Question

A spaceship carrying a light clock moves at a speed of 0.960c relative to an observer on Earth. If the clock on the ship advances by 1.00 s as measured by the space travelers aboard the ship, how long did that advance take as measured by the observer on Earth?

A) 0.96 s
B) 1.2 s
C) 2.6 s
D) 3.6 s
E) 5.8 s

Answer 1

Answer:

(A) The time taken as measured by the observer on Earth is 0.96 s.

Explanation:

Given;

relative speed of spaceship to an observer on Earth, v = 0.960c

$(\frac{speed \ of \ spaceship}{speed \ of \ Earth}=\frac{V}{C} = 0.96 )$  

time of the spaceship as it advances, t = 1.00 s

time of advance observed on Earth, t = ?

Speed = distance / time

Distance = speed x time

Assuming constant distance in the spaceship advancement;

$V_{space}*t_{space} = V_{Earth}*t_{Earth}\\\\\frac{V_{space}}{ V_{Earth}} = \frac{t_{Earth}}{t_{space}}\\\\ 0.96 = \frac{t_{Earth}}{1}\\\\t_{Earth} = 0.96 \ s$

Therefore, the time taken as measured by the observer on Earth is 0.96 s.

The correct option is "A" 0.96 s