A passenger on an interplanetary express bus traveling at v = 0.94c takes a seven-minute catnap by his watch. How long does the nap last from your vantage point on a fixed planet?

t =tₒ/sqrt(1 -β²) = 7/ sqrt(1 -0.94²) =

= 20.5 min

To determine the duration of the catnap from the perspective of an observer on a fixed planet, we need to consider the concept of time dilation. Time dilation occurs when an object or a person moves at a significant fraction of the speed of light (v), resulting in a difference in the perception of time between the moving object and a stationary observer.

The formula for time dilation can be expressed as:

Δt' = Δt / √(1 - (v^2/c^2))

Where:
Δt' is the time interval measured by the observer on the fixed planet,
Δt is the time interval measured by the passenger on the interplanetary express bus,
v is the velocity of the bus relative to the speed of light (c).

Given that the passenger's catnap lasted 7 minutes (Δt = 7 minutes) and the velocity of the bus is 0.94c, we can calculate the duration of the nap from the perspective of the observer on the fixed planet.

Δt' = 7 minutes / √(1 - (0.94c)^2/c^2)

First, we calculate the fraction (v^2/c^2):

(0.94c)^2 = 0.8836c^2

Next, we substitute this value back into the equation:

Δt' = 7 minutes / √(1 - 0.8836c^2/c^2)

Since c^2/c^2 equals 1, the equation simplifies to:

Δt' = 7 minutes / √(1 - 0.8836)

Calculating the square root:

Δt' = 7 minutes / √(0.1164)

Δt' = 7 minutes / 0.3411

Δt' ≈ 20.49 minutes

Therefore, from the perspective of an observer on the fixed planet, the nap duration would be approximately 20.49 minutes.

To determine the duration of the nap from the vantage point of a fixed planet, we can use time dilation.

The equation for time dilation is:
Δt' = Δt / √(1 - (v^2 / c^2))

Where:
Δt' = time interval measured from the fixed planet's frame of reference
Δt = time interval measured by the passenger on the interplanetary express bus
v = velocity of the bus relative to the speed of light (0.94c)
c = speed of light (~3.00 x 10^8 meters per second)

Given that the passenger's nap duration is 7 minutes, we can substitute the values into the equation and solve for Δt':

Δt' = 7 minutes / √(1 - (0.94c)^2 / c^2)

First, let's calculate the value of (v^2 / c^2):

(0.94c)^2 = 0.8836c^2

Then, substituting this value into the equation:

Δt' = 7 minutes / √(1 - 0.8836c^2 / c^2)
Δt' = 7 minutes / √(1 - 0.8836)
Δt' = 7 minutes / √(0.1164)
Δt' = 7 minutes / 0.3409

Simplifying further:

Δt' = 20.5 minutes

Therefore, from the vantage point of a fixed planet, the nap would last approximately 20.5 minutes.