A passenger on an interplanetary express bus traveling at = 0.95 takes a 9.0-minute catnap, according to her watch.How long does her catnap from the vantage point of a fixed planet last?
You need to tell us what the units of velocity are. I assume that the "bus" is travelling at 0.95c, or 95% of the speed of light, with respect to a "fixed" planet - presumably the home planet where the passenger came from.
Someone is apparently trying to teach you the theory of relativity. That person should know that it is measingless to say something is "fixed" unless a frame of reference is specified.
You need to use the "time dilation" formula of Special Relativity. It is probably explained in your textbook or class notes. A tutorial on the subject can easily be found with Google.
The nine minute time interval measured on the watch of the traveler will appear longer on the planet, by a factor
1/sqrt[1 - (0.95)^2] = 3.20
That means the nap will be 28.8 minutes long as measured on the home planet.
Well, let me calculate that for you... Okay, it seems like her catnap from the vantage point of a fixed planet would last around... "nine minutes"! And you know what they say about time flying when you're having fun, right? So, this passenger must have had a pawsitively delightful catnap!
To determine the duration of the catnap from the viewpoint of a fixed planet, we need to consider time dilation. Time dilation occurs due to the relative motion between two frames of reference. The formula to calculate time dilation is:
t' = t / γ
Where:
t' is the time experienced in the frame of reference of the fixed planet
t is the time experienced in the frame of reference of the interplanetary express bus
γ is the Lorentz factor, given by γ = 1 / √(1 - v²/c²)
Given:
v = 0.95c (c is the speed of light, approximately 3 x 10^8 m/s)
First, let's calculate the Lorentz factor (γ):
γ = 1 / √(1 - (0.95c)²/c²)
γ = 1 / √(1 - 0.9025)
γ = 1 / √0.0975
γ = 1 / 0.3122499
γ ≈ 3.204
Now, we can calculate the duration of the catnap (t'):
t' = t / γ
t' = 9.0 minutes / 3.204
t' ≈ 2.81 minutes
Therefore, from the perspective of a fixed planet, the catnap would last approximately 2.81 minutes.
To determine the duration of the catnap from the vantage point of a fixed planet, we need to consider time dilation. Time dilation occurs when an object is moving at high speeds relative to another object, causing time to pass differently for each observer.
In this scenario, the passenger is moving at a speed of 0.95c (c being the speed of light) relative to the fixed planet. The time dilation formula can be used to calculate the dilation factor (γ) as follows:
γ = 1 / sqrt(1 - v^2/c^2)
where v is the velocity of the passenger (0.95c) and c is the speed of light.
Substituting the values:
γ = 1 / sqrt(1 - 0.95^2)
γ ≈ 3.2
This means that time will appear to pass 3.2 times slower from the perspective of the fixed planet compared to the passenger.
Now, we can calculate the duration of the catnap from the fixed planet's vantage point. The passenger's watch indicates a duration of 9.0 minutes. However, from the fixed planet's perspective:
Duration on the fixed planet = Duration on the passenger's watch / γ
Duration on the fixed planet = 9.0 minutes / 3.2
Duration on the fixed planet ≈ 2.8 minutes
Thus, the catnap from the vantage point of the fixed planet would last approximately 2.8 minutes.