Answer to Question #211357 in Physics for Weds

Question

A spacecraft is moving relative to the earth.An observer on the earth finds that, according to her clock, 3601 s elapse between 1pm and 2pm on the spacecraft’s clock.What is the spacecraft’s speed relative to the earth?

Answer 1

The time dilation equation says

$$\Delta t=\frac{\Delta \tau}{\sqrt{1-\frac{v^2}{c^2}}}$$

Hence,

$$v= c\sqrt{1-\left(\frac{\Delta \tau}{\Delta t}\right)^2}=c\sqrt{1-\left(\frac{3600}{3601}\right)^2}=0.024c$$