# Optimization of location to find where 2 particles are closest.

A particle moves west at a speed of $25km/hr$ from the origin when $t=0$. Another particle moves north at a speed $20km/hr$ and stops at the origin when $t=1$, where $t$ is in hours. What will the minimum distance be between the $2$ particles and at what time $t$ does such minimum occur.

I have studied these questions for a while but i need help deriving an equation relating the two particle's movements to minimize them. Thanks for any help.

Note: The particles are co-planar.

• Write down parametric equations for each particle's position as a function of $t$. Subtract them. Use the usual Pythagoreal formula to get their distance as a function of $t$. Note that it will be somewhat easier to minimize the square of their distance than the distance itself. Nov 12 '15 at 10:47
• so far i have gotten this: $x_1(t)=-25t$ and $x_2(t)=20t-20$ Nov 12 '15 at 10:50

The particles can be viewed as one moving along the negative x-axis away from the origin, the other moving towards the origin along the negative x-axis. The equations for position can be given as $r_1=(-25t,0)$ and $r_2=(0, 20t-20)$
This distance between them is the hypoteneuse of the right angles triangle formed by the x coordinates of the first particle and the y coordinate of the second. This means the distance between them is $$\left( 25^2 t^2 + (20t-20)^2 \right)$$ If you find the minimum value of this for $0 \leq t \leq 1$, you have your answer.
• Ok i have gotten this so far, $d^2 = 25t^2+(20t-20)^2$, which expands to the quadratic $425t^2-800t+400$ Nov 12 '15 at 10:53