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How fast would a 6.0 m-long sports car have to be going past you in order for it to appear only 5.5 m long?
Question by OpenStax is licensed under CC BY 4.0.
$1.2 \times 10^8 \textrm{ m/s}$
Solution Video

OpenStax College Physics Solution, Chapter 28, Problem 13 (Problems & Exercises) (1:43)

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Video Transcript

This is College Physics Answers with Shaun Dychko. A sports car is 6 meters long and we assume that this measurement is done at rest with respect to the sports car. So a tape-measure is, you know, placed down beside it while it's not moving. And then the question asks us, how fast would the sports car have to go in order for it to appear 5.5 meters long as it goes past? So this 5.5 meters is the relativistic length or the contracted length. And so that's <i>L</i> is gonna be 5.5 and <i>L naught</i>—the proper length— is 6 meters. We know that the contracted length is gonna be the proper length times square root of 1 minus <i>v</i> squared over <i>c</i> squared and now we'll have to solve this for <i>v</i>. We are going to divide both sides by the proper length—<i>L naught</i>— and then square both sides. So we have <i>L</i> squared over <i>L naught</i> squared equals 1 minus <i>v</i> squared over <i>c</i> squared. And then we'll move this term to the left side and move this term to the right side by adding or subtracting, as the case maybe, and we have <i>v</i> squared over <i>c</i> squared equals 1 minus <i>L</i> squared over <i>L naught</i> squared. And then multiply both sides by <i>c</i> squared and you have <i>v</i> squared is <i>c</i> squared times 1 minus <i>L</i> squared over <i>L naught</i> squared and then take the square root of both sides So we have the speed—<i>v</i>—then is the speed of light times the square root of 1 minus the contracted length squared divided by the proper length squared. So that's speed of light times square root of 1 minus 5.5 meters squared divided by 6 meters squared giving a speed of 1.2 times 10 to the 8 meters per second.