Motion with two constant acceleration phases

[madahmad1]

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

 

[madahmad1]

nevermind the question is not about impulse, it is dealing with power. Anybody got any idea how to solve it? initial velocity is zero, acceleration is 1, acceleration 2 is -2, time is 60 s. I am guessing we use integrals, and derive v/d from acceleration, correct?

 

[madahmad1]

any help please?

 

[madahmad1]

i think we have to take the integral of velocity/time, correct?