Another short resolution question

 In a cross-country race, runner A is 3 km ahead of runner B. However, while runner A runs 4.5 km, runner B runs 6 km. How far does runner B have to run to catch up with runner A?

A) 9

B) 12

C) 3

D) 21

So we have the formula that works with time and space an is S= S0 + Vt or even that V = ΔS/t, but we gonna use the 1st

Considerations about the solutions, I want the the solution that SA = SB; I make a simplification they diddn't give the velocity but let's think no matters what is the grandeza of the time the time will deleted so think no matters if the A runs one Km each century the B will run in the same at century, the differences is that they will be encounter in a particular distance no matters if this takes seconds or millenius, is the lowest run of the world they walk a meter each 100 years, no matters if takes 300 years or 5 seconds they will encounter 12km after, Ok?

So we using the equations we need a incognita and the only thing we gonna find is the time that is 2 but this is case 2 years? 2 minutes? 2000 years?

So: 3 + 4,5t = 6t -> 6t - 4,5t = 3 ->t = 2  no matters what is 2

And know I have the 2 equations wich I gonna choose, of course the one that the answer is direct the SB

SB = 6 x 2 -> SB = 12Km

So B runs 12Km in 400 years 2 hours, in milliseconds to encounter A

So no matters in how much time, they will encounter 12Km for B and 12-3 for A.

Rute Bezerra de Menezes Gondim