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Motion with two constant acceleration phases

  1. Mar 25, 2008 #1
    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
     
  2. jcsd
  3. Mar 26, 2008 #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?
     
  4. Mar 27, 2008 #3
    any help please?
     
  5. Mar 30, 2008 #4
    i think we have to take the integral of velocity/time, correct?
     
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