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Multiple Choice
Carol is in the same reference frame with a clock. Bianca is flying past Carol and her clock at a high speed. Bianca sees Carol's clock ticking at one quarter the rate that Carol sees. How fast is Bianca flying relative to Carol?
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Verified step by step guidance
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Start by understanding the concept of time dilation in special relativity, which states that time observed in a moving reference frame is slower compared to a stationary reference frame. This is described by the time dilation formula: \( t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( t' \) is the dilated time, \( t \) is the proper time, \( v \) is the relative velocity, and \( c \) is the speed of light.
In this problem, Bianca observes Carol's clock ticking at one quarter the rate that Carol sees. This means \( t' = 4t \). Substitute this into the time dilation formula: \( 4t = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \).
Simplify the equation by dividing both sides by \( t \): \( 4 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \).
To solve for \( v \), first square both sides of the equation to eliminate the square root: \( 16 = \frac{1}{1 - \frac{v^2}{c^2}} \).
Rearrange the equation to solve for \( v^2 \): \( 1 - \frac{v^2}{c^2} = \frac{1}{16} \). Then, \( \frac{v^2}{c^2} = 1 - \frac{1}{16} \). Solve for \( v \) to find the relative speed of Bianca with respect to Carol.