Problem 2 (The Ripple Problem)

John is paddling across a lake in his canoe directly ahead of him a fish jumps out of the water creating a circular ripple which moves outward at a constant rate. Exactly 10 seconds after the fish jumps John paddles across the ripple coming towards him and after 20 seconds he crosses the ripple for the second time. How long after the fish jumps does John cross the location where the fish jumped ?

Solution

We show below the path John follows in the canoe going from left to right and the circular ripple at the position where John crosses it for the first time, and the position when he crosses it for the second time. The first time he crosses the ripple he is going against the ripple, and the second time he is going in the same direction of the ripple.

To solve this problem, we use the well-known equation that distance is velocity times time. We start by calling d the distance the fish is from John when the fish jumps, T the unknown time when John reaches the location where the fish jumps, and r and j the velocities of the (circular) ripple in the water and John's velocity in the canoe, respectively. We first draw graphs showing the position of John's canoe and the location of the ripples, one ripple moving towards John and the other the same direction as John.

Now, since John reaches the first ripple in 10 seconds, we set the location of the incoming ripple equal to John's position, or

d - 10 r = 10 j

But we also know that the radius of the ripple when John reaches the ripple for the first time to be 10 r and that the radius of the ripple when John reaches the ripple the second time to be 20 r, and so between 10 seconds and 20 seconds John covers the distance between the two ripples, he travels a distance of 10 r + 20 r = 30 r feet. But we also know that John travels with velocity j, and so between 10 and 20 seconds we know he travels 10j feet. Hence, we have 30 r = 10 j or j = 3 r, which means that John travels three times faster than the ripple. We now substitute this value into the previous equation, getting

d - (30/3) j = 10 j

or

d = (40/3)j

But we know that John crosses the place where the fish jumps when . Hence, we have

T = d/j = 40/3 = 13 1/3 seconds.

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Last modified on Friday, February 12, 1999