Step-by-step explanation:
The time dilation formula is given by
[tex]F(t) = \dfrac{t}{\sqrt{1-v^2}}[/tex]
where t is the time measured by the moving observer and F(t) is the time measured by the stationary earth-bound observer and v is the velocity of the moving observer expressed as a fraction of the speed of light.
a) If the observer is moving at 80% of the speed of light and observes an event that lasts for 1 second, a stationary observer will see the same event occurring over a time period of
[tex]F(t) = \dfrac{1\:\text{s}}{\sqrt{1-(0.8)^2}}[/tex]
[tex]\:\:\:\:\:\:\:=\dfrac{1\:\text{s}}{\sqrt{1-(0.8)^2}} =\dfrac{1\:\text{s}}{\sqrt{1-(0.64)}}[/tex]
[tex]\:\:\:\:\:\:\:=1.67\:\text{s}[/tex]
This means that any event observed by this moving observer will be seen by a stationary observer to occur 67% longer.
b) Given:
t = 1 second
F(t) = 2 seconds
We need to find the speed of the observer such that an event seen by this observer will occur twice as long as seen by a stationary observer. Move the term containing the radical to the left side so the equation becomes
[tex]\sqrt{1-v^2} = \dfrac{t}{F(t)}[/tex]
Take the square of both sides, we get
[tex]1 - v^2 =\dfrac{t^2}{F^2(t)}[/tex]
Solving for v, we get
[tex]v^2 = 1 - \dfrac{t^2}{F^2(t)}[/tex]
or
[tex]v = \sqrt{1 - \dfrac{t^2}{F^2(t)}}[/tex]
Putting in the values for t and F(t) we get
[tex]v = \sqrt{1 - \dfrac{(1\:\text{s})^2}{(2\:\text{s})^2}}[/tex]
[tex]v = \sqrt{1 - \dfrac{1}{4}} = \sqrt{0.75}[/tex]
[tex]\:\:\:\:=0.866[/tex]
This means that the observer must moves at 86.6% of the speed of light.
