"Two questions.
1. A HSC student doing her physics trial exam is sitting in her exam room on earth. Outside, she notices a spaceship travelling at 0.95c relative to her. If the astronaut on the spaceship times a certain event to take 1 hour, how long will it take from the student's point of view?
2. The same HSC student times a certain event to take 1 hour. How long will it take the astronaut on the spaceship to time the same event who is travelling 0.95 c relative to her."
1. Use of time dilation formula: T(v)= T(o)/√1-(v/c)^2 the one in bold is all inside the square root
T(v)= observer outside the frame of reference (hsc student)
T(o)= observer in the frame of rest or within the frame of reference (astronaut)
therefore T(v)= 1/√1-(0.95c/c)^2 (but your velocity is in terms of c thus the c can cancel out)
so T(v)= 1/√1-(0.95)^2
=3.20 hours
2. This time you just sub in 1 hour in T(v) and solve for T(o)