Two boats A and B are on an east-west shore a distance d apart. Both travel at the same speed.---------------

Question

(A ) travels due north, and at any given time B travels directly towards A. Eventually, B's course will approach due north. At that time, how far apart will the boats be?
I did this graphically and got about .4d. Can you show me how you solved this, and what kind of curve is B's course? Is it something easy, or some unobvious series?
I will appreciate any help you can give me.

Answer 1

LivingDownSouth's answer is not right.
He's tried to simplify it too much by simplifying the curve to a quarter of a circle.

If I have time later, I'll try to determine the actual curve ... and provide the answer for you.

Additional information entered later:
I don't have a complete solution to the problem, but the following reduces it to a differential equation.

For simplicity, assume that the initial distance between the boats is 1 unit, that A is at (0,0) and B is at (-1,0), and that both travel at a speed of 1 (= 1 distance unit per time unit).

Then A's position at any time t is (0,t).
B's velocity consists of an x (eastward) component and a y (northward) component. these two velocities can be expressed as derivatives in the following way:
dx/dt = -x/sqrt(x^2 + t^2)
[Remember that x is a negative number between -1 and 0, so -x in the numerator is a positive number, indicating movement in the positive (eastward) direction.]
dy/dt = t/sqrt(x^2 + t^2)

If you can solve these two equations (i.e., manipulate them and take the integrals in order to get rid of the derivative), then you will have expressions for x and y in terms of t. There will be a constant of integration, but you eliminate that by the initial condition (position at time 0 is (-1,0).

To find the eventual distance between the boats, take the expression t - y and determine what its value will be for a very large value of t. In this expression, t represents boat A's distance from the shore, and y (a function of t) represents boat B's distance from the shore. The difference is the distance separating the two boats.

Hope you can use this info.

Answer 2

You got me stuck on this math problem so I will give it a shot. Since A and B are going at the same speed I believe that Bs course would be 1/4 of a circle. Since the circumference of a circle (C) is C=2x3.14xd or 6.28d, then 1/4 of that would be 1.57d. Once B traveled that quarter of a circle it would be right behind A going North. Since B traveled 1.57d so did A; however, B has only gone the radius of the circle (d) North. Therefore the difference between the two is:

1.57d - 1d = .57d

That is my best shot. Good luck.