1. ## Time Travel Problem

Problem:
Driver starts out in the morning on a drive and leaves exactly on the minute. After going 10 miles, she looks down at her watch and notes that the hour and minute hands coincide – both pointing in same direction. She traveled the 10 miles at an average velocity of 44 mph. What time did she leave on her drive?

At a speed of 44 MPH, the 10 mile point was reached in 7 min and 20 seconds. There are 11 times in the AM when the watch hands will coincide. I can calculate from any hour the point between that hour and the next (~65 min interval) when the hands will align. So it seems to me that the departure time could be within any of these intervals by finding the hands align time and back up 7min, 20 sec. I don’t see how you arrive at a precise departure time. Am I correct in this assumption, and could this situation be expressed algebraically?

2. ## Re: Time Travel Problem

I noticed a couple things about this question one of which I think you may have missed... and that is that the driver started exactly on the minute. This means that the second hand was exactly on the 12. Taking that into account you should be able to rule out some times. I can not wrap my head around this one completely though either. But there are several things to consider. First what exact times exactly do the hour hand and minute hand point in the same direction. I can tell you that they line up at 12 am. But that's about all I can tell you. But you are definitely incorrect in saying that it is 65 minute intervals. To prove this lets move from 12 o'clock 65 minutes now it is 1:05. At 1:05 the minute hand and hour hands do not align. The minute hand is on the 1 at 1:05, however the hour hand is already 1/12th of the way towards the 2. Keep in mind the hour hand was exactly on the 1 at 1:00 not 1:05. More to come. I got some thinking to do. I like this question.

3. ## Re: Time Travel Problem

Man, I would really love to be able to figure this one out myself without resorting to help with the net. I am about 100 percent sure that there is a geometrical equation or formula that would be able to help with this immensely. I can tell you that the 11 times that the hands transpose will be when they are at the 11 angles around the circle of the clock. Man its been so long since I took geometry. I sure hope someone checks this out and can point is in the right direction before it drives me nuts and I go to wikipedia for the answers.

4. ## Re: Time Travel Problem

The angles should be around 30.5ish degrees at 1:06 ish, around 61 degrees at like 2:11 ish around 91 degrees at 3:16.5ish, around 122 degrees at 4:22ish, and so on? I am doing all sorts of calculations in my head and am probably going to be of no help in the rest of this discussion. Until I go to wikipedia that is, but I really want to resist that temptation as long as possible. Maybe some of these ramblings will spark something jimdec and you will figure out what to do. Here is to hoping!

5. ## Re: Time Travel Problem

This is a fairly simple problem if you follow the right steps.

10 miles at 44 mph, that means the driver took 10/44 hours = 150/11 minutes.

The hour and minute hands coincide at 12:00. They coincide a total of 11 times per 12 hours, so the angle formed will be 360/11 degrees. The next time the hour and minute hands coincide will be 1 1/11 hours after 12:00, or 720/11 minutes after 12:00. Therefore, the next time they coincide will be 1440/11 minutes after 12:00, and so on.

Therefore the only times where the hour and minutes coincide are (each number represents # of minutes after 12:00):
720/11
1440/11
2160/11
2880/11
3600/11
4320/11
5040/11
5760/11
6480/11
7200/11
7920/11

If t is the time that the driver left, we want $t + \frac{150}{11} = \frac{720k}{11}$ to have an integer solution for t (this represents the driver leaving "exactly on the minute").

Multiplying both sides by 11, we get $11t + 150 = 720k$. Modulo 11, $150 \equiv 7$, therefore $720k \equiv 7 (mod 11)$. Turns out that $5760 \equiv 7 (mod 11)$, so the driver arrived at 5760/11 minutes after 12:00 (8 8/11 hrs after 12:00, approx. 8:44 am), and the driver left at (5760-150)/11 = 5610/11 = 510 minutes after 12:00 (exactly 8:30 am).

6. ## Re: Time Travel Problem

Tobeelijah1 & Richard1234:
Thanks so much for your interest in this problem and your help. Richard1234, Let me absorb this a bit and maybe I'll need to come back. Thanks again.

7. ## Re: Time Travel Problem

Hi Richard1234,
Could you elaborate a little more on the transistion from 720k congruent to 7 (mod11) and 5760 congruent to 7 (mod11). It seems that k must equal 8 in order for 760k (760*8) to be congruent to 7 ; thus 760 *8 = 5760 congruent to 7, which is your next derivation, that 5760 is congruent to 7. So how do you arrive at k=8 to make that calculation? How did you get to " turns out that 5760 is congruent to 7? From that point on I do follow your reasoning and calculations, but am confused as to how you concluded that 5760/11 was the correct arrival time. Thanks again for your effort and help.

8. ## Re: Time Travel Problem

Hi jimdec23,

We know that $11t + 150 = 720k$. Since $11t \equiv 0 (mod 11)$ and $150 \equiv 7 (mod 11)$, it turns out that 720k must be equivalent to 7 (mod 11). If k = 8, 720k = 5760, and this is congruent to 7 mod 11 (because 5753 = 11*523). I had to brute force a little bit of this to arrive at k = 8.

9. ## Re: Time Travel Problem

Richard,
Aha! Now I understand. One other question. Are we looking for k (=8) as the multiplier for 720, to represent the hour after 12:00 from which we will determine the coincidence of the clock hands? I hope I'm making myself clear and not being a nuisance. Thanks so much for all your help.

10. ## Re: Time Travel Problem

Originally Posted by jimdec23
At a speed of 44 MPH, the 10 mile point was reached in 7 min and 20 seconds.
HOW do you figure that?

11. ## Re: Time Travel Problem

Wilmer,
Big error on my part. Didn't notice that I had divided 44mph by 60 min rather than 60 min by 44mph. Embarrassing!

12. ## Re: Time Travel Problem

@jimdec23, the k isn't really that important. We are trying to find the $t$ such that $11t + 150$ is a multiple of 720. t = 5610/11 = 510 is the number of minutes after 12:00.