Question

The Starship Enterprise speeds past an asteroid at 0.920c. If an observer on the asteroid sees 10.0 seconds pass on her watch, how long would that time interval be if measured by an observer on the Enterprise?

Answer 1

25.52 seconds

Step-by-step explanation:

Speed of space ship = 0.92 c = v

c = Speed of light

Time observed from asteroid = Δt = 10 seconds

Time dilation

`\Delta t'=\frac{\Delta t}{\sqrt{1-\frac {v^2}{c^2}}}\\\Rightarrow \Delta t'=\frac{10}{\sqrt{1-\frac {0.92^2c^2}{c^2}}}\\\Rightarrow \Delta t'=(10)/(√(1-0.92^2))\\\Rightarrow \Delta t'=25.52\ s`

∴ Time interval be if measured by an observer on the Enterprise would be 25.52 seconds