Question

Help me out with these 2 questions for 15 points.

Answer 1

Step-by-step explanation:

The time dilation formula is given by

$F(t) = \dfrac{t}{\sqrt{1-v^2}}$

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

$F(t) = \dfrac{1\:\text{s}}{\sqrt{1-(0.8)^2}}$

$\:\:\:\:\:\:\:=\dfrac{1\:\text{s}}{\sqrt{1-(0.8)^2}} =\dfrac{1\:\text{s}}{\sqrt{1-(0.64)}}$

$\:\:\:\:\:\:\:=1.67\:\text{s}$

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

$\sqrt{1-v^2} = \dfrac{t}{F(t)}$

Take the square of both sides, we get

$1 - v^2 =\dfrac{t^2}{F^2(t)}$

Solving for v, we get

$v^2 = 1 - \dfrac{t^2}{F^2(t)}$

or

$v = \sqrt{1 - \dfrac{t^2}{F^2(t)}}$

Putting in the values for t and F(t) we get

$v = \sqrt{1 - \dfrac{(1\:\text{s})^2}{(2\:\text{s})^2}}$

$v = \sqrt{1 - \dfrac{1}{4}} = \sqrt{0.75}$

$\:\:\:\:=0.866$

This means that the observer must moves at 86.6% of the speed of light.