if an elevator starts from rest and rises with an acceleration of +0.29m/s^2. If each floor is 3m high,
a) how long does it take to pass the 12th floor?
b) how fast is the elevator going when it passes the 5th floor?
distance=1/2 a t^2
time=sqrt(2*3*12/.29)
vf=at=.29*time where time is sqrt(2*3*5 /.29)
To solve these problems, we can use the equations of motion. The general equation of motion for an object under constant acceleration is:
\(d = ut + \frac{1}{2}at^2\)
where:
d = distance traveled
u = initial velocity
a = acceleration
t = time
To solve for time (t), we can rearrange the equation as follows:
\(t = \sqrt{\frac{2d}{a}}\)
Now let's solve each part of the problem:
a) To find the time it takes to pass the 12th floor, we need to calculate the distance traveled. Each floor is 3m high, so the 12th floor is at a height of 12 × 3m = 36m.
Substituting the values into the equation:
\(t = \sqrt{\frac{2 \times 36}{0.29}}\)
Using a calculator, we find \(t ≈ 10.95s\). Therefore, it takes approximately 10.95 seconds to pass the 12th floor.
b) To find the speed of the elevator when it passes the 5th floor, we need to calculate the time it takes to reach the 5th floor and then use that time to calculate the speed.
The distance traveled to reach the 5th floor is \(5 \times 3m = 15m\).
Using the equation \(t = \sqrt{\frac{2d}{a}}\) and substituting the values:
\(t = \sqrt{\frac{2 \times 15}{0.29}}\)
Using a calculator, we find \(t ≈ 2.13s\). Therefore, it takes approximately 2.13 seconds to pass the 5th floor.
Now that we have the time, we can find the speed using the equation:
\(v = u + at\)
As the elevator starts from rest, the initial velocity (u) is 0. Substituting the values:
\(v = 0 + 0.29 \times 2.13\)
Calculating this, we find \(v ≈ 0.617m/s\). Therefore, the elevator is going approximately 0.617 m/s when it passes the 5th floor.