# Closest approach of 2 objects

1. Oct 10, 2013

### Rct33

1. The problem statement, all variables and given/known data
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

3. 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

2. Oct 10, 2013

### CompuChip

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).

3. Oct 10, 2013

### haruspex

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.

4. Oct 10, 2013

### Staff: Mentor

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.

5. Oct 10, 2013

### Rct33

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

6. Oct 10, 2013

### haruspex

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.

7. Oct 10, 2013

### Rct33

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

8. Oct 10, 2013

### Staff: Mentor

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:
$$d=\frac{aV_1\sin \alpha}{\sqrt{A}}$$

9. Oct 10, 2013

### Rct33

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

10. Oct 10, 2013

### Staff: Mentor

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

11. Oct 10, 2013

### Rct33

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

12. Oct 10, 2013