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

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

Step-by-step explanation:

Given;

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

`((speed \ of \ spaceship)/(speed \ of \ Earth)=(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)\\\\(V_(space))/( V_(Earth)) = (t_(Earth))/(t_(space))\\\\ 0.96 = (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