A spaceship resting on Earth has a length of \(35.2 \mathrm{m} .\) As it departs on a trip to another planet, it has a length of \(30.5 \mathrm{m}\) as measured by the Earthbound observers. The Earthbound observers also notice that one of the astronauts on the spaceship exercises for 22.2 min. How long would the astronaut herself say that she exercises?

Using the length contraction formula, we can find the relative velocity of the spaceship concerning Earth. The length contraction formula is given by: \(L = L_0 \sqrt{1 - \frac{v^2}{c^2}}\) where \(L_0\) = proper length of the spaceship (At rest) = 35.2 m. \(L\) = contracted length of the spaceship (Observed by Earthbound) = 30.5 m. \(v\) = velocity of the spaceship relative to Earth, \(c\) = speed of light = \(3 \times 10^8\) m/s. Our goal is to solve for \(v\). First, let's solve for the term inside the square root: \(\sqrt{1 - \frac{v^2}{c^2}} = \frac{L}{L_0}\) Squaring both sides, we have: \(1 - \frac{v^2}{c^2} = \left(\frac{L}{L_0}\right)^2\) Now, we can solve for \(v\): \(v^2 = c^2\left(1 - \left(\frac{L}{L_0}\right)^2\right)\) \(v = c \sqrt{1 - \left(\frac{L}{L_0}\right)^2}\) Calculate the value of \(v\) using the given lengths.

The Lorentz factor (gamma) is given by the formula: \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) Using the calculated value of \(v\) from step 1, we can find the Lorentz factor (gamma).

Now that we have the Lorentz factor, we can determine the time dilation effect. The time dilation formula is given by: \(t' = \gamma t\) where \(t'\) is the dilated time (astronaut's perceived exercise time) and \(t\) is the Earthbound observer time. We are given that the Earthbound observers measure the exercise time as 22.2 minutes. We need to convert this time into seconds: \(t = 22.2 \times 60 = 1332\) seconds Now, using the calculated value of \(\gamma\) and the value of \(t\), we can calculate the astronaut's perceived exercise time (in seconds) by plugging these into the time dilation formula. \(t' = \gamma t\) Finally, convert the value of \(t'\) back to minutes. This would be the time the astronaut herself says she exercises.