Distance = Rate*Time Word Problem?

Question

In how many minutes after 5 o'clock will the hands of a clock first be together?

I have no idea how to set up an equation for this word problem except maybe x/12 for the hour hand. Please help!

Answer 1

To be precise this is a
time = distance / rate
problem.

Both hands travel for the same amount of time.

d = distance traveled by the hour hand
d + 5 = distance traveled by the minute hand
x = rate of hour hand
12x = rate of minute hand
t = time

t = d/x = (d + 5)/(12x)
12xd = (d + 5)x
12xd = dx + 5x
11xd = 5x
11d = 5
d = 5/11

The hands will be together in 5/11 hour or 300/11 minutes. That's 27 3/11 minutes or 27 minutes, 16 4/11 seconds.

Answer 2

What would "X" be in your equation? Which hands of the clock? If you are talking about the minute and hour hands, this problem can be set up as a D = RT problem, but the motion is rotary, involving radians, not straight-line miles! As another answer states, just use your brain! The minute hand makes a complete circle every hour, so the answer has to be "less han an hour". As for when, it really depends upon the clock. In a "perfect world", the minute hand moves 360 degrees in an hour while the hour hand moves 360 degrees in twelve hours! If "X" is the rate of the minute hand, then X/12 is the rate of the hour hand! They will be together when the (Original Location + D = RT) are equal. Set the rest up yourself.

Answer 3

The distance, h, the hour hand goes is 1/12 of the distance, m, the minute hand goes. Hour hand starts at 25 min after the top of the clock. When the hands are together, they are at the same point measured as minutes after the top, so

25 + h = m
25 + m/12 = m
25 = (11/12)m
300/11 = m
27 3/11 = m

Answer 4

Here is how to set it up:

let m = number of minutes after 5pm. The minute hand is at position m, the hour hand's position in terms of minutes is 25 + m/12, this is because although the hour hand starts at 5 (the 25 minute mark), it moves at 1/12 the speed of the minute hand.

So let:
m = 25 + m/12
and solve for m to get:
.
.
.
.
.
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11m = 300 which means m = 300/11.

Answer 5

Well, the hands will be together at approximately 5:27. So the answer would be about 27 minutes. I don't think this is actually a D=R x T problem.

Answer 6

OK here is my work.
First we need some facts:
1. One revolution on the clock is 360 degrees or for our exact purposes...2pi radians.
2. The mh has a speed of 2pi rad/hr and an initial position of 0 rad.
3. The hour hand (hh) has a speed of 2pi rad/24hrs or pi/6 rad/hr and an initial position of 5pi/6 rad
Now we want to find the time it takes (t) to make the mh and hh have the same position.

set up as initial position plus rate times t for the mh and hh:

mh: 0 + 2pi(t)
hh: 5pi/6 + (pi/6)t

now set equal and solve for t:
0 + 2pi(t) = 5pi/6 + (pi/6)t
2pi(t)-(pi/6)t = 5pi/6 (group like items)
12pi(t)/6 - pi(t)/6 = 5pi/12 (like denominators)
11pi(t)/6 = 5pi/6
11pi(t) = 30pi/6
11t = 30/6
t = 30/66 (in hours) = .454545... hours = 27.2727...minutes = about 27 minutes and 16 seconds.

We add this time to 5 o-clock and we get 5:27:16

That was fun!!!!!!!!

Answer 7

it depends on what ur watch is , If it works very exact , then this is the answer ;
let 5:y' be the time when the hands first meet then we have
y/12+25=y then 11y=12*25 hence y=300/11

Answer 8

You don't need an equation, just use your brain.
12 Midnight. (Don't think it's 6:30)

Answer 9

5:27? i ono i was never good at math.