# Shortest Distance Problem

1. Mar 1, 2006

### GregA

Firstly I apologise if my problem here seems a bit daft but I have got nobody other than myself or you guys to tell me if I'm doing things correctly or not.

The question:
Two cyclists are riding one along each of two perpendicular roads that meet at A. At one instance both cyclists are 500m from A and both are approaching A. If the speed of one of the cyclists is 8m/s, and the shortest distance between them is 50m find the two possible speeds of the other cyclist.

My attempt to solve:
http://img295.imageshack.us/img295/6594/question53ju.jpg [Broken]
The first diagram on the left represents how far P is away from Q initially and the second shows the line QR (the velocity of Q relative to P.
In the second diagram I know that the line PR is perpendicular to QR for this length to be the shortest distance, also since I know two lengths of this triangle I find the angle RQP to be 4.055degs.
In a triangle of vectors for velocity (I was in a rush and didn't transfer this from my work book, sorry) the angle that the line representing Q's relative velocity makes with it's true velocity is 45 + or - 4.055degs
Using 8/tan theta, the answers I finally end up with being 9.22m/s or 6.94m/s

My problem however is that the books answers are given as 6.9km/h or 9.23km/h (I take the km/h bits to be typo's)...and though this looks reasonably close to my answers, firstly one answer needed only to be given to 2sf whilst the other 3sf...secondly I could take potshots at this question and still arrive at an answer that is pretty close to mine or the book's, and so I cannot be sure that my working is correct. Have I done something wrong or should I chill out and move on to my next question?...If I am wrong can somebody give me a clue?

Last edited by a moderator: May 2, 2017
2. Mar 1, 2006

### Staff: Mentor

I'm having trouble understanding the object of the problem. Give the two speeds of the second rider so that what happens?

3. Mar 1, 2006

### 0rthodontist

??? This problem is incomplete.

4. Mar 1, 2006

### GregA

Thanks Orthodontist & Berkeman, I was in a rush to get this question posted that I didn't quote the book fully...I've edited my first post:
Here's the full question:

Two cyclists are riding one along each of two perpendicular roads that meet at A. At one instance both cyclists are 500m from A and both are approaching A. If the speed of one of the cyclists is 8m/s, and the shortest distance between them is 50m find the two possible speeds of the other cyclist.

Last edited: Mar 1, 2006
5. Mar 1, 2006

### Staff: Mentor

Ah, that makes a lot more sense now. I wouldn't bother with the vector thing myself. You have the equation of motion of the one rider (speed and direction), so you can write the equation of the separation distance as a function of the y-directed rider's velocity. Then just solve for the minima of the equation, and you'll get the two cases where the 2nd rider passes the intersection before and after the 1st rider. I haven't done the math yet, but it seems pretty straightforward. The main catch is writing the separation equation carefully, with the distance being the diagonal...

6. Mar 2, 2006

### GregA

Thanks for the reply Berkeman...Unfortunately for me I don't think I'm ready to answer this question yet with that method ...I have done a bit of calculus but it wasn't very rigorous and did not involve vectors (Am starting from scratch again using the Thomas's Calculus book) and the mechanics book I'm studying from right now will start to make use of it (in a different context) about 100-150 pages away from where I am right now. ...damn!

What I have tried (and it's ugly..and probably has nothing to do with what you suggested):
If I call the point of intersection 'A', call time 't', and call 'q' the speed of Q...
then the length PA = 500 - 8t
the length QA = 500 - qt
the diagonal^2 (50^2) = PA^2 + QA^2
2500 - (500 -8t)^2 = (500 - qt)^2
-247500 + 8000t -64t^2 = 250000 - 1000qt +(qt)^2
64t^2 + (qt)^2 - 8000t - 1000qt + 497500 = 0
I have 2 variables here and I'm betting that tackling this problem correctly would leave me with just 1...I am in uncharted waters here though!

7. Mar 2, 2006

### Staff: Mentor

Finding a minima or maxima of a function without differentiation will be a chore. One step that may help you out is to use Excel to plot the distance versus time for various speeds of the 2nd person. Play with the speeds of the 2nd person some to see what the value of the diagonal (separation distance) is at the closest approach. With some hand-tuning, you should be able to get the correct answer to a couple of sig-figs fairly soon.

I'm not sure why you've been given this minima problem before you have the tools to figure it out. Was it extra credit or something?

8. Mar 2, 2006

### GregA

Unfortunately I'm just learning at home right now...so the only extra credit in it for me is confidence with tougher questions...I do know how to differentiate functions with one variable (and have a foggy recollection of how to find the maximum or minimum value value of a function (ie: f'(t) = 0 = min or max... and is max if f''(t) = -ve or min if +ve)...but this, unless I have formed the equation completely wrong is with 2 variables (q and t) and I cannot see how to differentiate wrt any of those variables and yield a solution.
Incidently the topic is just relative vectors and closest approach, but despite my best efforts to solve in the first instance adapting what I have read to fit...the nature of the books answer suggests they used a different working than mine... which is very un-settling.

I have had to move on from this question but I will have a go at your suggestion and see what I come up with so thanks again

9. Mar 2, 2006

### Staff: Mentor

If you have a good 3-d plotting program like Mathematica, you could do the plot of d(t,v) and see how it shapes up. You are right that the distance d is a function of both time and the velocity of the 2nd person. From the 3-d plot, you will be able to see the regions where the minimum separation comes close to the 50m number. You could then write a simple program to home in on the final answer, by varying v and seeing if you are getting closer or farther from the answer. The function will be monotonic near the final answer points.