The speed of a moving walkway is about 2.2 feet per second. Walking on the moving​ walkway, it takes FredFred a total of 50 seconds to travel 55 feet with the movement of the walkway and then back again against the movement of the walkway. What is Fred'sFred's normal walking​ speed?

Let's denote Fred's normal walking speed as \( x \) feet per second. The speed of the moving walkway is given as 2.2 feet per second.

### Step 1: Define the total speed for each direction.

1. **With the walkway**: Fred’s speed when moving with the walkway is \( x + 2.2 \) feet per second.
2. **Against the walkway**: Fred’s speed when moving against the walkway is \( x - 2.2 \) feet per second.

### Step 2: Calculate the time taken in each direction.

Let:
- \( t_1 \) be the time taken to travel 55 feet with the walkway.
- \( t_2 \) be the time taken to travel 55 feet against the walkway.

The total time for both directions is 50 seconds:

\[
t_1 + t_2 = 50
\]

Now, express the time taken in each direction using distance and speed:

\[
t_1 = \frac{55}{x + 2.2} \quad \text{and} \quad t_2 = \frac{55}{x - 2.2}
\]

### Step 3: Set up the equation.

Substituting the expressions for \( t_1 \) and \( t_2 \):

\[
\frac{55}{x + 2.2} + \frac{55}{x - 2.2} = 50
\]

### Step 4: Solve the equation.

Multiply through by \((x + 2.2)(x - 2.2)\) to eliminate the fractions:

\[
55(x - 2.2) + 55(x + 2.2) = 50(x + 2.2)(x - 2.2)
\]

Simplify:

\[
55x - 121 + 55x + 121 = 50(x^2 - 4.84)
\]

\[
110x = 50x^2 - 242
\]

Rearrange into a standard quadratic form:

\[
50x^2 - 110x - 242 = 0
\]

Divide by 2 to simplify:

\[
25x^2 - 55x - 121 = 0
\]

### Step 5: Use the quadratic formula.

The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substituting \( a = 25 \), \( b = -55 \), and \( c = -121 \):

\[
x = \frac{-(-55) \pm \sqrt{(-55)^2 - 4(25)(-121)}}{2(25)}
\]

\[
x = \frac{55 \pm \sqrt{3025 + 12100}}{50}
\]

\[
x = \frac{55 \pm \sqrt{15125}}{50}
\]

\[
x = \frac{55 \pm 123}{50}
\]

Now, calculate the two possible values for \( x \):

1. \( x = \frac{55 + 123}{50} = \frac{178}{50} = 3.56 \)
2. \( x = \frac{55 - 123}{50} = \frac{-68}{50} = -1.36 \) (not feasible since speed cannot be negative)

### Final Answer:

Fred's normal walking speed is:

\[
\boxed{3.56 \text{ feet per second}}
\]

Would you like any further details or have any questions? Here are some related questions for practice:

1. How would the problem change if the walkway speed increased?
2. What would be Fred's time if he only walked with the walkway?
3. How would the result change if the walkway were 100 feet long?
4. How can the quadratic equation be verified using graphing methods?
5. What would happen if Fred's walking speed matched the walkway's speed?

**Tip:** Always double-check signs and ensure the correct interpretation of physical constraints (like speed cannot be negative) when solving quadratic equations in real-world contexts.

The speed of a moving walkway is about 2.2 feet per second. Walking on the moving walkway, it takes Fred a total of 50 seconds to travel 55 feet with the movement of the walkway and then back again against the movement of the walkway. What is Fred's normal walking speed?

This problem involves calculating Fred's walking speed using the relationship between distance, speed, and time in both directions of a moving walkway. The solution includes solving a quadratic equation, making it suitable for students in Grades 8-10.

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