Stephanie M. answered • 05/17/15
Tutor
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Degree in Math with 5+ Years of Tutoring Experience
Remember that distance = rate × time. We'll write one equation for Karen while she's walking with the walkway and one equation for Karen while she's walking against the walkway. Call her normal walking speed x.
WITH THE WALKWAY:
Karen walks the distance of the walkway, which is 50 ft.
Karen walks at a rate of x + 2.5, since she's walking normally and benefiting from the extra 2.5 ft/s.
Karen takes some portion of the 40 seconds to walk the walkway, but we're not told what portion. So, call time t.
That means we've got:
50 = (x + 2.5)t
AGAINST THE WALKWAY:
Karen again walks the distance of the walkway, which is 50 ft.
Karen walks at a rate of x - 2.5, since she's walking normally but being pulled backwards an extra 2.5 ft/s.
Karen again takes some portion of the 40 seconds to walk the walkway, but we don't know what portion. We do know that its the rest of the 40 seconds, so it takes her 40 - t seconds.
That means we've got:
50 = (x - 2.5)(40 - t)
Now, we've got a system of equations:
50 = (x + 2.5)t
50 = (x - 2.5)(40 - t)
Solve for t in the first equation, then plug that into the second equation and solve for x:
50/(x + 2.5) = t
50 = (x - 2.5)(40 - t)
50 = (x - 2.5)(40 - 50/(x + 2.5))
50 = (x - 2.5)(40(x + 2.5)/(x + 2.5) - 50/(x + 2.5))
50 = (x - 2.5)((40x + 100)/(x + 2.5) - 50/(x + 2.5))
50 = (x - 2.5)((40x + 100 - 50)/(x + 2.5))
50 = (x - 2.5)((40x + 50)/(x + 2.5))
50(x + 2.5) = (x - 2.5)(40x + 50)
50x + 125 = 40x2 + 50x - 100x - 125
50x + 125 = 40x2 - 50x - 125
125 = 40x2 - 100x - 125
0 = 40x2 - 100x - 250
0 = 4x2 - 10x - 25
Use the quadratic formula to solve for a positive value for x. You'll get x = 4.05 ft/s, Karen's normal walking speed
By the way, the problem is a bit unclear. I decided to make the assumption that the walkway is 50 feet in both directions, not that the total trip across it and back is 50 feet.
