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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.

User Icon for Step-by-Step Bot Step-by-Step Bot answered
1 year ago

To find the time taken for the cage to reach the top, we need to break down the different phases of the motion.

Phase 1: Acceleration

First, we need to calculate the time taken during the acceleration phase. We can use 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 taken during acceleration phase

Rearranging the equation, we have:

t = (v - u) / a

t = (9 - 0) / 1.5
t = 6 seconds

Phase 2: Constant velocity

During this phase, the velocity remains constant at 9 m/s. To find the time taken during this phase, we can use the equation:

t = distance / velocity

In this case, the distance is equal to the height of the cage, which is 120 m. Thus:

t = 120 / 9
t ≈ 13.33 seconds

Phase 3: Retardation

In this final phase, the cage experiences retardation with an acceleration of -6 m/s^2. The initial velocity is 9 m/s, and the final velocity is 0 m/s. We can use the equation:

v = u + at

where:
v = final velocity = 0 m/s
u = initial velocity = 9 m/s
a = acceleration = -6 m/s^2
t = time taken during retardation phase

Rearranging the equation, we have:

t = (v - u) / a

t = (0 - 9) / -6
t = 1.5 seconds

Total time taken:

To find the total time taken for the cage to reach the top, we sum up the times taken during each phase:

Total time = time during acceleration phase + time during constant velocity phase + time during retardation phase
Total time = 6 seconds + 13.33 seconds + 1.5 seconds
Total time ≈ 20.83 seconds

Therefore, the cage takes approximately 20.83 seconds to reach the top.

User Icon for Explain Bot Explain Bot answered
1 year ago

To find the time taken for the cage to reach the top, we can break the motion into two parts: the accelerated motion and the uniform motion.

First, let's calculate the time taken for the cage to reach a velocity of 9 m/s during the accelerated motion. We can use the equation:

v = u + at,

where
v = final velocity, which is 9 m/s,
u = initial velocity, which is 0 m/s (as the cage starts from rest),
a = acceleration, which is 1.5 m/s^2,
and t = time taken.

Rearranging the equation to solve for t:

t = (v - u) / a.

Substituting the values:

t = (9 - 0) / 1.5 = 6 seconds.

So, the cage takes 6 seconds to reach a velocity of 9 m/s.

Next, let's calculate the time taken during the uniform motion (when the velocity is constant) until the cage nears the top. Since the velocity is constant, we can use the equation:

s = vt,

where
s = distance traveled during uniform motion, which is the remaining height of 120 m,
v = velocity during uniform motion, which is 9 m/s (since it remains constant),
and t = time taken.

Rearranging the equation to solve for t:

t = s / v.

Substituting the values:

t = 120 / 9 = 13.33 seconds (rounded to two decimal places).

So, the cage takes 13.33 seconds during the uniform motion.

Now, let's find the total time taken. During the accelerated motion, the cage takes 6 seconds, and during the uniform motion, it takes 13.33 seconds. Therefore, the total time taken for the cage to reach the top is:

6 seconds (accelerated motion) + 13.33 seconds (uniform motion) = 19.33 seconds (rounded to two decimal places).

Hence, the time taken for the cage to reach the top is approximately 19.33 seconds.