Motion with two constant acceleration phases

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The discussion revolves around calculating the maximum speed and shaft depth of a mine hoist cage that descends with two phases of acceleration. The cage accelerates at 1 m/s² initially and then decelerates at 2 m/s², taking a total of 60 seconds for the descent. Participants suggest using integrals to derive velocity from acceleration, while also mentioning the impulse equation, although it is deemed irrelevant to the problem. The correct answers are a maximum speed of 40 m/s and a shaft depth of 1200 m. Clarification on the calculations and the application of kinematic equations is sought to solve the problem accurately.
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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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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?
 
any help please?
 
i think we have to take the integral of velocity/time, correct?
 
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