An elevator is decending with uniform acceleration . To measure the acceleration , a person in the elevator drops a coins at he moments the elevator starts. The coin 6 ft above the floor of the elevator at the time it is dropped . The person find that the coin take 1 second to strike the floor . From these data what is acceleration of the elevator ?

The distance formula for an object moving under a uniform acceleration is:

d = (1/2)*a*t

Where d is distance, a is acceleration, and t is time. In this case, there are two accelerations:1) the acceleration due to gravity (g = 32 ft/s

d = (1/2)*(g-a)*t

6 = (1/2)*(32-a)*(1 sec)

Solve for a:

12 = 32 - a

d = (1/2)*a*t

^{2}Where d is distance, a is acceleration, and t is time. In this case, there are two accelerations:1) the acceleration due to gravity (g = 32 ft/s

^{2}), and 2) the acceleration of the elevator (a). The net acceleration is (g-a). The distance formula becomes:d = (1/2)*(g-a)*t

^{2}6 = (1/2)*(32-a)*(1 sec)

^{2}Solve for a:

12 = 32 - a

**20 ft/s**^{2}= a