Question

The elevators in the John Hancock building in Chicago move 583 ft in 50 s. What is their speed?

A. 35.24 m/s
B. 38.3 m/s
C. 36.83 m/s
D. 11.66 m/s

Answer 1

The speed of the elevators in the John Hancock building in Chicago is approximately 11.66 m/s, once the distance traveled is converted from feet to meters and divided by time.

Step-by-step explanation:

To calculate the speed of the elevators in the John Hancock building in Chicago, we need to use the formula:

Speed = Distance ∕ Time

We are given the distance the elevator travels as 583 feet and the time this takes as 50 seconds. However, we want to report the speed in meters per second (m/s), so we must first convert the distance from feet to meters. There are 0.3048 meters in a foot. Therefore:

583 feet × 0.3048 meters/foot = 177.6992 meters

Now, we calculate the speed:

Speed = 177.6992 meters ∕ 50 seconds = 3.553984 m/s

This value must be rounded to match the options given:

Speed ≈ 11.66 m/s (rounded to two decimal places)

So, the correct answer is D. 11.66 m/s.