Question

An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.

Answer 1

Given:
Distance between the coin and the floor of the elevator before the coin is dropped = 6 ft
Let a be the acceleration of the elevator.
It is given that the coin reaches the floor in 1 second. This means that the coin travels 6 ft distance.
The initial velocity is u for the coin and  zero for the elevator.
Using the equation of motion, we get:
Equation for the coin :

\[s_c = ut + \frac{1}{2}a' t^2 \]

Here,
a' = g − a ( a' is the acceleration felt by the coin.)
g = Acceleration due to gravity
g = 9.8 m/s² = 32 ft/s²
On substituting the values, we get:

\[s_c = \frac{1}{2}\left( g - a \right) \left( 1 \right)^2 \]

\[ = \frac{1}{2}\left( g - a \right)\]

Therefore, we can write:

\[6 = \frac{1}{2} \times \left( 32 - a \right)\]

\[12 = 32 - a\]

\[ \therefore a = 20 {\text{ fts} }^{- 2}\]

Hence, the acceleration of the elevator is 20 ft/s².