Question

Two particles, A and B, are in motion in the xy-plane. Their coordinates at each instant of time t(t is greater than or equal to 0) are given by the following (click to see). Find the minimum distance between A and B...

Answer 1

\[x _{A}=t, y_A=4t, x_B=13-t, y_B=t\]

Answer 2

Difference of x = |13-2t|
Difference of y = |3t|

Pythagorean Theorem
\[Distance = \sqrt{(13-2t)^2+(3t)^2}= \sqrt{13t^2-52t+169}=\sqrt{13(t-2)^2+117}\]

Since (t-2)^2 is always bigger or equal to 0, Distance is bigger or equal to \[\sqrt{117}\]
Therefore, minimum distance is \[\sqrt{117}=3\sqrt{13}\]