An elevator is descending with a uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 1.5 m above the floor of the elevator at the time of being dropped. The person observes that the coin strikes the floor after 1 s. What is the acceleration of the elevator to the nearest integer?
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first we must find the acceleration of the coin in the lift using the equation S=ut+1/2*at^2 where-
S = displacement,
u = initial velocity,
t = time taken and
a = acceleration.
Substituting the values in the formula as u = 0, t = 1, and S=1.5, we can find acceleration to be equal to 3 m/s^2. We know that acceleration due to gravity is equal to 9.8 or the closest integral value 10 m/s^2. The motion of the lift relative to the ground is downwards hence the relative acceleration is found as above to be 3. Knowing the acceleration we can calculate acceleration of the lift using the formula-
relative acceleration = Acceleration on body 1 - Acceleration on body 2 (with +ve sign on the acceleration if in the same direction) . Relative acceleration = 10 - 3.
Using this formula we get acceleration to be equal to 7 m/s^2