minimum time plan is like this:

1) m meets c at t1=k/24, when a and c(m) are at a distance L1=k-14k/24-2k/24=k/3.

2) m drives c backwards t2 hours and drop him in between a and b. now a and c(m) are at distance L2=L1-t2(14+2)=k/3+16t2.

3) it takes t3=L2/16=k/48-t2 to meet a, when a, b and c are at (t1+t2+t3)*2=k/8, (t1+t2+t3)*8=k/2, and k-14k/24-14t2+10t1=5k/6-14t2 distance from X respectively, i.e., they are L3=7k/8, L4=4k/8 and L5=k/6+14t2 away from Y.

4) m picks up a and drives him forward to Y, and arrive at Y when b and c also just arrive at Y (This takes t4=L3/14=k/16.). This means L3:L4:L5=14:8:10. L3:L4=14:8 is always satisfied. L4:L5=(k/2): (k/6+14t2)=8:10

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So the answer:

1) If we don't require integer solutions, then let k 0 and and the minimum time can be as small as we wish! (This is a direct result from "property 2" from here:http://www.mathhelpforum.com/math-he...rd-183226.html)

2) For integer solutions, we need 2k/24 (pickup point for c), 2k/16 (pickup point for a) ,k and t=t1+t2+t3+t4=k/8 to be integers. Hence k=48 and total time T=48/8=6.

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Second time edit: Red letters