# Time when two particles will meet [closed]

Particle P moves with constant speed of 13m/s always aiming at O . As P starts moving , O moves with constant velocity of 5m/s towards positive y-axis. Time taken by particle P to catch O will be? I tried solving this by realtive motion concept but i am not able to find the path it will follow. Please help me out with this problem....

• Related – Farcher Mar 26 '19 at 6:43
• Hi Tan and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. – John Rennie Mar 26 '19 at 6:56
• See also Wikipedia, Wolfram, and Math.SE. – Qmechanic Mar 26 '19 at 6:56
• @JohnRennie i just want to know the idea how to approach this type of problems – Tan_R Mar 26 '19 at 7:25

The solution to this type of problem is called a "pursuit curve" and is generally found by constructing and solving a differential equation for the co-ordinates of the pursuer as a function of time. In this particular case the pursuit curve is called a radiodrome.

If the co-ordinates of $$O$$ at time $$t$$ are $$(x_O(t), y_O(t))$$ (we assume that the functions $$x_O(t)$$ and $$y_O(t)$$ are known) and the co-ordinates of $$P$$ at time $$t$$ are $$(x_P(t), y_P(t))$$ then the condition that $$P$$ always aims at $$O$$ is

$$\frac{dy_P}{dx_P} = \frac{y_O-y_P}{x_O-x_P}$$

and the condition that $$P$$ travels at constant speed $$s_P$$ is

$$\left(\frac{dy_P}{dt}\right)^2 + \left(\frac{dx_P}{dt}\right)^2 = (s_P)^2$$

A useful text about pursuit curves in general is Paul Nahin's Chases and Escapes.

Note that the calculation of the pursuit curve is more complex that the related problem where $$P$$ anticipates that $$O$$ is travelling at a constant velocity and calculates an interception course, which will be a straight line. On the interception course $$P$$ is not aiming at $$O$$ itself but at the point where $$O$$ will be at a future time.