OM - B - Task 5


A motorboat set off upstream at 9:00, and at the same moment, a ball was thrown from the motorboat into the river. At 9:15, the motorboat turned around and started moving downstream. At what time did the motorboat catch up with the ball?
On standing water, the motorboat would have returned to the ball within the next 15 minutes, i.e., at 9:30. The same would happen on the river, as the current equally carries the motorboat and the ball. The motorboat will catch up with the ball at 9:30.

If someone did not come up with the above simple reasoning, they could solve the problem using equations.

At first glance, it seems that there is not enough data in the problem, as the boat's own speed, i.e., its speed on standing water, and the speed of the river current were not specified.

Not being discouraged by this, let's denote by $ x $ the sought number of hours from 9 o'clock to the moment when the motorboat caught up with the ball. In addition to this, let's introduce the speed of the current $ u $ expressed in kilometers per hour and the speed $ v $ of the motorboat on standing water, also expressed in kilometers per hour. To set up the equation, we need to express the distance traveled by the ball and the distance traveled by the motorboat using these quantities.

In $ x $ hours, the ball drifted with the river current $ xu $ kilometers.

The motorboat first traveled for $ \frac{1}{4} $ of an hour against the current, its speed was then $ (v - u) $ kilometers per hour, the distance traveled was thus $ \frac{1}{4}(v - u) $ kilometers; in the remaining time $ (x - \frac{1}{4}) $ hours, the motorboat traveled with the current, so at a speed of $ (u + v) $ kilometers per hour and traveled $ (x - \frac{1}{4})(u + v) $ kilometers, covering in the opposite direction first the distance $ \frac{1}{4}(v - u) $ kilometers to the starting point, and then the distance $ xu $ kilometers traveled by the ball to the point where the motorboat caught up with the ball.

Therefore






When we transform this equation by opening the parentheses and collecting the terms containing $ x $ on one side and the remaining terms on the other side, we get the equation






Since the motorboat's own speed $ v $ is certainly not zero, we can divide both sides of the equation by $ v $, obtaining






Thus, the motorboat will catch up with the ball after $ \frac{1}{2} $ of an hour from the moment of departure, i.e., at 9:30.
It turned out that the unknown quantities $ u $ and $ v $ were needed only to set up the equation. When performing the calculations, these quantities disappeared, i.e., as is said in mathematics, they were eliminated. Such auxiliary quantities are sometimes called parameters.