# Catching a train

1. Apr 13, 2004

### jennypear

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!

2. Apr 13, 2004

### jennypear

nevermind I got it. I forgot to add the initial 3.78m the train traveled before the person started running.

3. Apr 13, 2004

### Chen

Why did you decide that Xo of the train is 3.78m and its Vo is 1.26m/s? Instead of calculating how much the train has passed in 6 seconds (which is what I suspect you have done, and maybe also where your error lies), I would just give it a 6 seconds fore on the person by changing "t" in its equations with "t + 6".

So we have:

$$x_{train} = x_{person}$$
$$v_{train} = v_{person}$$

$$\frac{1}{2}a(t + 6s)^2 = vt$$
$$a(t + 6s) = v$$

Substituting v we get:

$$\frac{1}{2}a(t + 6s)^2 = a(t + 6s)t$$

And indeed t = 6s. Now just plug that into the second equation above:

$$v = a(t + 6s) = a(6s + 6s) = 2.52 m/s$$

4. Apr 13, 2004

### outandbeyond2004

I find your explanation bewildering. I am restating the problem as I understand it.

At time = 0 the end of the train is at x = 0 and v = 0. Six seconds later, xend = 0 + 0*t + (0.21)*t^2/2 = 3.78 meters
vend0 = left to you to calc.

At this point, xpass = 0, vpass = something, call it vpass

xpass = vpass*t

By the time the passenger catches the train,

xpass = xend
vpass = vend

So,
vpass*t = 3.78 + vend0*t + 0.21*t*t/2
vend = vpass = 0.21*t

I believe that's all you need -- solve for t then solve for vpass.