# Closest approach of 2 objects

Rct33

## Homework Statement

The problem describes an aircraft taking off from a point on a runway with constant speed $V_{1}$, climbing at a constant angle $\alpha$, at the point of takeoff, a car drives towards the aircraft a distance $a$ away with speed $V_{2}$. I simply have to find the closest distance between the two objects

## The Attempt at a Solution

The x component of the distance is given by $(a-(V_{1}\cos(\alpha)+V_{2}))t$
The y component of the distance is given by $V_{1}\sin(\alpha)$

Therefore the distance is given by $\sqrt{((a-(V_{1}\cos(\alpha)+V_{2}))t)^2+(V_{2}\sin(\alpha))^2}$

Which I need to minimise, expanding the brackets and simplifying as much as I can gives the distance as:
$\sqrt{a^2-2aV_{2}t-2aV_{1}t\cos(\alpha)+V_{1}^2t^2+V_{2}^2t^2+2V_{1}V_{2}t^2\cos(\alpha)}$

Kinda have no idea what to do next or if I even went in the right direction so any pointers would be great, thanks

Homework Helper
Looks like you have the right idea, but you may want to check your brackets (e.g. are you sure that you meant to have ##a \cdot t - \cdots## rather than just ##a - \cdots## in the x-component)?

So you have an expression for the distance d(t) as a function of time. Now, in general, how do you minimise a function?

To make life easier for yourself, note that d(t)² has a minimum wherever d(t) has a minimum - i.e. you can forget about the square root and minimise ##v_x^2 + v_y^2## instead.

(PS I would call the vertical component z, not y, but that's probably a matter of personal preference).

1 person
Homework Helper
Gold Member
There are a few typos along the way, but your final expression is right so I presume they're transcription errors. You need to find the smallest value of that expression as time varies. What does that suggest in calculus terms?
Of course, the smallest value of the distance occurs at the same time as the smallest value of the square of the distance, so that should simplify things a little.

1 person
Mentor
There are a few typos along the way, but your final expression is right so I presume they're transcription errors. You need to find the smallest value of that expression as time varies. What does that suggest in calculus terms?
Of course, the smallest value of the distance occurs at the same time as the smallest value of the square of the distance, so that should simplify things a little.

As follow up to haruspex, the thing under the square root sign is a quadratic. The minimum or maximum of a quadratic occurs at t = -B/2A.

1 person
Rct33
As follow up to haruspex, the thing under the square root sign is a quadratic. The minimum or maximum of a quadratic occurs at t = -B/2A.

So $t=\frac{aV_{2}+aV_{1}\cos(\alpha)}{V_{2}^2+2V_{2}V{1}\cos(\alpha)+V_{1}^2}$?

Would I have to plug that into $\sqrt{a^2-2aV_{2}t-2aV_{1}t\cos(\alpha)+V_{1}^2t^2+V_{2}^2t^2+2V_{1}V_{2}t^2\cos(\alpha)}$ to get the distance? Seems awfully complicated

Homework Helper
Gold Member
So $t=\frac{aV_{2}+aV_{1}\cos(\alpha)}{V_{2}^2+2V_{2}V{1}\cos(\alpha)+V_{1}^2}$?

Would I have to plug that into $\sqrt{a^2-2aV_{2}t-2aV_{1}t\cos(\alpha)+V_{1}^2t^2+V_{2}^2t^2+2V_{1}V_{2}t^2\cos(\alpha)}$ to get the distance? Seems awfully complicated
It might help to solve it in the abstract first. Suppose d2= Ax2+Bx+C. It's minimised at x=-B/2A, giving d2=C-B2/4A.

1 person
Rct33
It might help to solve it in the abstract first. Suppose d2= Ax2+Bx+C. It's minimised at x=-B/2A, giving d2=C-B2/4A.

Well it still ain't pretty, but I have d, thanks alot for your help

Mentor
Well it still ain't pretty, but I have d, thanks alot for your help
It should be prettier than you are implying. Are you sure you manipulated it into the most favorable final form. I got the following result:
$$d=\frac{aV_1\sin \alpha}{\sqrt{A}}$$

Rct33
It should be prettier than you are implying. Are you sure you manipulated it into the most favorable final form. I got the following result:
$$d=\frac{aV_1\sin \alpha}{\sqrt{A}}$$

Do you get that result from simplifying
$$a^2+\frac{(2aV_{2}+2aV_{1}\cos(\alpha))^2}{4(V_{2}^2+2V_{2}V{1}\cos( \alpha)+V_{1}^2)}$$?

Mentor
Do you get that result from simplifying
$$a^2+\frac{(2aV_{2}+2aV_{1}\cos(\alpha))^2}{4(V_{2}^2+2V_{2}V{1}\cos( \alpha)+V_{1}^2)}$$?
No, from simplifying
$$a^2-\frac{(2aV_{2}+2aV_{1}\cos(\alpha))^2}{4(V_{2}^2+2V_{2}V{1}\cos( \alpha)+V_{1}^2)}$$

Rct33
No, from simplifying
$$a^2-\frac{(2aV_{2}+2aV_{1}\cos(\alpha))^2}{4(V_{2}^2+2V_{2}V{1}\cos( \alpha)+V_{1}^2)}$$

is b not equal to $$-2aV_{2}-2aV_{1}\cos(\alpha)$$? Thats why I canceled the negative in d2=C-B2/4A

Mentor
is b not equal to $$-2aV_{2}-2aV_{1}\cos(\alpha)$$? Thats why I canceled the negative in d2=C-B2/4A