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[PLAIN]http://img573.imageshack.us/img573/9443/1132y.jpg [Broken]

So I did what I thought was right, but I'm a little confused on whose frame they are talking about when they say she arrived at the back of the train at the same time the back of the train leaves the tunnel..I think it makes a difference..

but anyway

a) I said the length of the train in the ground frame is L'=L/γ= (1-(0.6c/c)

Now the time in the ground's frame for the back of the train to pass the tunnel is t=(L+L')/v=(9L/5)/(3c/5)=3L/c, and since the person reaches the back of the train at the same time the train leaves the tunnel, it takes t=3L/c.

b) Now the person travels a distance L in this time, so v

c)t

Those are my answers, but I am not sure if I used the frames correctly.

Thanks for any help :).

So I did what I thought was right, but I'm a little confused on whose frame they are talking about when they say she arrived at the back of the train at the same time the back of the train leaves the tunnel..I think it makes a difference..

but anyway

a) I said the length of the train in the ground frame is L'=L/γ= (1-(0.6c/c)

^{2})^{1/2}*L=4L/5.Now the time in the ground's frame for the back of the train to pass the tunnel is t=(L+L')/v=(9L/5)/(3c/5)=3L/c, and since the person reaches the back of the train at the same time the train leaves the tunnel, it takes t=3L/c.

b) Now the person travels a distance L in this time, so v

_{w.r.t ground}=L/(3L/c)=c/3.c)t

_{train}=t_{ground}/γ=(4/5)*(3L/c)=12L/5c...either that or t=3L/c...I am a little confused as to in whose frame she arrives at the back of the train at the same time as it leaves the tunnel.Those are my answers, but I am not sure if I used the frames correctly.

Thanks for any help :).

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