Difference between revisions of "2002 AMC 10B Problems/Problem 24"

Revision as of 13:48, 26 February 2023

Problem 24

Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?

$\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 7.5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 15$

Solution 1

$[asy] unitsize(1.5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair O=(0,0), A=(0,-20), B=(0,-10), C=(10sqrt(3),-10); real r=20; path ferriswheel=Circle(O,r); draw(ferriswheel); draw(O--A); draw(O--C); draw(B--C); draw(A--C); pair[] ps={A,B,C,O}; dot(ps); label("$O$",O,N); label("$A$",A,S); label("$B$",B,W); label("$C$",C,SE); label("$10$",(O--B),W); label("$10$",(A--B),W); label("$20$",(O--C),NE); [/asy]$

We can let this circle represent the ferris wheel with center $O,$ and $C$ represent the desired point $10$ feet above the bottom. Draw a diagram like the one above. We find out $\triangle OBC$ is a $30-60-90$ triangle. That means $\angle BOC = 60^\circ$ and the ferris wheel has made $\frac{60}{360} = \frac{1}{6}$ of a revolution. Therefore, the time it takes to travel that much of a distance is $\frac{1}{6}\text{th}$ of a minute, or $10$ seconds. The answer is $\boxed{\mathrm{(D) \ } 10}$ . Alternatively, we could also say that $\triangle ABC$ is congruent to $\triangle OBC$ by SAS, so $AC$ is 20, and $\triangle AOC$ is equilateral, and $\angle BOC = 60^\circ$

Solution 2 (trig)

The path that the rider takes along the Ferris wheel can be represented by a sinusoidal graph, where $x$ represents the time in seconds. Since $x=0$ is at the crest of the graph and not at the midline, we will use a cosine graph. Therefore, we will use the form: $f(x) = A\cos(Bx - C) + D.$

The graph starts at the lowest point at $0$ feet, then goes up to reach the highest point at $40$ feet, then comes back down. Therefore, the amplitude is $20$ (and negative since it starts at the bottom, not the top), and the vertical shift is $20$ . There is no horizontal shift since the lowest point is at $x=0$ . It takes $60$ seconds to make one full revolution (the period), so $B = \frac{2\pi}{60} = \frac{\pi}{30}.$ Now we have all the parts we need for the equation of our graph, and we can set it equal to the height we want, $10.$

$10 = -20\cos\left(\frac{\pi}{30}x\right) + 20.$

We get to $\frac{1}{2} = \cos\left(\frac{\pi}{30}x\right)$ and remember that the problem is looking for the first instance of $f(x) = 10$ , so $\frac{\pi}{30}x = \frac{\pi}{3}.$ Solving, we get that $x = \boxed{\mathrm{(D) \ } 10}$ .

~jp06132

Solution 3 (Quick Method)

Seeing as the radius is $20$ feet, a vertical height of $10$ feet means that the rider must be half the length of the radius above the ground. Noticing that the height (Diameter) is $40$ feet and dividing the Ferris wheel into 4 quarters, the rider must travel half of a quarter starting from the bottom to be $10$ feet high. Traveling half of a quarter of the ride is equivalent to traveling an eighth of the entire ride, so the time in seconds is represented by the expression $\frac{1}{8}(60)$ , and therefore our answer is $\boxed{\mathrm{(D) \ } 10}$ .

~Darth_Cadet

@@ Line 38: / Line 38: @@
 ~jp06132
+==Solution 3 (Quick Method)==
+Seeing as the radius is <math> 20 </math> feet, a vertical height of <math> 10 </math> feet means that the rider must be half the length of the radius above the ground. Noticing that the height (Diameter) is <math> 40 </math> feet and dividing the Ferris wheel into 4 quarters, the rider must travel half of a quarter starting from the bottom to be <math> 10 </math> feet high. Traveling half of a quarter of the ride is equivalent to traveling an eighth of the entire ride, so the time in seconds is represented by the expression <math> \frac{1}{8}(60) </math>, and therefore our answer is <math> \boxed{\mathrm{(D) \ } 10}</math>.
+~Darth_Cadet
 == See also ==
 {{AMC10 box|year=2002|ab=B|num-b=23|num-a=25}}
 {{MAA Notice}}

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search