Minimum time required to reach a point A police inspecter P is at point $ (\acute{x},\acute{y})$ and a thief X is at point $(x, y)$. X has a constant velocity $V_x\hat{i} + V_y\hat{j} $, where $\hat{i}$ and $\hat{j}$ are unit vectors in $X$ and $Y$ direction respectively. Maximum speed of P is S. 
What is the minimum time in which P will catch X? (assuming thatthe acceleration and deceleration of P are instantaneous)
 A: Let me make a few variable substitutions to make life easier:
Initial coordinates of the thief: $(x_0, y_0)$, initial coordinates of the officer: $(0,0)$. Assume the thief to have velocity $(t_x,t_y)$. The officer has speed $(p_x,p_y)$ with maximum speed $p_m^2 \geq p_x^2 + p_y^2$.
Since the thief has a constant velocity (i.e. does not run away from the officer), it appears logical that the best way for the officer to catch the thief is to run in a straight towards a particular point along the thief’s trajectory. Hence we have $p_x = \textrm{ const } = p_y$. This has the nice side effect that we don’t have to differentiate anything to find the minimum time $t$ at which the two meet, since they will only meet once. One can then set up equations as following:
$$ p_x t = x_0 + t_x t \qquad (1) $$
$$ p_y t = y_0 + t_y t \qquad (2) $$
$$ p_m^2 = p_x^2 + p_y^2 \qquad (3) $$
Solving $(3)$ for $p_x$, substituting in $(1)$, solving for $p_y$ and then substituting in $(2)$ yields a single equation with only one unknown, $t$. Solving and substituting back into $(1)$ and $(2)$ gives $p_{x,y}$. 
