Question

A space cowboy wants to eject from his spacecraft 1.00×105 km after passing a space buoy, as seen by spectators at rest with respect to the buoy. To do this, the cowboy sets a timer on his craft that will start as he passes the buoy. He plans to cruise by the buoy at 0.217c.How much time should he allow between passing the buoy and ejecting?options:1.50 s1.70 s1.30 s1.60 s1.55 s

Answer 1

option D.

Step-by-step explanation:

Given,

distance, d = 1 x 10⁵ km

speed , v = 0.217 c

time of dilation , T₀= ?

Using the formula of time dilation

`T=\frac{T_(0)}{\sqrt{1-(v^(2))/(c^(2))}}`

`T_(0)=T \sqrt{1-(v^(2))/(c^(2))}`

`=\left((d)/(v)\right) \sqrt{1-(v^(2))/(c^(2))}`

`=\left((1.00 * 10^(8))/(0.217 * 2.99 * 10^(8))\right) \sqrt{1-((0.217 c)^(2))/(c^(2))}`

`=1.54 * 0.976`

`=1.50\ s`

Time he should be allowed between passing the buoy and ejecting is equal to 1.50 s.

The correct answer is option D.