This is a nice variation on a racing problem by Geoffrey Mott-Smith from 1954.
“On one side of the playground some of the children were holding foot-races, under a supervisor who handicapped each child according to age and size. In one race, she placed the big boy at the starting line, the little boy a few paces in front of the line, and she gave the little girl twice as much headstart over the little boy as he had over the big boy. The big boy won the race nevertheless. He overtook the little boy in 6 seconds, and the little girl 4 seconds later.
Assuming that all three runners maintained a uniform speed, how long did it take the little boy to overtake the little girl?”
See the Handicap Racing for solution.
Let BB, LB and LG be the speeds of the big boy, the little boy and the little girl respectively. Also x and 2*x be the initial distances as mentioned.
The relative speed of big boy with respect to little boy be (BB-LB) and that of big boy with respect to little girl be (BB-LG).
(BB-LB)*6=x
and (BB-LG)*(4+6)=3*x
30*BB – 30*LB = 5*x …(1)
and 30*BB – 30*LG = 9*x ….(2)
Subtracting (1) from (2),
30*(LB-LG)=4*x
or LB-LG=(2/15)*x. , which is relative speed of little boy with respect to little girl.
To cover 2*x distance, the time required will be (2*x)/(2/15)*x=15 seconds.