Question

An intergalactic rock star bangs his drum every 1.50 s. A person on earth measures that the time between beats is 2.70 s. How fast is the rock star moving relative to the earth?

Answer 1

v = 83.1 % of speed of light

Step-by-step explanation:

given,

T_e is the earth time = 2.7 s

T_s is the ship time = 1.5 s

we know,

`T_s = T_e * \gamma`

where c is the speed of light

v is the speed of the rock star moving

`T_s = T_e* \sqrt{1-(v^2)/(c^2)}`

`1.5= 2.7* \sqrt{1-(v^2)/(c^2)}`

`\sqrt{1-(v^2)/(c^2)} =0.556`

squaring both side

`1-(v^2)/(c^2)=0.3086`

`v^2=0.6914c^2`

v = 0.831 c

v = 83.1 % of speed of light