[OOM: Back home, Izana decides to take action.]
Today's math problem: If a seed ship, traveling
y km per hour, is traveling away from a target which may or may not be in the same position it was at one week ago, how fast and far must a rescue mission travel in order to both rescue one or two persons (
a+b=c kilograms) and return to the seed ship which has continued in its original velocity. Do not forget to account for the gravitational pull of any asteroids or other bodies with sufficient mass.
Izana is sitting at a booth with a datapad (which looks perhaps more like a sheet of clear plastic with glowing kanji and graphics), notes, specifications for garde units, and a calculator (really their communication device in its computation mode). Izana is very focused and has no plans to go anywhere until they work out their plan. After all, time is frozen back home and were they to work out their plan there, the distances would just get farther apart.
Still, as focused as they are, a break is likely a good thing.
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