By Jaan Kalda (TalTech).
A spaceship takes off from the Earth at t=0, and henceforth keeps the modulus of its proper acceleration equal to g; here and in what follows, t denotes the spaceship’s proper time. Thus, the astronauts on board will always feel a constant free fall acceleration g. However, the direction of the proper acceleration is changed four times at equal intervals, by turning the engines counterclockwise by 90^\circ. So, the proper acceleration is:
parallel to the x-axis when 0\le t< \tau;
parallel to the y-axis when \tau\le t< 2\tau;
antiparallel to the x-axis when 2\tau\le t< 3\tau;
antiparallel to the y-axis when 3\tau\le t< 4\tau.
It turns out that the spaceship’s speed relative to the Earth takes exactly the same value v at t=\tau, and t=4\tau. Find this value v.
Please submit the solution to this problem via e-mail to email@example.com. The first intermediate results for Problem 4 will be published on 5th March, 13:00 GMT. The first hints will appear here on 12th March 2023, 13:00 GMT. After the publication of the first hint, the base score is reduced to 0.9 pts. For full regulations, see the “Participate” tab.