Problem 118 - College Algebra Essentials

Question

Use Einstein's special-relativity equation $$R_{a}=R_{f} \sqrt{1-\left(\frac{v}{c}\right)^{2}}$$ described in the Blitzer Bonus on page \(47,\) to solve this exercise. You are moving at \(90 \%\) of the speed of light. Substitute \(0.9 c\) for \(v,\) your velocity, in the equation. What is your aging rate, correct to two decimal places, relative to a friend on Earth? If you are gone for 44 weeks, approximately how many weeks have passed for your friend?

Step-by-Step Solution

Answer

The aging rate \(R_{a}\) is found to be approximately 0.44 when calculated. So, when the 'moving person' is gone for 44 weeks, approximately \(44*0.44 = 19.36\) weeks have passed for the friend on Earth.

Step 1: Input given values into the equation

Replace the given values for \(v\), which is 0.9c (90% of the speed of light), into the equation. This will give: \[R_{a}=R_{f} \sqrt{1-\left(\frac{0.9c}{c}\right)^{2}}\]

Step 2: Simplify the equation

Perform the operation inside the bracket first as according to the order of operation (BIDMAS/BODMAS). Given that the velocity, \(v\), is 90% the speed of light, \(c\), \(v/c = 0.9\). So the equation simplifies to: \[R_{a}=R_{f} \sqrt{1-(0.9)^{2}}\]

Step 3: Solve for the aging rate

Calculate the square of 0.9, subtract it from 1 and find the square root. As per the problem, \(R_{f} = 1\) (friend's aging rate is always one). So, calculate the value for \(R_{a}\). This gives the resulting aging rate.

Step 4: Calculate the time passed for the friend

We know that the 'moving person at 90% speed of light' is gone for 44 weeks. So, for the friend on Earth, elapsed time in weeks would be \(44*R_{a}\). Calculate the value.