Relativity problem -- A person walking in a train that is traveling at 3c/5

[LCSphysicist]

Homework Statement

A train of proper length L and speed 3c/5 approaches a tunnel of
length L. At the moment the front of the train enters the tunnel, a person
leaves the front of the train and walks (briskly) toward the back.
She arrives at the back of the train right when it (the back) leaves the
tunnel.

Relevant Equations

.

"(a) How much time does this take in the ground frame?
(b) What is the person’s speed with respect to the ground?
(c) How much time elapses on the person’s watch?"

I solved it, but i am with a doubt yet wrt to the letter b. We can find that the train length in the ground frame is 4L/5, so, why is not the person's speed equal to (4L/5)/T? The right is, actually, L/T.

 

[Ibix]

You haven't specified what $T$ is, so it's a bit difficult to know. I guess that it's the time in the ground frame that it takes the person to walk from one end of the train to the other?

In the train frame, the person started at the front of the train and finished at the end of the train. In the ground frame, where did she start (relative to some point at rest in this frame) and where did she finish (again, relative to some point at rest in this frame)? How far did she travel, then, and how long did it take?

 

[anuttarasammyak]

During her running she stays in the tunnel. In the ground frame she moves from one to another ends of the tunnel with constant speed of tunnel length L/T where T is the answer of (a) which should be
\frac{\frac{4}{5}L+L}{\frac{3}{5}c}
via your contraction calculation.