Word Problem (How do I start the problem?)

I'm sorry again... I am really having diffilculty doing word problems as you can tell

Consider a regular 12 hour clock, and measure time in hours starting from noon. Let 'm' denote the angle that the minute hand makes with the top position (12). Similarly, let 'h' be the angle that the hour hand makes with the vertical position. In this case, suppose we measure the angle clockwise, so that the angles involved are positive.

At 2:30pm we have

'h'= ________ degrees.
At that time the minute hand has completed two and half rotations, and so the angle

'm'= _________ degrees. (Enter a number between 720 and 1080.)
In general, at time (expressed in hours so that e.g. 2:30 is 2.5 hours)

'h'= _______ degrees, and

'm'= ________ degrees.

There is a time 't' between '2' and '3' pm at which the minute hand is on top of the hour hand. At that time the hour hand forms an angle 'x' of ______ degrees with the vertical.

Hint: How many degrees are in a full circle? How many divisions are there in the circle of a clock-face?

Eliz.

Well full circle is 360 degrees and I figured out what what these two parts are

'h'= ___70_____ degrees.
At that time the minute hand has completed two and half rotations, and so the angle

'm'= ___900______ degrees. (Enter a number between 720 and 1080.)

Click to expand...

Now the next part of the question where it says

expressed in hours so that e.g. 2:30 is 2.5 hours

Click to expand...

what does that mean?

That means that 30 minutes is 1/2 hours so 2:30 is 2+.5 = 2.5 hours. 15 minutes is 1/4 hour so 2:15 = 2.25 hrs.
You should have gotten that the hour hand moves 360/12 = 30° each hour.
2.5*30° = ?
(taint 70)
You got the 2.5*360 = 900.

Oh okay Got'ca Thanks for all the help

This similar problem should enable you to solve your own.

<< At what time after 4:00 will the minute hand and the hour hand of a clock first be in the same position? >>

1--The hour hand moves 1 minute every .50º or 1M/.5º.
2--The minute hand moves 1 minute every 6º or 1M/6º.
3--When the two hands are coincident, the hour hand has moved Xº and the minute hand has moved (120 + X)º.
4--So, in T minutes, the two movements are (120 + X)1/6 = X/.5.
5--This leads to 60 + .5X = 6X or X = 10.909º.
6--Therefore, the hour hand has moved 10.909º, the minute hand 130.909º, and the minute hand has moved 20 + 10.909090(1/6) = 21.818 minutes past 4 o'clock.

Alternatively:
1--Let the speed (rate) of the hour hand be "r" and the speed (rate) of the minute hand be "12r" in minutes per minute.
2--Let the distance the hands move be measured by the minutes of the hour.
3--We are looking for the distance (in minutes) that the minute hand must move to overtake, and be coincident with, the hour hand.
4--Let this distance (or time in minutes) be "x."
5--Let the distance the hour hand must move in the same time period be (x - 20), the hour hand already being located at the "20 minutes after the hour" point.
6--As time is equal to distance divided by speed, the time for the minute hand to reach the point of coincidence is given by x/12r.
7--Similarly, the time for the hour hand to reach the same point is given by (x - 20)/r.
8--SInce the two hands each move the same amount of time, we can write x/12r = (x -20)/r or x = (12/11)20 = 21 9/11 = 21.8181 minutes.
9--Therefore, the two hands coincide at 04:21:49.09 or 4h-21m-49.09s after 12.
10--It is worth noting that in the expression x = (12/11)20, the quantity 20 represents the number of minutes that the hour hand is ahead of the minute hand to start out with.
11--Therefore, the value of 12/11 represents the number of minutes required for the minute hand to overtake the hour hand, per minute of head start.
12--For example, if we were seeking the time after 9 o'clock that the two hands overlapped, we need only multiply (12/11)45 = 49.0909 to derive the time of 09:49.0909 or 09h-49m-5.45s after 12.