A train pulls away from a station with a constant acceleration of 0.37 m/s2. A passenger arrives at a point next to the track 6.1 s after the end of the train has passed the very same point. What is the slowest constant speed at which she can run and still catch the train?

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

## Comments

I think that the displacement of train after 6.1 seconds will be (0.5(.37)(6.1)^2 = 6.88385 m