How far above the ground is the tack?

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Discussion Overview

The discussion revolves around a problem involving the calculation of how far above the ground a tack is after being picked up by a bike wheel. Participants analyze the geometry and trigonometry involved in determining the height of the tack based on the wheel's diameter and the distance traveled. The scope includes mathematical reasoning and problem-solving related to geometry and circular motion.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • Devon calculates the circumference of the bike tire as 84.78 inches and determines the total distance traveled to be 1200 inches, leading to an initial answer of 0.000876 inches for the height of the tack.
  • Another participant suggests using more digits of pi for accuracy, recalculating the number of revolutions to be 14.147 and proposing a final answer of 5.368 inches.
  • A later reply acknowledges a mistake in the previous calculations, suggesting that the correct relationship should involve a different angle calculation, leading to a revised value of 55.4 degrees.
  • One participant reiterates the circumference calculation and questions the small angle derived from the initial setup, emphasizing the need for accurate calculations.
  • There is a mention of the wheel making 14 revolutions and a fraction of a revolution, with the fraction calculated to be approximately 53.136 degrees.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and methods used to determine the height of the tack, with no consensus reached on the final answer. Multiple competing approaches and corrections are presented throughout the discussion.

Contextual Notes

Some calculations depend on the precision of pi used, and there are unresolved issues regarding the correct relationships between angles and distances in the context of the problem.

veronica1999
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Devon's bike has wheels that are 27 inches in diameter. After the front wheel picks up a tack, Devon rolls another 100 feet and stops. How far above the ground is the tack?

I did all my work carefully but the answer seems weird.

circumference of tire :84.78 inches
total distance traveled is 1200 inches

1200 /84.78 = 14.154

84.78 : 0.154 = 360: X

X = 0.653

13.5cos0.653 = 13.4991

13.5 - 13.4991 = 0.000876

answer : 0.000876
 
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First, it would help to use more digits of pi. Then 1200/(27$\pi$)=14.147 up to 3 decimal digits.

veronica1999 said:
84.78 : 0.154 = 360: X
This should be X / 360 = 0.147 (this is a fraction of the full circle). Then X = 52.958 degrees and the final answer is 5.368 inches.
 
Last edited:
Thank you! :D
Now I see my mistake.

Instead of 84.78 : 0.154 = 360 : X
the equation should have been 84.78 :13.056 = 360 :X
X = 55.4

1200/27pi = 14.147

84.78 : 12.46 = 360 :X
x= 52.958
 
Hello, veronica1999!

Devon's bike has wheels that are 27 inches in diameter.
After the front wheel picks up a tack, Devon rolls another 100 feet and stops.
How far above the ground is the tack?

I did all my work carefully but the answer seems weird.

Circumference of tire: 84.78 inches . This is wrong.
Total distance traveled is 1200 inches

1200 / 84.78 = 14.154

84.78 : 0.154 = 360 : X . This is wrong, too.

X = 0.653 . That is a VERY small angle, isn't it?

13.5cos0.653 = 13.4991

13.5 - 13.4991 = 0.000876

Answer: 0.000876
$\text{Circumference: }\:27\pi \:\approx\:84.82 $

$ \text{Then: }\:\dfrac{1200}{84.82} \:=\:14.1476\text{ revolutions} $

$\text{The wheel makes 14 revolutions and a fraction of a revolution.}$

$\text{That fraction is: }\:0.1476 \times 360^o \:=\:53.136^o $

$\text{Got it?} $
 

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