Motion with two constant acceleration phases

Click For Summary

Homework Help Overview

The problem involves a mine hoist cage that descends a shaft with two phases of constant acceleration: an initial acceleration of 1 m/s² followed by a deceleration of 2 m/s². The total descent time is given as 60 seconds, and participants are tasked with determining the cage's maximum speed and the depth of the shaft.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevance of different equations, initially mentioning impulse but later suggesting the problem relates to power. There is speculation about using integrals to derive velocity from acceleration, and questions arise about the correct approach to take.

Discussion Status

The discussion is ongoing, with participants exploring various mathematical approaches and clarifying the nature of the problem. Some guidance has been offered regarding the use of integrals, but no consensus has been reached on a specific method to solve the problem.

Contextual Notes

Participants note the initial conditions, including zero initial velocity and the specified accelerations, but there is uncertainty about the correct equations and methods to apply in this context.

madahmad1
Messages
42
Reaction score
0
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
 
Physics news on Phys.org
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?
 
any help please?
 
i think we have to take the integral of velocity/time, correct?
 

Similar threads

Replies
28
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
3
Views
2K
Replies
1
Views
4K
Replies
3
Views
996
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K