At 3 o clock, the hour hand coincides with the minute hand

The speed of the hour hand is one-twelfth the speed of the minute hand, because the hour hand travels five divisions in an hour, and the minute hand moves sixty.

Solution: If the minute hand goes x, then the hour hand goes x&12

x&12+15=x

x=180&11

Answer: At 3:180 & 11, the hour hand coincides with the minute hand.

Note: "& is a semicolon".

First of all, it is a catch-up problem, the minute hand and the hour hand start 15 stops apart, that is, 15 minutes of time, and the minute hand is moving at the same time as the hour hand is moving, and the minute hand goes 60 stops, which is 60 minutes, and the hour hand goes 5 stops.

Then the speed of the minute hand is 1 block per minute, and the speed of the hour hand is 1 12 blocks (5 60 stops) per minute

Let the x-minute hand catch up with the upper hour hand, and the equation can be obtained:

1*x=15+1/12*x

x-1/12x=15

x=180/11

x=infinite loop).

The answer is: the minute hand coincides with the hour hand at 3 o'clock.

Think: This problem is actually the problem of catching up in the itinerary problem, 3 hours and minutes point to 12, and the hour hand points to 3. The minute hand is 5 3 15 divisions away from the hour hand.

The minute hand travels 1 block per minute, and the hour hand moves 1 block per minute. To coincide the minute hand with the hour hand, the minute hand should travel 15 notches longer than the hour hand. According to the catch-up time = distance difference in the catch-up problem, the speed difference can be used.

Solution: 15 (1) = 16 (min).

Answer: At 3:16, the hour hand coincides with the minute hand.

This is a matter of time calculation.

Then we should first think of solving the problem of distance and speed.

A clock has 60 cells and 12 cells in each cell.

The speed of the hour hand is 5 60 divisions per minute, which is 1 12 divisions.

The speed of the minute hand is 60 60 or 12 squares per minute.

Then the difference in their speed can be concluded is 11 to 12 minutes per minute.

Let's take a look at the distance.

When it is 3 o'clock, the minute hand is at 12 o'clock and the hour hand is at 3 o'clock.

At this time, the distance between the minute hand and the hour hand is 3*5=15 divisions.

What is this distance? This distance is the distance at which the minute hand is more than the hour hand when the hour and minute hands coincide.

Now that we have the extra distance and the extra speed, it's not hard to calculate the time.

15 (11 12) = 15 * 12 11= Since there is no full space, the minute hand is still at 16 squares.

After 180 11 minutes the two needles coincided for the first time.

The hour hand travels 360° per hour 12 = 30°

The hour hand moves 30° 60= per minute

The minute hand travels 360° per minute 60 = 6°

Now it's 3 o'clock, which is 90°, 90° (6°.)

That is, after 180 11 minutes, the two needles coincide for the first time.

The minute hand rotates 360° in one hour, so the angular velocity of the minute hand = 360° 60 minutes = 6° minutes;

The hour hand turns 30° in one hour, so the angular velocity of the hour hand = 30° 60 minutes = minutes;

At 3 o'clock, the minute hand points to 12 and its initial position is denoted as 0°, i.e. =0°; The hour hand points to 3 and its initial position is denoted as 90°, ie.

=90°；Now the opening is to keep time: let the two hands coincide after t minutes, then there is the equation:

+6t= +, i.e. 6t=90+; t=90 points=16'''

That is, at 3:16 and 2, the two hands coincide for the first time.

The minute and hour hands meet 11 times in 12 hours, an average of 1 meeting every 12 11 hours. The third encounter (coincidence) at more than 3 o'clock is (12 11) x 3 = 3 and 3 11 o'clock, about 3:16 minutes and 22 seconds.

6-1 2=11 2 (degrees).

90 11 2 = 16 and 11 4 (points).

Answer: 3:16 and 11 4 minutes.

Example: 5/4 = 5 4).

180 degrees per 90

3:180/11 coincides.

1) Overlap 3:00 at right angles means that the difference between the hour and minute hand at 3 o'clock is 15 divisions (60 divisions for a clock), the minute hand moves one block per minute, and the hour hand moves one block in 12 minutes.

Suppose the hour hand coincides with the minute hand at the X minute after 3:00. Get the equation.

x=x/12+15

12x=x+180

11x=180

x=180 11 is approximately equal to minutes.

That is, after 180 minutes after 3 o'clock, the minute hand coincides with the hour hand.

2) at a right angle, that is, 180 degrees, that is, the minute hand and the hour hand are 15 blocks apart to become a flat angle. The minute hand is more than 15 stops larger than the hour hand.

x-15=15+x/12

12x-180=180-x

11x=360

x=360 11 is approximately equal to minutes.

That is, after 3 o'clock 360 11 minutes, the minute hand and the hour hand are at right angles.

3) Flat angle is 180 degrees, that is, the minute hand and the hour hand are 30 squares different before it becomes a flat angle. The minute hand is more than 30 stops larger than the hour hand.

Or assume that the hour hand of the x minute after 3:00 coincides with the minute hand. Get the equation.

x-30=15+x/12

12x-360=180+x

11x=540

x=540 11 is approximately equal to minutes.

That is, after 3 o'clock 540 11 minutes, the minute hand is at a flat angle to the hour hand.

The speed of the minute hand is 1 block minute, and the speed of the hour hand = 1 12 blocks minutes.

Let a minute later coincide for the first time.

1-1/12）×a=15

11/12a=15

a = 180 11 16 minutes 22 seconds.

At 3:16:22, the first coincidence.

Let x points coincide for the first time.

The angle of the minute hand is 6x (the minute hand rotates 6 degrees per minute), and the angle of the hour hand is x 2 (the hour hand turns 1 2 degrees per minute), and the pin is only 6x-90 = x 2

Solution x = 180 11

So the deficit overlapped for the first time after 180 and 11 points.

After 180 11 minutes the two needles coincided for the first time.

The hour hand travels 360° per hour 12 = 30°

The hour hand moves 30° 60= per minute

The minute hand travels 360° per minute 60 = 6°

Now it's 3 o'clock, which is 90°, 90° (6°.)

That is, after 180 11 minutes, the two needles coincide for the first time.

3 in 15 points, 4 in 20 points; If there is a difference of 5 squares between 3 and 4, then the minute goes for 1 minute, and the hour hand goes 5 60=1 12 squares, and if x minutes coincide, then x=15 (where 3 o'clock is located) + x 12, then x=16

3:16 a.m.

The speed at which the minute hand travels at an angle per minute is v1 360 60 (degrees min) 6 (degrees min).

If the speed of the hour hand travels at an angle per minute is v2 (360 12 60 degrees minutes), then from 3 o'clock sharp, let t minutes elapse, and the hour and minute hands coincide for the first time.

v1*t v2*t+90, that is, 6t, the solution gives t 180 11=16 and 4 11, that is, 3:16 and 4 11 The hour hand and the minute hand coincide for the first time.

Think about it from 2 o'clock sharp:

In this case, the hour hand is in front and the minute hand is in the back, and the angle is 60 degrees.

If you want to coincide, it is the distance difference between the minute hand and the time hand (refer to chasing and question solving): 60 degrees;

Speed difference: the minute hand rotates 6 degrees per minute, the hour hand rotates every minute;

The elapsed time is: 60 (6 and 10 11 (minutes), i.e. the time of chase (coincidence) is 2:10 and 10 11 minutes.