The problem describes Moody walking on an escalator. The duration of his ride is influenced by his walking pace and the escalator's velocity. The following variables are established:
The problem provides two scenarios:
- When Moody walks at his standard pace, his combined velocity is \( W + E \). He traverses the escalator in 30 seconds. Therefore, the equation is: \[ (W + E) \times 30 = 1 \]
- When Moody walks at double his standard pace, his combined velocity is \( 2W + E \). He completes the ride in 20 seconds. The equation is: \[ (2W + E) \times 20 = 1 \]
From the first equation: \[ 30(W + E) = 1 \] This simplifies to: \[ 30W + 30E = 1 \] Solving for \( E \): \[ E = \frac{1 - 30W}{30} \] Substitute this expression for \( E \) into the second equation: \[ (2W + \frac{1 - 30W}{30}) \times 20 = 1 \] Simplify the expression inside the parentheses: \[ 2W + \frac{1 - 30W}{30} = \frac{1}{20} \] To eliminate the fraction, multiply the entire equation by 30: \[ 60W + 1 - 30W = \frac{30}{20} \] Combine like terms: \[ 30W + 1 = 1.5 \] Isolate \( 30W \): \[ 30W = 0.5 \] Solve for \( W \): \[ W = \frac{0.5}{30} = \frac{1}{60} \]
Substitute the calculated value of \( W \) into the expression for \( E \): \[ E = \frac{1 - 30 \times \left( \frac{1}{60} \right)}{30} \] Perform the calculation: \[ E = \frac{1 - 0.5}{30} = \frac{0.5}{30} = \frac{1}{60} \]
If Moody remains stationary, his effective velocity is solely the escalator's velocity, \( E \). The time required to cover the escalator's length is calculated as:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{1}{E} = \frac{1}{\frac{1}{60}} = 60 \text{ seconds} \]
If Moody stands still on the escalator, the ride duration will be 60 seconds.
The correct answer is \( \boxed{60} \) seconds.