Solving Velocity Formula: Runner A & B Crossing Paths

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Runner A starts 6.0 km west of a flagpole, moving east at 9.0 km/h, while Runner B begins 5.0 km east, running west at 8.0 km/h. The key to solving when their paths cross is determining the time it takes for the sum of their distances to equal 11 km. By setting up a coordinate system and using the equation V = d/t, the relationship between their velocities and distances can be established. Ultimately, the solution reveals the distance of each runner from the flagpole at the moment they meet.
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[SOLVED] Velocity Formula

Homework Statement


"Runner A is initially 6.0km west of a flagpole and is running with a constant velocity of 9.0km/h due east. Runner B is initially 5.0km east of the flagpole and is running with a constant velocity of 8.0km/h due west. What will be the distance of the two runners from the flagpole when their paths cross? (Leave answer in km)"

Homework Equations


Solve this only using V = d / t


The Attempt at a Solution


I've tried too many.
 
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well first you need to find when their paths cross. This is when the sum of their individual distances is equal to 11 (the distance between them). Once you have this time you can figure out how far each person ran in that time because you have their velocities.
 
Tricky.
Here's what I did.
Set up a co-ordinate axis with your runners on the x-axis and flag pole at origin.
Now express the distance each runner runs in terms of the original distance from the flag pole given and the equal final distance d_f.
Once you have this see if it doesn't pop out at you.
 
D1 would be the position of Runner A; D2 would be the position of Runner B.

|D2 - D1| = 11

I don't know what to do from here.
 
if you think about it runner a runs 9kms in an hour and runner b runs 8kms in an hour.

9t+8t=11 gives you a way to find when they pass each other
 
Okay, I've figured it out. I solved for D in the equation I gave and got the answer. Before, for some unknown reason, I simply could not figure it out. Thanks!
 
Last edited:
Both ways work, spoon's may be a bit easier:-p.
 
<--- said:
Both ways work, spoon's may be a bit easier:-p.

It is essentially the same thing.
 

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