machines
A winding drum raises a cage through a height of 120 m. the cage has, at first, an accelerationof 1.5 m/s^2 until the velocity of 9 m/s is reached, after which the velocity is constant until the cage nears the top, when the final retardation is 6 m/s^2. Find the time taken for the cage to reach the top
during acceleration for 6s (9/1.5): s = 0.75 6^2 = 27m
during deceleration for 1.5s (9/6): s = 9*1.5 - 3*1.5^2 = 6.75m
total time: 6 + (120-27-6.75)/9 + 1.5 = 17 1/12 seconds
To find the time taken for the cage to reach the top, we can analyze the motion of the cage in three different phases:
1. Acceleration phase: The cage starts from rest and accelerates at 1.5 m/s^2 until it reaches a velocity of 9 m/s.
2. Constant velocity phase: The cage maintains a constant velocity of 9 m/s.
3. Retardation phase: The cage decelerates with a retardation of 6 m/s^2 as it nears the top.
Let's calculate the time taken for each phase separately:
1) Acceleration phase:
We can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Given:
Initial velocity, u = 0 m/s (as the cage starts from rest)
Final velocity, v = 9 m/s
Acceleration, a = 1.5 m/s^2
Using the equation v = u + at and rearranging it to solve for time t:
9 = 0 + (1.5)t
9 = 1.5t
t = 9/1.5
t = 6 seconds
So, the acceleration phase takes 6 seconds.
2) Constant velocity phase:
In this phase, the velocity remains constant at 9 m/s. Since the velocity is constant, the time taken is irrelevant in this phase.
3) Retardation phase:
The retardation phase starts when the cage is near the top. The final retardation is given as -6 m/s^2, which indicates deceleration.
We can again use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case retardation), and t is the time taken.
Given:
Initial velocity, u = 9 m/s
Final velocity, v = 0 m/s (as the cage comes to rest at the top)
Retardation, a = -6 m/s^2
Using the equation v = u + at and rearranging it to solve for time t:
0 = 9 + (-6)t
0 - 9 = -6t
-9 = -6t
t = -9/-6
t = 1.5 seconds
So, the retardation phase takes 1.5 seconds.
Now, to find the total time taken, we add the times taken for each phase:
Total time = Acceleration phase time + Retardation phase time
Total time = 6 seconds + 1.5 seconds
Total time = 7.5 seconds
Therefore, the time taken for the cage to reach the top is 7.5 seconds.
To find the time taken for the cage to reach the top, we can break down the motion into three different phases: acceleration, constant velocity, and retardation.
First, let's find the time taken during the acceleration phase using the equation:
v = u + at
where:
v = final velocity = 9 m/s
u = initial velocity = 0 m/s
a = acceleration = 1.5 m/s^2
t = time
Rearranging the equation, we have:
t = (v - u) / a
t = (9 - 0) / 1.5
t = 6 seconds
So, it takes 6 seconds for the cage to reach a velocity of 9 m/s during the acceleration phase.
Next, during the constant velocity phase, the velocity remains the same. Therefore, the time taken for this phase is simply the distance divided by the constant velocity.
The distance covered during the constant velocity phase can be found using the equation:
s = ut
where:
s = distance = ?
u = initial velocity = 9 m/s
t = time
From the previous calculation, we know that the acceleration phase takes 6 seconds. The total time for the cage to reach the top is not given, so we can assume that the constant velocity phase takes the remaining time.
Therefore, the total time (t_total) can be represented as:
t_total = t_acceleration + t_constant_velocity
t_total = 6 + t_constant_velocity
Now, using the known height of the lift (120 m) and the constant velocity (9 m/s) during the constant velocity phase:
s = ut
120 = 9t_constant_velocity
Rearranging the equation, we have:
t_constant_velocity = 120 / 9
t_constant_velocity = 13.33 seconds (approximately)
Finally, we can find the total time taken for the cage to reach the top by summing up the times for both phases:
t_total = t_acceleration + t_constant_velocity
t_total = 6 + 13.33
t_total ≈ 19.33 seconds
Therefore, it takes approximately 19.33 seconds for the cage to reach the top.