How Near Can They Be?

If the initial position of Red is considered as the origin O, then the distance between Red and Black after any time t can be expressed as $\sqrt{(5t - 10)^{2} + (10t)^{2}}$ , according to Pythagoras' theorem. The above expression can be reordered as $\sqrt{(\sqrt{125}t - \frac {50}{\sqrt{125}})^{2} + 80}$ . Then, for this distance to be minimum, $(\sqrt{125}t - \frac {50}{\sqrt{125}})^{2}$ has to be 0, which yields an acceptable t = 0.4. So the minimum distance between the two boys is $\sqrt{80}$ , or $4\sqrt{5}$ m.

Let Red be at the origin initially, and let Black be at the point (0,-10). From the point of view of Black, Red is moving along a straight line governed by the equation x+2y=0. Now using formula for perpendicular distance of a point from a line, we get d=4*sqrt(5)

d(t) = (10-5t)/(cos(tan^(-1)(10t/10-5t)))

Solved for minimum using graphing calculator because I couldn't take the derivative of the function, which is how I would attempt to solve for a minimum value.