Rate - Time - Distance

[Guest]

From the same point A, B and C start together. They are moving at a, b, and c miles per hour around a circular lake n miles in circumference. C is going in the opposite direction from A and B. Determine when A and B will be together and A and C will be together.

Formula used: Rate * Time = Distance Distance/Rate = Time Distance/Time = Rate


Rate Time Distance
A a an

B b bn

C c cn


How do I proceed from this point ?

[Guest]

A & B will move apart at a "relative speed to each other" equal to the difference of their actual speeds. They will meet when relatively they have moved apart by the distance "n" one revolution of the track.

A & C will move around the track coming closer at a "relative speed to each other" equal to the sum of their actual speeds. They will meet when relatively they have moved so that the sum of each distance is equal to "n" one revolution of the track.

And so on for multiple laps of the track.

[Guest]

I am not clear on your solution of "relative speed". Please give an example. Thanks.

[Guest]

---------An Example..........
If the lake circumference was 24 miles
and A goes at 8 miles per hour
and B goes at 4 miles per hour
Both go is same direction...

A does a lap in 3 hours and B will only have done 1/2 a lap at that stage
B does a lap in 6 hours and A will have done 2 laps at that stage.
So from B point of view.... B sees A maving away from him at (8-4) = 4 miles per hour. A is travelling 4 miles per hour faster relative to B.
So A will be (4 mph x 6 hours) = 24 miles (1 lap) ahead of B when B finishes his first lap and A will have actually finished his second lap and both A and B will have met again at the start.

[Guest]

---------A 2nd Example..........
If Lake circumference was 24 miles
and A goes at 8 miles per hour
and B goes at 2 miles per hour
Both go is same direction...

A does a lap in 3 hours and B will only have done 1/4 a lap at that stage
B does a lap in 12 hours and A will have done 4 laps at that stage.
So from B point of view.... B sees A maving away from him at (8-2) = 6 miles per hour. A is travelling 6 miles per hour faster relative to B.
So A will be (6 mph x 12 hours) = 72 miles (3 laps) ahead of B when B finishes his first lap and A will have actually finished his fourth lap and both A and B will have met again at the start. But A and B will also have met on 2 occasions before that..... At a (1/16 of a lap) past the 1/4 mark on A's 2nd lap, then at a (1/8 of a lap) past the 1/2 mark on the 3rd lap. At end of 4th lap both will be back at the start.

In the first example their relative speed was equal to the slower actual speed.......

In this 2nd example their relative speed is faster that the slower speed.....

[Guest]

"A & C will move around the track coming closer at a "relative speed to each other" equal to the sum of their actual speeds. They will meet when relatively they have moved so that the sum of each distance is equal to "n" one revolution of the track."


So if A is 8 mph and C is 6 mph then A is traveling at 14 mph relative to C. Correct ?

[Guest]

Yes and C is travelling at 14 mph relative to A.

[Guest]

So A and C will meet at 24 miles. Still correct ?

[Guest]

Further to earlier post.....
"But A and B will also have met on 2 occasions before that..... At a (1/16 of a lap) past the 1/4 mark on A's 2nd lap, then at a (1/8 of a lap) past the 1/2 mark on the 3rd lap. At end of 4th lap both will be back at the start".

I only approximated here to indicate the A and B met at some tine after 1/4 lap and again at some time after 1/2 lap.
( because when A does the 1/4 lap he is behind, B has then done 1/16 lap past 1/4, then when A does this 1/16, B has moved on a bit further, etc etc etc.)

The meeting point can be worked out more accurately......

The ratio of their speeds is 4:1
Let x = dist B travels past the 1/4 after the start of A 2nd lap
When they meet or when A overtakes B, A will have travelled 4 times what B has travelled.
And A will have travelled (6 + x)

So 4x = 6 + x

3x = 6

x = 2

So B is overtaken by A when B has travelled 2 miles past the 1/4 mark.
And A has travelled (6 + 2) = 8 miles of lap 2.....equals 1/3 of lap2.

Similary at the 3rd lap for A, A overtakes B at 4 miles past the 1/2 lap mark.

Calc. from....
Let y be the dist past 1/2 mark when A overtakes B.....
then..... 4y = 12 + y
3y = 12
y = 4

So A has travelled 12 + 4 = 16 miles of lap 3 .... 2/3 of lap3....when overtaking B.

At the end of A's 4th lap both are again at the start point.

So for each time A "meets of overtakes" B, A has travelled (1 + 1/3) = 1 1/3 laps = 4/3 laps and B has travelled 1/3 of a lap....Actual distance.
But A has only travelled 1 lap relative to B.....Thay are never any more than 1 lap apart (24 miles).
And their relative speed is 6 miles per hour. So they meet every 4 hours.
And from above B takes 12 hours for 1 lap so in 4 hours B has travelled 1/3 of a lap.
And from above A takes 3 hours to do 1 lap so in 4 hours has travelled 1 1/3 laps.

[Guest]

"So A and C will meet at 24 miles. Still correct ? "

I assume you are talking about relative distances here. A and C start together in opposite directions. A will see C somewhere around the track coming towards him at 14 miles per hour. The length of the track is 24 miles so Yes they will meet after travelling a total of 24 miles relative to one another. "A " will have travelled part of it and "B" will have travelled part of it. But where is this in reality.

[Guest]

I didn't understand what is meant by relative speed and asked for an example. I was just using the 24 as distance, 8 and 6 for A and C. When you explained A and B, I was wondering about A and C. I did not know how to solve the problem.

[Guest]

Example.....travel opposite directions....

So if A is 8 mph and C is 6 mph then A is traveling at 14 mph relative to C.
And track is 24 miles long.....when do they meet?

They meet when they have travelled 24 miles relative to one another.
A travels a bit and C travels a bit.
They will each have the same time of travel as both start together.
Their speeds will be added because travel is in oposite directions meeting each other.

Let t = time until both meet

8t + 6t = 24
14t = 24
t = 1.714 hours until they meet

In time t ..... A travels 8 x 1.714 = 13.714 miles

In time t ..... C travels 6 x 1.714 = 10.286 miles

So every 1.714 hours they will meet

So "C" will meet "A" twice each lap .....

At a distance of 10.286 from start and also at 2 x 10.286 = 20.572 miles on 1st lap in direction of "C" travelling.

OR .... distance 13.714 from start and also 2 x 13.714 = 27.428 miles. So "A" has completed a lap and is 3.428 into 2nd lap in direction of "A" travelling.

The meeting point will move around the track each lap....
.................................

[Guest]

I understand it now. Thanks for your detailed explanation.