Kinematics problem -- A lift ascends from rest with uniform acceleration

AI Thread Summary
A lift ascends with a uniform acceleration of 4 m/s², covering a total distance of 28 meters in 8 seconds before coming to rest with the same rate of retardation. The problem requires determining the time spent moving at a uniform velocity and calculating that velocity. A hint suggests that the time spent accelerating is equal to the time spent decelerating. The final uniform velocity is given as 4 m/s, and participants are encouraged to show their work rather than request complete solutions. The discussion emphasizes the use of kinematic equations to solve for unknowns in the problem.
D.Man Hazarika
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Homework Statement



A lift ascends from rest with uniform acceleration of 4m/s², then it moves with uniform velocity and finally comes to rest with uniform retardation of 4m/s². If the total distance covered during ascending by the lift is 28m and the total time for ascending 8s respectively, then find the time for which the lift moves with uniform velocity. Also find its uniform velocity.

Homework Equations

The Attempt at a Solution


I am trying to find the time for acceleration, retardation and uniform velocity separately by using equations of motion...but somehow I am not being able to solve it...the information looks inappropriate.
(The answer given is 4m/s)
 
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You need to set up a system of equations. You have the time and acceleration.
 
D.Man Hazarika said:

Homework Statement



A lift ascends from rest with uniform acceleration of 4m/s², then it moves with uniform velocity and finally comes to rest with uniform retardation of 4m/s². If the total distance covered during ascending by the lift is 28m and the total time for ascending 8s respectively, then find the time for which the lift moves with uniform velocity. Also find its uniform velocity.

Homework Equations

The Attempt at a Solution


I am trying to find the time for acceleration, retardation and uniform velocity separately by using equations of motion...but somehow I am not being able to solve it...the information looks inappropriate.
(The answer given is 4m/s)
Well, you should show your best attempt at solution.

Hint: since the elevator starts and stops with the same magnitude of acceleration, namely 4 m/s2, it would be reasonable to assume that the amount of time spent accelerating to the unknown constant velocity is the same amount of time required to come to a stop. You should be able to write some kinematics equations knowing the total time and distance to solve for the unknown constant velocity.
 
Please solve it
 
D.Man Hazarika said:
Please solve it
Sorry, that's your job. The Rules of PF prohibit members from providing full solutions to HW problems in the HW forums.

We've given you some hints on how to find a solution, so take the next step and show us what you can do with this problem.
 
Ok, I solved it, now you solve it.
 
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