Answer
See answers.
Work Step by Step
a. The length you measured is the contracted length. Use equation 26-3a to calculate the rest length.
$$\mathcal{l}_o=\frac{\mathcal{l}}{\sqrt{1-v^2/c^2}}=\frac{4.80m}{\sqrt{1-(0.720)^2}}=6.92m$$
The height, because it is perpendicular to the motion, does not change. The height at rest is still 1.35 m.
b. The time interval measured by your friend’s one watch is the proper time interval. Use equation 26–1a.
$$\Delta t_0=\Delta t\sqrt{1-v^2/c^2}=(20.0s) \sqrt{1-(0.720)^2}=13.9s$$
As expected, your friend’s watch “is slow” because it is moving.
c. Your friend sees you moving at the same relative speed of 0.720c.
d. The time interval measured by your watch is the proper time interval. This situation is symmetric to the one described in part b, so your watch would have advanced by only 13.9 s.
As expected, your single watch “is slow” as reported by your friend because your friend would say it is moving.
