An elevator starts from rest at the ground floor of the building. It experiences an acceleration a=0.500m/s^2 and reaches a velocity of 5.48m/s. What is the floor reached by the moving elevator, knowing that each floor us 3.00m high?
fast elevator ... unclear question ...
according to the answer key its 10 floors (11 floors)
at which floor does the elevator reach final velocity ...
you have the answer
the elevator is traveling almost two floors per second, so it "reaches" well above the 11th floor ... I repeat ... unclear
To find the floor reached by the moving elevator, we need to calculate the distance it travels.
We can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
We know the initial velocity (u) is 0 m/s because the elevator starts from rest. The final velocity (v) is given as 5.48 m/s, and the acceleration (a) is given as 0.500 m/s^2. We need to find the distance traveled (s).
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Let's substitute the given values:
s = (5.48^2 - 0^2) / (2 * 0.500)
s = 30.0104 / 1.000
s = 30.0104 m
Since each floor is 3.00 m high, we can now determine the floor reached by dividing the distance traveled by the height of each floor:
floor = s / height of each floor
floor = 30.0104 m / 3.00 m
floor = 10.0035
Therefore, the moving elevator reaches the 10th floor.