Question

Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.)

Question by OpenStax is licensed under CC BY 4.0
Final Answer

Length: 377 ft.377 \textrm{ ft.} or 4530 in.4530 \textrm{ in.}

Width: 280 ft.280 \textrm{ ft.} or 3300 in.3300 \textrm{ in.}

Solution video

OpenStax College Physics, Chapter 1, Problem 5 (Problems & Exercises)

OpenStax College Physics, Chapter 1, Problem 5 (PE) video thumbnail

In order to watch this solution you need to have a subscription.

Start free trial Log in
vote with a rating of votes with an average rating of .

Calculator Screenshots

  • OpenStax College Physics, Chapter 1, Problem 5 (PE) calculator screenshot 1
  • OpenStax College Physics, Chapter 1, Problem 5 (PE) calculator screenshot 2
Video Transcript
This is College Physics Answers with Shaun Dychko To convert 115 meters into feet we need to have a conversion factor that has meters on the bottom of the fraction so that the meters will cancel between there and the meters that are on top of this fraction... which might sound confusing since you don't see a fraction, but there's always a fraction that's assumed. You can always assume that things can be over 1. In which case you have meters on the top of this fraction and just a 1, which doesn't do much, on the bottom. And then this conversion factor needs to have feet on the top so that when the meters cancel we're left with units of feet. And then we fill in the numbers, and there are 3.281 feet in every 1 meter (assume there's a number 1 there even if no number is written). And that works out to 377 feet, where I've rounded our final answer to 3 significant figures since we had 3 significant figures in the number to start with, and we're doing multiplying. And likewise for width, we have 85 meters times the conversion factor 3.281 feet per meter, and we end up with 280 feet. And this has two significant figures. This zero is just a placeholder zero, and it's not considered significant. And there are two significant figures because 85 has 2 significant figures, and our operation here is multiplying. And then for inches we have the number of feet in the length times 12 inches per foot. So that the feet cancel, leaving us with inches. And we're using 377.315 feet instead of 377 feet because we want to avoid intermediate rounding error. You always use un-rounded numbers in calculations and you only round things just to display as a final answer. But you don't use this rounded final answer in any subsequent calculations. You always use un-rounded numbers in your calculations. So we have the length working out to 4530 inches, again a number with 3 significant figures here. And then for width we have the unrounded 278.885 feet times by 12 inches per foot, and we end up with 3300 inches. Two significant figures there as well.