Question

A supertrain of proper length travels at a speed of 0.86c as it passes through a tunnel having proper length 88.0 m. How much longer is the tunnel than the train or vice versa as seen by an observer at rest with respect to the tunnel?

Answer 1

Answer:

$L' = 104.6\ m$

Explanation:

given,

speed of length of travel = 0.86 c

length of tunnel = 88 m

Assume length of super train  = 205 m

An observer at rest in the tunnel's reference frame will observe the train's length to be contracted:

    L' = L₀ / γ

L' is the length the observer measures,

L₀ is the proper length of the train

$\gamma = \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$

$L' = L_0\sqrt{1-\dfrac{v^2}{c^2}}$

$L' = L_0\sqrt{1-\dfrac{(0.86c)^2}{c^2}}$

$L' = 205\times \sqrt{1-0.7396}$

$L' = 205\times \sqrt{0.2604}$

$L' = 205\times 0.5103$

$L' = 104.6\ m$