Let's break down the problem:

1. When Moody walks at his normal speed on the escalator, it takes him 30 seconds to complete the ride.
2. When he walks at twice his normal speed, it takes him only 20 seconds.
3. We want to find the time needed if Moody stands still on the escalator.

Analysis

Let:

• $S$ be the speed of the escalator.
• $W$ be Moody's normal walking speed.
• $L$ be the length of the escalator.

From the problem, when Moody walks at his normal speed: $\text{Time} = \frac{L}{W + S} = 30$

When he walks at twice his normal speed: $\text{Time} = \frac{L}{2W + S} = 20$

We can solve these two equations to find $L$, $W$, and $S$.

Solving the Equations

1. From the first equation: $L = 30(W + S)$

2. From the second equation: $L = 20(2W + S)$

Setting them equal: $30(W + S) = 20(2W + S)$

Simplifying: $30W + 30S = 40W + 20S$ $10S = 10W$ $S = W$

Finding the Time when Moody Stands Still

If Moody stands still, his speed is zero, so the time to finish riding the escalator is simply:

$\text{Time} = \frac{L}{S}$

Substitute $L = 30(W + S)$ and $S = W$:

$\text{Time} = \frac{30(W + W)}{W}$ $\text{Time} = 60 \, \text{seconds}$

Thus, the time needed if Moody stands still is:

60 seconds

Would you like a more detailed explanation or have any follow-up questions?

Relative Questions

1. How can you generalize this formula for different speeds?
2. What if Moody walked at three times his normal speed?
3. How would the time change if the escalator was twice as long?
4. What happens if the escalator's speed increases?
5. Can this approach be used for variable-speed escalators?

Tip

When dealing with relative motion, always set up equations for each scenario to solve for unknown variables efficiently.