Question 277012: In order to access the outer part of Terminal 1 at O'Hare International Airport, you must walk quite some distance in a tunnel that travels under part of the airport. To make the walk less difficult, there is a moving walkway that travels at 2 feet per second. Suppose that Hana can travel 152 feet while walking on the walkway in the same amount of time it taker her to travel 72 feet while walking on the pavement without the aid of the moving sidewalk. How fast does Hana walk?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! rate * time = distance.
on the walkway, hana can walk 152 feet in the same time it takes her to walk 72 feet when she's not on the walkway.
let x = the rate per second that Hana walks when she is not on the walkway.
since the walkway moves at 2 feet per second, then Hana's rate on the walkway is x + 2 feet per second.
let T = the amount of time it takes Hana to walk both distances.
The first equation is:
x * T = 72 feet.
Th second equation is:
(x+2) * T = 152 feet.
From the first equation, you get x = 72/T
Substitute that for x in the second equation to get:
(72/T + 2) * T = 152
Simplify by removing parentheses to get:
72/T * T + 2*T = 152
Simplify further to getr:
72 + 2*T = 152
Substract 72 from both sides of the equation to getr:
2*T = 152-72 = 80
Divide both equations by 2 to get T = 40 seconds.
Now that we have T, we can solve for x.
From the first equation, we get:
40 * x = 72
Divide both sides of the equation by 40 to get:
x = 72/40 = 1.8 feet per second.
From the second equation, we get:
(x+2) * 40 = 152
Solve for x to get:
x = 1.8 as it should, so the answer is confirmed to be good.
If x = 1.8 feet per second, then x + 2 = 3.8 feet per second.
72/1.8 = 40
152/3.8 = 40
Answer is confirmed again because T is equal to 40 in both cases as it should based on the problem statement.
Your answer is:
Hana walks at 1.8 feet per second.
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