Time difference word problem help

[seb3256]

I have no idea how to go about setting up this problem. Could someone explain what to do and why?

Problem:

Your watch loses 1.5 sec every hour. You have a friend whose watch gains 1 sec every hour. The watches show the same time now. After how many more seconds will they show the same time again?

 

[Unknown008]

I have no idea how to go about setting up this problem. Could someone explain what to do and why?

Problem:

Assuming it's a watch with hands...

The 'long' way round is to just use speed, distance time.

Speed of first watch is: 354 degrees/hour.
Speed of second watch is: 369 degrees/hour.

After a certain time t, the difference between the two 'distances' is 360 degrees again (or 0, but working in multiples of 360 gives the next times the second hands coincide).

369t - 354t = (360)(60)(12)

Solve for t to get the time in minutes. Then convert to seconds.
This would give you the time elapsed until the two second hands are together again.
360 degrees is when they have 1 minute difference, multiply by 60 to get one hour difference, and by 12 to get 12 hours difference. Note that on a digital watch, you will have to use 24 instead.

Well, an easier way around is to use relativity.

Assume that you sat on the tip of the seconds hand on your watch. Your own watch would, from there appear to have stopped (it moves while it moves you too). How fast would the other watch appear to run?

An example for this: You are in a car and is driving backwards. A car passes right besides you and is moving forward. If you are going backwards at 1 km/h, and the other car is going forward at 1 km/h, you being in your car would get the impression that your car is not moving, and that the other car is moving at 2 km/h.

Anyway, to your example, that means that the other watch would appear to get (1.5+1) = 2.5 seconds ahead after each hour. We can say that the speed of the first watch is 360 degrees/hour and the second watch is 362.5 degrees/hour. That said, the other watch has to get 12 hours ahead to show the same time. So, you set up:

2.5t = (12)(360)(60)

Which comes back to the same result.

EDIT: Took the first as 1 second back instead of 1.5 and the second watch as 1 second ahead instead of 1.5.

 

[burakaltr]

I have no idea how to go about setting up this problem. Could someone explain what to do and why?

Problem:

1 Day is 24 hours and is divided between a.m and p.m.

00:00+1x = 24:00-1.5x

2.5 x = 24*60*60 secs = 09:36 a. m.


x=24*60*60/2.5 = 34560 seconds

at 9:36 a.m. they show the same time which corresponds to 34560 seconds later


OPTIONAL ( For the same time once more ) :
09:36+x=24:00+09:36-1.5x

2.5x=24:00

x=09:36

09:36+09:36=19:12 = 07:12 p.m.


EDIT : The above times determined, ie : 9:36 a.m. and 7:12 p.m. only apply if the watches start both at 00:00.

 

[burakaltr]

Your watch loses 1.5 sec every hour. You have a friend whose watch gains 1 sec every hour. The watches show the same time now. After how many more seconds will they show the same time again?


Let us again start at 00:00

24:00-1.5 (01:00)x = 00:00 + 1 x(01:00)

2.5x(01:00)=24:00

x=24:00 / [(01:00) * 2.5] = 24*60*60 / 2.5

34560 secs later , the same time

AND

If the watch is clock-like, with arms, one arm showing the hour, the other the mins

12:00-1.5x(1:00)=00:00+1x(1:00)

12:00=2.5 x

and leads to half of the prior solution


17280 secs