How to Calculate Time Dilation with Spaceship Speeds | Expert Help

[cyberdx16]

Time Dilation help please

Homework Statement

A spaceship is moving at 0.75 c (0.75 times the speed of light) with respect to the earth. An observer on the spaceship measures the time interval between two events on Earth as 38 hrs. If the spaceship had been moving with a speed of 0.92 c with respect to the earth, what would the time interval between the events have been?

Δt0.92 c = _____hrs

Homework Equations

Δt=Δtp/sqrt1-v^2/c^2

The Attempt at a Solution

v=.75(3e8)=2.25e8m/s
t=1/368e5sec
v=d/t
d=3.078e13m

Δtp=3.078e13/v

v=.92(3e8)=2.76e8


Δtp=3.078e13/2.76e8
=1.115e5sec
=30.97hr

but this is not the correct ans

 

[cyberdx16]

nvm got it

 

[ogb p]

wer. Can someone please help me figure out what I am doing wrong?

To calculate time dilation in this scenario, we need to use the equation Δt = Δtp/√(1-v^2/c^2), where Δt is the time interval as measured by the observer on the spaceship, Δtp is the time interval as measured by an observer on Earth, v is the speed of the spaceship, and c is the speed of light. Using this equation, we can find the time interval as measured by the observer on the spaceship when it is moving at a speed of 0.92c.

First, we need to calculate the time interval as measured by the observer on the spaceship when it is moving at a speed of 0.75c. This can be done by substituting the given values into the equation:

Δt = 38 hrs
v = 0.75c
c = 3e8 m/s

Δt = Δtp/√(1-v^2/c^2)
Δt = 38 hrs/√(1-0.75^2)
Δt = 38 hrs/√(0.4375)
Δt = 38 hrs/0.6614
Δt = 57.52 hrs

This is the time interval as measured by the observer on the spaceship when it is moving at a speed of 0.75c.

Now, we can use this value to find the time interval as measured by the observer on the spaceship when it is moving at a speed of 0.92c. This can be done by substituting the new speed value into the equation:

Δt = Δtp/√(1-v^2/c^2)
Δt = 57.52 hrs/√(1-0.92^2)
Δt = 57.52 hrs/√(0.1464)
Δt = 57.52 hrs/0.3826
Δt = 150.36 hrs

Therefore, the time interval between the events as measured by the observer on the spaceship when it is moving at a speed of 0.92c is 150.36 hours. This is the correct answer.

It is important to note that time dilation is a consequence of special relativity, which states that time is