Urgent Math Help Needed!

[http.amyxo]

I really didn't want to turn to this - it was my last resort. I really have no clue as to how I can answer this question! I can't even start it!
Basically, I am to create a poster about a problem.
The problem is:

John's Train Is On Time.
A train leaves the station on time. After it has gone 8 miles the driver looks at his watch and sees that the hour hand is exactly over the minute hand.
The average speed of the train over the 8 miles was 33mph.
When did the train leave the station?

I need to answer:
what the problem is,
What steps I took to solve the problem
Key formulas I used to solve the problem
The solution
And any extra questions I need to ask about the homework.

I AM NOT ASKING FOR SOMEONE TO DO MY HOMEWORK.
I am asking for help.
I desperately need to know how to do this, I have no clue!

Thank you <3

 

[Ishuda]

I really didn't want to turn to this - it was my last resort. I really have no clue as to how I can answer this question! I can't even start it!
Basically, I am to create a poster about a problem.
The problem is:

John's Train Is On Time.
A train leaves the station on time. After it has gone 8 miles the driver looks at his watch and sees that the hour hand is exactly over the minute hand.
The average speed of the train over the 8 miles was 33mph.
When did the train leave the station?

I need to answer:
what the problem is,
What steps I took to solve the problem
Key formulas I used to solve the problem
The solution
And any extra questions I need to ask about the homework.

I AM NOT ASKING FOR SOMEONE TO DO MY HOMEWORK.
I am asking for help.
I desperately need to know how to do this, I have no clue!

Thank you <3

Since the hour hand travels one revolution per hour and minute hand travels 60 revolutions per hour, what formula connects the speed of the hour hand and the minute hand? Since distance (number of revolutions) is average speed times time, when does the minute hand and hour hand line up? Given that time, how far did the train travel?

For example, suppose person A is running around a circular track once every minute [it's a small track] and person B is running around the track every 20 minutes [person B is carrying a small elephant]. So the distance person A is running in each minute is given by
D_a = 1 (rev/min) * t (min) = t
For person B it is
D_b = (1/20) (rev/min) * t (min) = (1/20) t
Now suppose person A has completed n times around the track. Then person B has completed the integer part of (n)/20 times around the track. That is, if A has 20 times around the track [and is in their 21st], B has completed 1 time around the track [and is in their second].

For simplicity, assume we are still in the first time around the track for B, they have both been 'running for a bit of time' and A has just caught up to B again. Well A has completed the integer part of t times around the track and is the fractional part of t around the track. If B is at the same part of a revolution around the track, then B is (1/20) times the fractional part of t. Let
t=n+\(\displaystyle \delta\); \(\displaystyle \delta\, \lt\, 1,\, n\, \lt\, 20\),
Then
\(\displaystyle D_a\, =\, n (rev)\, +\, \delta (rev)\)
\(\displaystyle D_b\, =\, \delta (rev)\, =\, \frac{1}{20}\, t\) (rev)
or
\(\displaystyle t =20\, \delta\)
So how far has the train traveled in that time if its average speed was 33 mph?

 

[stapel]

Since the hour hand travels one revolution per hour and minute hand travels 60 revolutions per hour....

Perhaps you mean that the hour hand travels one-twelfth of a revolution per hour, the minute hand travels one whole revolution, and the second hand travels sixty revolutions?

 

[Ishuda]

Perhaps you mean that the hour hand travels one-twelfth of a revolution per hour, the minute hand travels one whole revolution, and the second hand travels sixty revolutions?

Well only if I want to be correct \(\displaystyle \underset{\text{and i do want to be correct. maybe standing in the corner will improve my correctness (and spacing)}}{.}\)

EDIT: Wait a second here. Yes, the hour hand travels 1/12 of a revolution in one hour [assuming a 12 hour clock of course and not a 24 hour clock as some clocks are]. How so some ever, the minute hand travels 60 revolutions [even if it is a 24 hour clock] and the second hand travels travels around once every minute [again, even if it is 24 hour clock]. So I was only partially wrong and since I have stood in the corner for the hole 55 minuets [which i s a whole lot longer than 55 minutes], I can afford to be wrong again at least one more time without having to stand in the corner. BTW: Standing in the corner doesn 't seem to have improved my s p a c in g.

 

[stapel]

How so some ever, the minute hand travels 60 revolutions....

How many revolutions does the minute-hand make between striking the hour (that is, pointing directly upward at the "12") and a quarter after (that is, fifteen minutes after the hour)? Does it make fifteen revolutions, so at a quarter after the minute-hand is once again pointing directly upward?

Hint: Think about how a minute-hand marks the time, from the hour, to a quarter past, to half past, to a quarter of, and finally to the next hour.

 

[Ishuda]

How many revolutions does the minute-hand make between striking the hour (that is, pointing directly upward at the "12") and a quarter after (that is, fifteen minutes after the hour)? Does it make fifteen revolutions, so at a quarter after the minute-hand is once again pointing directly upward?

Hint: Think about how a minute-hand marks the time, from the hour, to a quarter past, to half past, to a quarter of, and finally to the next hour.
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Oh darn, I knew there was a reason for the minuets. I'll now have to use my freebie.