Question

A passenger on an interplanetary express bus traveling at v 5 0.99c takes a 5-minute catnap, according to her watch. Show that her catnap from the vantage point of a fixed planet lasts 35 minutes.

Answer 1

:

35 minutes (See proof below)

Explanation:

Given

Velocity = v = 0.99c

Time = ∆t =5 minutes

Lorentz Factor is the factor by which time, length, and relativistic mass change for an object while that object is moving and it is given as

y = 1/√(1 - v²/c²) where v = velocity = 0.99c and c = time constant = 1

So,

y = 1/√(1 - (0.99c)²/c²)

y = 1/√(1 - 0.9801c²/c²)

y = 1/√(1 - 0.9801)

y = 1/√0.0199

y = 7.088812050083359

Calculating her catnap;

Catnap Time = ∆t * y

= 5 minutes * 7.088812050083359

= 35.44406025041679 minutes

= 35 minutes ----- Approximated

Answer 2

Answer:

Explanation:

Time = Time of catnap ÷ gamma

gamma = square root(1 - v^2/c^2)

V (speed of light) = 0.99

c (time constant) = 1.00

g = square root(1- 0.99^2/1)

g = square root (1-.99^2)

g = square root ( 1 - 0.9801)

g = 0.141

t' = 5 minutes ÷ 0.141 = 35.46 minutes