# Finding the distance between two moving objects

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?

• "Middle term"? What does that mean? What is the domain of $\;t\;$ ? Perhaps you mean the vaule of the distance when $\;t\;$ is at the middle of its domain? Commented May 15, 2016 at 7:48
• I have corrected a typo in your expression for the positive of $\ a \$ and the omission of the $\ t \$ in the last term of your "distance-squared" function. The second and third terms are the result of applying the distance formula: the squared-distance is a quadratic function. (The third term is not the "distance traveled for each object".) All three terms would appear even if the two objects were heading straight for one another. Commented May 15, 2016 at 9:11
• The middle term consists of velocity by time, while the last term is acceleration by time. Commented May 15, 2016 at 9:25
• @N.S.JOHN: Acceleration by "time squared". Commented May 15, 2016 at 10:32
• @Alexm. I missed that. Thanks Commented May 15, 2016 at 10:32

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:

$d^2 = 181000 - 77000t + 8500t^2$