Time Dilation spacecraft watches

[crazynut52]

I have the attached equation
but when I plug it, I don't get the same answer as in the back of the book...

Here is the question

Two observers, A on earth, B in a spacecraft whose speed is 2.0X10^8 m/s both set their watches to the same time, when the ship is abreast the earth.

How much time must elapse by A's reckoning before the watches differ by 1.00 seconds?

 

[NEWO]

just a quick thought

you do know that the speed of light is normalised to 1

hence if you are traveling at 2x10^8 ms^-1 you are traveling 2/3 times the speed of light hence your speed is 2/3 relative to light

does this help in anyway

 

[crazynut52]

i understand it is 2/3 the speed of light but am I using the right equation?
using this equation I found that one second passes for A as 1.34 seconds passes to person B. But not sure how the book gets 3.92s as the answer

 

[NEWO]

I can't see the equation yet because it hasn't been moderated however if you are using the following equation

1/sqrt(1-\v^/c^2)


then I get the same answer as you 1.34s

 

[crazynut52]

yes I am using the same equation as you, but the answer in the back of the book is 3.93, not 1.34

 

[Retrogra_de_ing]

Hey,
For that question, my answer is 2.93 for A. If the question asks B's time, then I guess it's 3.93... I don't know... xD

 

[Janus]

i understand it is 2/3 the speed of light but am I using the right equation?
using this equation I found that one second passes for A as 1.34 seconds passes to person B. But not sure how the book gets 3.92s as the answer

Here's the point, you are looking for how much time has passed for A when the difference between A and B is one sec. In the example you just gave, A reads 1.34 sec and B reads 1 sec, for a difference of 0.34 sec.

Hint: you are looking for the value of T₀ when T₁= T₀-1

 

[Gamma]

or in a different notation,

$$\Delta t_A - \Delta t_B = 1 sec$$

You are asked to find $$\Delta t_A$$

 

[crazynut52]

I get it now. Thanks guys I appreciate the help. This semster of modern physics is going to be tough.