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

Question

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 ?

Answer 1

Hi Tony,

 

To find the acceleration of the elevator, we must first take into account the acceleration due to gravity, -32 f/s^2. Since the elevator is decending, we will denote it's acceleration as -a_e. The acceleration due to gravity is in the same direction as the acceleration of the elevator, so we use a negative sign for both. Denote the total acceleration as 

 

a=-32-a_e

 

Using the position equation x(t)=(1/2)at^2+vt+x₀, we have:

 

x(t)=(1/2)*(-32-a_e)t^2+0t+6

 

Since it takes one second to hit the ground, x(1)=0. Plugging this in to our equation, we see that

 

x(1)=0

 

      =(1/2)*(-32-a_e)*1^2+6

 

so we need to solve

 

(1/2)*(-32-a_e)+6=0

 

Moving the six over, we have

 

(1/2)*(-32-a_e)=-6

 

Multiplying by two we have

 

-32-a_e=-12

 

Solving the above for a_e yields

 

a_e=-20 f/s^2

 

Answer 2

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

 

d = (1/2)*a*t²

 

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/s²), and 2) the acceleration of the elevator (a).  The net acceleration is (g-a).  The distance formula becomes:

 

d = (1/2)*(g-a)*t²

 

6 = (1/2)*(32-a)*(1 sec)²

 

Solve for a:

 

12 = 32 - a

 

20 ft/s² = a