|
|
(21 intermediate revisions by 12 users not shown) |
Line 1: |
Line 1: |
− | ==Problem==
| + | a |
− | A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5 miles per hour, how many hours will it take to cover the one-mile stretch?
| |
| | | |
− | Note: 1 mile = 5280 feet
| + | ==Solution== |
− | | + | IF YOU READ THIS YOU ARE GAY |
− | ==Video Solution for Problems 21-25== | |
− | https://www.youtube.com/watch?v=6S0u_fDjSxc
| |
− | | |
− | ==Solution(s)==
| |
| ===Solution 1=== | | ===Solution 1=== |
− | There are two possible interpretations of the problem: that the road as a whole is <math>40</math> feet wide, or that each lane is <math>40</math> feet wide. Both interpretations will arrive at the same result. However, let us stick with the first interpretation for simplicity. Each lane must then be <math>20</math> feet wide, so Robert must be riding his bike in semicircles with radius <math>20</math> feet and diameter <math>40</math> feet. Since the road is <math>5280</math> feet long, over the whole mile, Robert rides <math>\frac{5280}{40} =132</math> semicircles in total. Were the semicircles full circles, their circumference would be <math>2\pi\cdot 20=40\pi</math> feet; as it is, the circumference of each is half that, or <math>20\pi</math> feet. Therefore, over the stretch of highway, Robert rides a total of <math>132\cdot 20\pi =2640\pi</math> feet, equivalent to <math>\frac{\pi}{2}</math> miles. Robert rides at 5 miles per hour, so divide the <math>\frac{\pi}{2}</math> miles by <math>5</math> mph (because <math>t = \frac{d}{r}</math> and time = distance/rate) to arrive at <math>\boxed{\textbf{(B) }\frac{\pi}{10}}</math> hours.
| + | GYATTTTTT |
− | | |
− | ===Solution 2===
| |
− | If Robert rides in a straight line, it will take him <math>\frac{1}{5}</math> hours. When riding in semicircles, let the radius of the semicircle <math>r</math>, then the circumference of a semicircle is <math>\pi r</math>. The ratio of the circumference of the semicircle to its diameter is <math>\frac{\pi}{2}</math>, so the time Robert takes is <math>\frac{1}{5} \cdot \frac{\pi}{2}</math>, which equals to <math>\boxed{\textbf{(B) }\frac{\pi}{10}}</math> hours.
| |
| | | |
| ==See Also== | | ==See Also== |
| {{AMC8 box|year=2014|num-b=24|after=Last Problem}} | | {{AMC8 box|year=2014|num-b=24|after=Last Problem}} |
| {{MAA Notice}} | | {{MAA Notice}} |
Latest revision as of 16:41, 28 September 2024