Question 266995
When Caitlin starts running towards Asha, Asha is 100 meters from Caitlin.
 As Caitlin runs, Asha moves directly away from Caitlin at one-third of Caitlin’s speed. 
How far has Caitlin run when she first catches up to Asha?
:
Briefly, C has to run 100 m more than A in the same amt of time
:
Let s = A's running speed
then
3s = C's running speed
:
Let d = distance run by A
then
(d+100) = distance run by C
:
Write a time equation: Time = dist/speed
:
C's running time = A's running time
{{{((d+100))/(3s)}}} = {{{d/s}}}
Cross multiply
s(d+100) = 3s(d)
sd + 100s = 3sd
:
Divide thru by s and you have:
d + 100 = 3d
100 = 3d - d
100 = 2d
d = 50m is the distance run by A
then
50 + 100 = 150m is the distance run by C to catch A
:
Check this, assume A's speed is 10 m/sec, C's speed is 30 m/sec
Check the times of each, they should be equal
150/30 = 5 min
50/10 = 5 min