There is an escalator moving upward from floor 1 to floor 2. When Anna walked up the escalator, she counted 10 steps. When Anna ran down the escalator at 2 times her walking speed, she counted 30 steps. How many steps are visible on the escalator when it is stationary?

To solve this problem, we need to establish relationships between the number of steps Anna takes, the speed of the escalator, and the total number of visible steps on the escalator.

### Let’s define the variables:
- \( S \) = Number of visible steps on the escalator when stationary.
- \( v \) = Speed of the escalator in steps per second.
- \( a \) = Anna's walking speed in steps per second.

### Case 1: Anna walking up the escalator
When Anna walks up, she takes 10 steps. During this time, the escalator also moves and covers some steps. The time it takes for Anna to walk these 10 steps is \( \frac{10}{a} \) seconds. During this time, the escalator moves \( v \times \frac{10}{a} \) steps.

Thus, the total distance (in steps) covered by Anna (10 steps) plus the distance covered by the escalator must equal the total number of steps \( S \) on the escalator:
\[
S = 10 + v \times \frac{10}{a} \quad \text{(Equation 1)}
\]

### Case 2: Anna running down the escalator
When Anna runs down the escalator at twice her walking speed, she takes 30 steps. Her speed now is \( 2a \). The time it takes to cover these 30 steps is \( \frac{30}{2a} = \frac{15}{a} \) seconds. During this time, the escalator also moves, but in the opposite direction (downward), covering \( v \times \frac{15}{a} \) steps.

Since Anna is moving against the escalator, the total steps covered by her running down minus the steps covered by the escalator should equal the total number of steps \( S \):
\[
S = 30 - v \times \frac{15}{a} \quad \text{(Equation 2)}
\]

### Solving the System of Equations:
Now, let's solve the system formed by Equation 1 and Equation 2.

#### Step 1: From Equation 1, express \( S \) in terms of \( v \) and \( a \):
\[
S = 10 + \frac{10v}{a}
\]

#### Step 2: From Equation 2, express \( S \) in terms of \( v \) and \( a \):
\[
S = 30 - \frac{15v}{a}
\]

#### Step 3: Set the two expressions for \( S \) equal to each other:
\[
10 + \frac{10v}{a} = 30 - \frac{15v}{a}
\]

#### Step 4: Solve for \( v \):
\[
\frac{10v}{a} + \frac{15v}{a} = 30 - 10
\]
\[
\frac{25v}{a} = 20
\]
\[
v = \frac{20a}{25} = \frac{4a}{5}
\]

#### Step 5: Substitute \( v = \frac{4a}{5} \) back into either Equation 1 or Equation 2 to find \( S \). Using Equation 1:
\[
S = 10 + \frac{10 \times \frac{4a}{5}}{a} = 10 + \frac{40}{5} = 10 + 8 = 18
\]

Thus, the number of visible steps on the escalator when it is stationary is **18 steps**.

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

1. What if Anna's running speed was 3 times her walking speed instead of 2 times?
2. How would the solution change if the escalator was moving downward while Anna walked up?
3. What if Anna took 15 steps instead of 10 while walking up the escalator?
4. How would the time taken to travel up the escalator change if the escalator speed doubled?
5. What if Anna walked at half her usual speed up the escalator? How many steps would she take?

**Tip:** When working with problems involving relative motion, like escalators and moving walkways, always break down the motion into the individual's speed and the object's speed to set up the correct equations.

Explore the solution to a problem involving an escalator where Anna counts steps while walking and running. Understand the relationships between her speed, the escalator's speed, and the total number of visible steps when the escalator is stationary. Suitable for high school students studying relative motion and equations of motion.

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