In this case there is a missile whose initial position is $A(30,40)$ with a velocity of $[50,30]$ and an asteroid whose initial position is $B(400,250)$ with a velocity of $[-20,-30]$. The position of each object is given by the equations $ \ a = [30,40] + t[50, 30] \ $ and $ \ b = [400,250] + t[-20,-30] \ $.
I need to explain why distance between these two at any time t is given by the equation $$d^2 \ = \ 181000 \ - \ 77000 \ t \ + \ 8500\ t^2 \ \ . $$ I know that the first term $18100$ is the distance between the two initial positions and that the final term is the distance traveled for each object but, what is the middle term?
as stated by @N.S.JOHN the terms specify the initial position, velocity with which they come together (since there's a negative sign) and at last the acceleration at which them come apart (since it's a positive sign).
To find those coeffiecients (181000, -77000, 8500) you'll need to do some vector algebra.
Let's assume that $a$ it's a vector-valued function of time that represents the movement of the missile so $\vec{a}(t) = [30, 40] + t[50,30]$. Let's represent the meteor movement $b$ in the same manner $\vec{b}(t) = [400, 250] + t[-20, -30]$. Since they are vectors in the $R^2$ vector space we can decompose each vector in their constituents components:
$\vec{a}(t) = a_x + a_y$ in which $a_x(t) = 30 + 50t$ and $a_y(t) = 40 + 30t$ corresponds to the horizontal and vertical movements of the missil respectively.
The same goes for the meteor:
$\vec{b}(t) = b_x + b_y$ in which $b_x(t) = 400 + 250t$ and $b_y(t) = -20 -30t$ corresponds to the horizontal and vertical movements of the meteor respectively.
Now we have all the pieces we need to measure the distance between both as a function of time:
$d(\vec{a}(t), \vec{b}(t)) = \sqrt{(a_x - b_x)^2 + (a_y - b_y)^2}$
$d^2(\vec{a}(t), \vec{b}(t)) = (a_x - b_x)^2 + (a_y - b_y)^2$
$d^2(\vec{a}(t), \vec{b}(t)) = (30 + 50t - (400 - 20t))^2 + (40 + 30t - (250 - 30t))^2$
$d^2(\vec{a}(t), \vec{b}(t)) = (370 - 70t)^2 + (210 - 60t)^2$
$d^2(\vec{a}(t), \vec{b}(t)) = 370^2 + 210^2 - 2 \times 370 \times 70t - 2 \times 210\times 60t + 70^2t^2 + 60^2t^2$
and finally:
$d^2(\vec{a}(t), \vec{b}(t)) = 181000 - 77000t + 8500t^2$
this you can simply (with some abuse of notation) to:
$d^2 = 181000 - 77000t + 8500t^2$