# Thread: It seems simple but I am struggling with this.

1. Ok John cycles from Norwich to London at 15 mph .
The distance is 100 miles.

Geraldine takes the car and ,at a steady speed of 45 mph leaves at the same time as John .
She arrives in London ahead of John ,turns around instantly and follows the same back path to Norwich.

She meets John en route.

How long after John has left Norwich does he encounter Geraldine?

Especially what is the formula for getting the answer.

And what is the formula if all we know is that Geraldine's speed is "r" times faster than John's?

thanks!

2.

3. Its pretty trivial, let be the distance form london to the point the two meet, be Geraldine's speed and be John's. so we can say:

Now multiply it by witch yeilds:

Now let be so

Solving this for gives us:

is the answer (time after leaving Norwich)

Good luck!

4. A good way to figure this out would be to consider the distances traveled by Geraldine and John. John will be traveling x, while geraldine will be traveling 200-x, since she will be traveling a full 100 miles (the distance from Norwich to London) and then the difference of johns position along the 100 mile track. Geraldine is traveling at , while John is traveling at

Now, we know that the times for travel will be the same, so all that's left is determine the distances traveled relative to eachother. We will call time c, so and also and since we know the times equal eachother, we can equate the two expressions.

If we know the speeds, or atleast that one speed is a specific fraction of the other speed, we can solve for x. Let's say the specific fraction is A. Now we have

5. Even more trival:

The Sum of the distances traveled by Geraldine and John is 2d ( if d is the distance from London to Norwich)

The sum of their speeds is V1+V2.

Ergo, the Time it takes for them to meet is

If v2 is 3*v1, then

6. thanks all .I was rustier than I thought!

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