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The service elevator in a high-rise building travels between the lowest underground parking level and the seventeenth story of the building. The lowest parking level is 15 meters below street level, and the seventeenth story is 51 meters above street level. If the elevator rises at a rate of 2 meters per second, how long, in seconds, could a person ride the elevator when starting from the lowest level? Assume the elevator makes no extra stops.

A. [tex]\( x \leq 21 \)[/tex]
B. [tex]\( x \leq 66 \)[/tex]
C. [tex]\( x \leq 18 \)[/tex]
D. [tex]\( x \leq 33 \)[/tex]



Answer :

GinnyAnswer
Let's solve the problem step by step:

1. Determine the total distance traveled by the elevator from the lowest underground parking level to the seventeenth story of the building.
- The lowest level is 15 meters below street level.
- The seventeenth story is 51 meters above street level.

To find the total distance traveled by the elevator, we add the distance below street level to the distance above street level:
[tex]\[ 15 \text{ meters} + 51 \text{ meters} = 66 \text{ meters} \][/tex]

2. Calculate the time it takes for the elevator to travel this distance.
- The elevator rises at a rate of 2 meters per second.

Using the formula for time, where [tex]\( \text{time} = \frac{\text{distance}}{\text{rate}} \)[/tex]:
[tex]\[ \text{time} = \frac{66 \text{ meters}}{2 \text{ meters/second}} = 33 \text{ seconds} \][/tex]

Therefore, the time it takes for the elevator to travel from the lowest level to the seventeenth story is 33 seconds.

Based on this calculation, the correct answer is:

D. [tex]\( x \leq 33 \)[/tex]

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