Question

Two twins are 25.0 years old when one of them sets out on a journey through space at nearly constant

speed. The twin in the spaceship measures time with an accurate watch. When he returns to Earth, he

claims to be 31.0 years old, while the twin left on Earth knows that she is 43.0 years old. What was the

speed of the spaceship?

Answer 1

Let $\Delta t_0$ be the time elapsed by the twin in the spaceship and $\Delta t$ be the time elapsed by the twin on the earth. Then due to time dilation

$\Delta t=\frac{\Delta t_0}{\sqrt{1-v^2/c^2}}$

where $v$ is the speed of the spaceship.

Here $\Delta t= (43.0-25.0)\ years = 18.0\ years\\ and\ \Delta t_0=(31.0-25.0)\ years = 6.0 \ years$

So,

$18.0=\frac{6.0}{\sqrt{1-v^2/c^2}}\\ \Rightarrow \sqrt{1-v^2/c^2}=1/3\\ \Rightarrow 1-v^2/c^2=1/9\\ \Rightarrow v^2/c^2=8/9\\ \Rightarrow v/c=\sqrt8/3\\ \Rightarrow v=0.94c$

Thus, the speed of the spaceship is 0.94c