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24 Feb 2022, 22:37
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20% (03:03) correct
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wrong
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Let 'r' be the speed in miles per hour at which a wheel, 11 feet in circumference travels. If the time for a complete rotation of the wheel is shortened by \(\frac{1}{4}\) of a second, the speed 'r' is increased by 5 miles per hour. Then 'r' is:
A. 9
B. 10
C. 10.5
D. 11
E. 12
A. 9
B. 10
C. 10.5
D. 11
E. 12
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06 Mar 2022, 21:11
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Let 'r' be the speed in miles per hour at which a wheel, 11 feet in circumference travels. If the time for a complete rotation of the wheel is shortened by \(\frac{1}{4}\) of a second, the speed 'r' is increased by 5 miles per hour. Then 'r' is:
A. 9
B. 10
C. 10.5
D. 11
E. 12
A. 9
B. 10
C. 10.5
D. 11
E. 12
Let's convert 5 mi/hr to ft/s:
5 mi/hr = 5*5280/(60*60) = 22/3 ft/s
Let the original speed be s ft/s
New speed = (s + 22/3) ft/s
Time to travel 11 ft initially = 11/(22/3) = 3/2 seconds
New time = 3/2 - 1/4 = 5/4 seconds
Thus:
11/(s + 22/3) = 5/4
=> 33/(3s + 22) = 5/4
=> 132 = 15s + 110
=> s = 22/15 ft/s
We need to convert this to mi/hr:
r = 22/15 * 3600/5280 = 1 mi/hr
Answer = 1 mile/hr
Please check the answer choices
Posted from my mobile device
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
