The cat mouse trajectory

A cat is initially at A and a mouse at B. The mouse starts moving perpendicular to AB with constant speed v.The cat moves towards the mouse with constant speed u, in order to catch it in such a way that it’s velocity vector always points towards the mouse. Find the trajectory traced by the cat.

#Mechanics

Note by Ayan Sharma
1 year, 6 months ago

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Comments

The trajectory can be obtained by solving the following set of coupled differential equations:

dxdt=u(vt−x)(vt−x)2+(L−y)2\frac{dx}{dt} = \frac{u(vt-x)}{\sqrt{(vt-x)^2 + (L-y)^2}} dydt=u(L−y)(vt−x)2+(L−y)2\frac{dy}{dt} = \frac{u(L-y)}{\sqrt{(vt-x)^2 + (L-y)^2}}

x(0)=y(0)=0x(0) = y(0) = 0

The way these equations are derived are as such:

  • At a general time tt the coordinates of the mouse are P1: (vt,L)P1: \ (vt,L) and that of the cat are P2: (x,y)P2: \ (x,y).
  • Say the velocity vector of the cat makes an angle θ\theta with the horizontal. Then the velocity vector is along the line joining the points P1P1 and P2P2.
  • So, this gives three equations:

tan⁡θ=L−yvt−x\tan{\theta} = \frac{L-y}{vt-x} dxdt=ucos⁡θ\frac{dx}{dt} = u\cos{\theta} dydt=usin⁡θ\frac{dy}{dt} = u\sin{\theta}

Eliminating θ\theta gives the equations listed above. This set of equations needs to be solved numerically. If you want more information like plots, etc., I will post if requested.

Karan Chatrath - 1 year, 6 months ago
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