Define "best time spacing". Do you want them to leave as close to each other in time as possible? Or do you want the cars to end up as close to each other in time as possible and still be greater than 10 seconds apart?
I have no idea what kind of math this falls under, but heres the problem:
Assuming all cars mentioned below are moving at constant speed.
Assume that the road looks like this:
X -> A,B,C -> D
Road A splits into three separate roads (A, B, C) of different distances, then merges at one point to be a single road D. The time to get from X->A->D is = 1 minutes, X->B->D = 2 minutes, X->C->D = 3 minutes".
If there are 3 parked cars at the beginning X with each car taking a respective road (A, B, C)
What is the best time spacing such that the cars are as closely spaced (Ex: Each car after car1 should leave P seconds after each other) that each car should move such that when the three cars are driving on road D, they are at least greater than 10 seconds apart? (Note that the best case is desired)
Problem Asks to find P.
All cars mentioned below are moving at the same constant speed.
The roads looks like this:
Code:X A D * * o * * * * * * * * * * * o * * * * * * * * B * * * * * * * * * * * C * * * * * * * *
Three cars will drive from to ,
each taking a different road, and continue on road
The time requires to drive road is one minute (60 seconds).
The time required to drive road is two minutes (120 seconds).
The time required to drive road is three minutes (180 seconds).
Each car starts seconds after each other.
When they are all driving on road D, they are to be at least 10 seconds apart.
Find the minimum
Car-1 drive on road and takes 180 seconds.
Car-2 must take 190 seconds to drive road
. . Hence, Car-2 must wait 70 seconds.
Car-3 must take 200 seconds to drive road
. . Hence, Car-3 must wait 140 seconds.
A different approach: Car 1 drives on C and takes 180 seconds. Car 2 takes road B and needs to wait 170 seconds. Therefore, Car 2 waits 50 seconds and ends up ahead of Car 1. Car 3 takes road C and needs to wait 160 seconds. Therefore, Car 3 waits 100 seconds and ends up ahead of Car 2.