An interesting problem in that there is more than one velocity the passenger can take up to his v(max), but what is the minimum velocity necessary to catch the train with a head start?
s = displacement
s(train) = (1/2)a(train)t^2 + v (train)t + 13.77m, (displacement at 6.1 sec)
s(person) = v(person)t
(1/2)a(train)t^2 + v(train)t + 13.77m = v(person)t
(1/2)a(train)t^2 - (v(train)-v(person))t + 13.77m = 0
solve for t
t = (v(person) -v (train) ± ((v train)-v(person))^2 - 4(.37m/sec)(1/2)(13.77m))^(1/2))/(.37m/sec) Solve the discriminant for positive value. V(train)- v(person) = ± 3.19 m/sec v (person) = v(train) + 3.19 m/sec = 2.26 m/sec + 3.19m/sec = 5.45 m/sec v(person)> 5.45m/sec