Motion with two constant acceleration phases

In summary, the cage of a mine hoist starts from rest and descends the top part of the shaft with a constant acceleration of 1m/s^2. It then decelerates at a constant 2m/s^2 for the remainder of the shaft until it stops at the bottom. The total time taken for the descent is 60s. Using the impulse equation, the maximum speed of the cage is 40 m/s and the depth of the shaft is 1200m. The initial velocity is 0 and the acceleration is 1 m/s^2 for the first part, and -2 m/s^2 for the second part. To solve, we can use integrals and derive v/d from the acceleration
  • #1
madahmad1
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1.The cage of a mine hoist starts from rest and descends the top part of the shaft with a constant acceleration of 1m/s^2. For the remainder of the shaft the cage decelerates at a constant 2m/s^2 so that it stops at the bottom of the shaft. The total time taken for the descent is 60s. Determine the cage`s maximum speed and the shaft depth. The Answer is 40 m/s , 1200m but i do not know how i got it. We have to use the impulse equation, which states that Force= mass x time
 
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  • #2
nevermind the question is not about impulse, it is dealing with power. Anybody got any idea how to solve it? initial velocity is zero, accleration is 1, acceleration 2 is -2, time is 60 s. I am guessing we use integrals, and derive v/d from acceleration, correct?
 
  • #3
any help please?
 
  • #4
i think we have to take the integral of velocity/time, correct?
 

1. What is motion with two constant acceleration phases?

Motion with two constant acceleration phases is when an object moves with a constant acceleration for a certain amount of time, then changes to a different constant acceleration for the rest of its motion.

2. How is motion with two constant acceleration phases different from uniform acceleration?

Motion with two constant acceleration phases is different from uniform acceleration in that the acceleration is not constant throughout the entire motion. It changes at a certain point in time, resulting in a different acceleration for the remaining time.

3. What is the equation for calculating velocity in motion with two constant acceleration phases?

The equation for calculating velocity in motion with two constant acceleration phases is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

4. How do you determine the total displacement in motion with two constant acceleration phases?

To determine the total displacement in motion with two constant acceleration phases, you can use the equation s = ut + (1/2)(v + u)t, where s is the total displacement, u is the initial velocity, v is the final velocity, and t is the total time.

5. Can motion with two constant acceleration phases occur in real-life scenarios?

Yes, motion with two constant acceleration phases can occur in real-life scenarios. For example, a car accelerating at a constant rate, then decelerating at a constant rate when approaching a stop sign, is an example of motion with two constant acceleration phases.

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