Question

A space cowboy wants to eject from his spacecraft 2.00×105km 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 0.347 c.

How much time should he allow between passing the buoy and ejecting?

Answer 1

1.80s
2.00s
1.60s
1.90s
1.85s

These are the answers.

I was trying to be simply, and use d=vt, and got hte answer t= 1.921, so I though the answer may be 1.90s. Anyone else?

Answer 2

UM, *becomes the first one to answer* im sure dude.

Answer 3

Have you covered in class the theory of Special Relativity.

Answer 4

Yes.

Answer 5

Says its not 1.92 though.

Answer 6

i mean 1.90, to which my answer is closest too

Answer 7

then let us calculate it

Answer 8

what is the unit of the other information?

Answer 9

Here's what I did:
t=d/v= 2*10^8/0.347C=2*10^8/1.041*10^8

Answer 10

what is C?

Answer 11

I assumed C to be the speed of light.

Answer 12

where c is 3.00*10^8

Answer 13

you multiply the values?

Answer 14

Then the appropriate equation to start with is:\[x'=\gamma (x-vt)\]Now just solve that for t:\[t=\frac{ 1 }{ v }(x-\frac{ x' }{ \gamma })\]where of course:\[\gamma=\frac{ 1 }{ \sqrt{1-\frac{ v ^{2} }{ c ^{2} }} }\]

Answer 15

To be honest, I haven't seen that formula in class. It's on my formula sheet tho

Answer 16

The tip-off here is that the velocity is not very much less than the speed of light. Almost certainly you are going to need to correct for time dilation t = To/sqrt[1-(v/c)^2] or length alteration due to relativistic effects or both.

Answer 17

yeah use psi's suggestion

Answer 18

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html#c3

Answer 19

what's xprime?

Answer 20

hello