- #1

tobix10

- 7

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## Homework Statement

A train of length L moves at speed 4c/5 eastward, and a train of length 3L moves at speed 3c/5 westward. How fast must someone run along the ground if he is to coincide with both the fronts-passing-each-other and backs-passingeach-other events?

## Homework Equations

Velocity addition formula

**[tex]u = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}[/tex]**

Lorentz contraction

**[tex] L = \frac{L_0}{\gamma}[/tex]**

Time

[tex] t = \frac{s}{v} [/tex]

Lorentz factor

[tex] \gamma (v) = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

## The Attempt at a Solution

I took the point of view of a running man (its frame). Initial point is when both fronts and a man are in line. Then I calculated speed of both trains in man's frame.

Train B is the one going westward and C eastward.

[tex] u_B = \frac{\frac{3}{5}c + v}{1+\frac{\frac{3}{5} v}{c}} [/tex]

[tex] u_C = \frac{-\frac{4}{5}c + v}{1-\frac{\frac{4}{5} v}{c}} [/tex]

Then calculate the length of trains in a frame of reference related to man

[tex] L_B = \frac{3L}{\gamma (u_B)}[/tex]

[tex] L_C = \frac{L}{\gamma (u_C)}[/tex]

I think that the passaging time of the trains should be equal

[tex] t_B = \frac{L_B}{u_B} = \frac{L_C}{u_C} = t_C [/tex]

but this equation gives solution v = c or v = -c so something is wrong.

I would expect that the velocity v is less than 4/5c.