Question

An intergalactic rock star bangs his drum every 1.30 s. A person on earth measures that the time between beats is 2.50 s. How fast is the rock star moving relative to the earth

Answer 1

v = 0.89 c = 2.67 x 10⁸ m/s

Step-by-step explanation:

The time dilation consequence of the special theory of relativity shall be used here, From time dilation formula we have:

t = t₀/√[1 - v²/c²]

where,

t = time measured by the person on earth = 2.50 s

t₀ = rest time of the intergalactic rock star = 1.30 s

v = relative speed of the rock star = ?

Therefore,

2.5 s = (1.3 s)/√[1 - v²/c²]

√[1 - v²/c²] = 1.3/2.5

√[1 - v²/c²] = 0.52

[1 - v²/c²] = 0.52²

[1 - v²/c²] = 0.2074

v²/c² = 1 - 0.2074

v²/c² = 0.7926

v/c = √0.7926

v = 0.89 c

where,

c = speed of light = 3 x 10⁸ m/s

v = (0.89)(3 x 10⁸ m/s)

v = 0.89 c = 2.67 x 10⁸ m/s