Let's solve the problem step by step as requested. --- ### (a) **How much time does this take in the ground frame?** **Step 1: Find the time it takes for the train to pass through the tunnel in the ground frame.** - **Tunnel length:** \(2L\) - **Train speed in ground frame:** \(v_{train} = c/2\) - The time for the train to traverse the tunnel (from front entering to back exiting) is: \[ t_{ground} = \frac{\text{length of tunnel} + \text{length of train (as seen from ground)}}{\text{speed of train}} \] But, let's clarify: The person runs from the front to the rear of the train, arriving at the back just as the rear leaves the tunnel. So, the relevant time is the time it takes for the train to traverse the tunnel, i.e., for the rear to move from entering to leaving the tunnel. The proper length of the train is \(L\), so in the ground frame (where the train is moving), the train is **length-contracted**: \[ L_{train,\,ground} = L \sqrt{1 - \frac{v^2}{c^2}} \] where \(v = c/2\): \[ L_{train,\,ground} = L \sqrt{1 - \frac{(c/2)^2}{c^2}} = L \sqrt{1 - \frac{1}{4}} = L \sqrt{\frac{3}{4}} = L \frac{\sqrt{3}}{2} \] Now, the rear of the train has to travel the entire tunnel length plus the train's own length (in ground frame), but since the front enters at \(t=0\), and the person starts running at that instant, we want the time until the **rear** leaves the tunnel. The rear is behind the front by the contracted train length. Thus: \[ t_{ground} = \frac{2L + L_{train,\,ground}}{v_{train}} = \frac{2L + (L \frac{\sqrt{3}}{2})}{c/2} \] \[ t_{ground} = \frac{2L + \frac{\sqrt{3}}{2}L}{c/2} \] \[ t_{ground} = \frac{(2 + \frac{\sqrt{3}}{2})L}{c/2} \] \[ t_{ground} = \frac{(4 + \sqrt{3})L}{c} \] **Explanation:** We calculated the total time for the rear of the train to traverse the tunnel (from when the front enters to when the rear exits), considering length contraction for the train. --- ### (b) **What is the person's speed with respect to the ground?** **Step 2: Use relativistic velocity addition.** - The person runs toward the rear at speed \(v\) **relative to the train**. - The train moves at \(c/2\) **relative to the ground**. The person's velocity **relative to the ground** (in the opposite direction to the train's motion) is: \[ u = v_{person,\,ground} = v_{train,\,ground} - \frac{v_{person,\,train}}{1 - \frac{v_{train,\,ground} \cdot v_{person,\,train}}{c^2}} \] But the person is running **toward the rear**, so in the ground frame: \[ u = v_{train,\,ground} - \frac{v_{person,\,train}}{1 - \frac{v_{train,\,ground} \cdot v_{person,\,train}}{c^2}} \] But actually, the correct formula for velocity addition (when both velocities are in the same direction; here, the person is running in the opposite direction) is: \[ u = \frac{v_{train,\,ground} - v_{person,\,train}}{1 - \frac{v_{train,\,ground} v_{person,\,train}}{c^2}} \] Assuming the person runs toward the rear at speed \(v\) **relative to the train**: \[ u = \frac{c/2 - v}{1 - \frac{(c/2)v}{c^2}} \] **Explanation:** We applied the relativistic velocity addition formula for the person running toward the rear of the moving train. --- ### (c) **How much time elapsed on the person's watch?** **Step 3: Use time dilation (proper time is the time measured by the person).** - The time taken in the ground frame is \(t_{ground}\) (**from part a**). - The person's speed **relative to the ground** is \(u\) (**from part b**). The time elapsed on the person's watch: \[ t_{person} = t_{ground} \sqrt{1 - \frac{u^2}{c^2}} \] **Explanation:** We used the time dilation formula, where \(t_{person}\) is the proper time as measured by the person running at speed \(u\) relative to the ground. --- ## **Final Answers** ### (a) **Time in ground frame:** \[ \boxed{t_{ground} = \frac{(4 + \sqrt{3})L}{c}} \] ### (b) **Person's speed with respect to ground:** \[ \boxed{u = \frac{c/2 - v}{1 - \frac{(c/2)v}{c^2}}} \] ### (c) **Time elapsed on person's watch:** \[ \boxed{t_{person} = t_{ground} \sqrt{1 - \frac{u^2}{c^2}}} \] where \(t_{ground}\) and \(u\) are given above.