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?(adsbygoogle = window.adsbygoogle || []).push({});

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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Catching a train

**Physics Forums | Science Articles, Homework Help, Discussion**