A train pulls away from a station with a constant acceleration of 0.21 m/s2. A passenger arrives at the track 6.0 s after the end of the train passed the very same point. What is the slowest constant speed at which she can run and catch the train? Xtrain=Xo+Vot+1/2(a*t^2) Xtrain=3.78m + 1.26t + .5*(.21m/s^2)t^2 Vtrain=Vo + at Vtrain=1.26m/s + (.21m/s^2)t Xperson=Vperson*t Vperson=X/t **at the time the person catches the train the Vtrain=Vperson & Xtrain=Xperson Vtrain=Vperson 1.26m/s + (.21m/s^2)t=X/t since the X's are the same substitute Xtrain for X on right side of equation. 1.26m/s+(.21m/s^2)t=[3.78m + (1.26m/s)t +(.105m/s^2)t^2]/t (1.26m/s)t + (.21m/s^2)t^2=3.78m + (1.26m/s)t +(.105m/s^2)t^2 (.105m/s^2)t^2=3.78m t^2=36s^2 t=6s plug it in X=Vot + 1/2*a*t^2 X=1.26m/s*(6s) + .105m/s^2(36s^2) X=7.56m+3.78m=11.34m **which in theory would make min velocity needed 11.34m/6s=1.89m/s but this isnt correct Can anyone see what I have done wrong w/this problem? Thanks!