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

In summary, the 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.
  • #1
D.Man Hazarika
7
0

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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  • #2
You need to set up a system of equations. You have the time and acceleration.
 
  • #3
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.
 
  • #4
Please solve it
 
  • #5
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.
 
  • #6
Ok, I solved it, now you solve it.
 
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Likes D.Man Hazarika

1. What is the equation for calculating the velocity of the lift at a given time?

The equation for calculating the velocity of the lift at a given time can be expressed as v = u + at, where v is the final velocity, u is the initial velocity (which in this case is 0 as the lift starts from rest), a is the acceleration, and t is the time.

2. How do you calculate the acceleration of the lift?

The acceleration of the lift can be calculated by using the equation a = (v - u)/t, where v is the final velocity, u is the initial velocity, and t is the time taken for the lift to reach that velocity.

3. Is the acceleration of the lift constant?

Yes, the acceleration of the lift is constant as it is ascending with a uniform acceleration, meaning that the rate of change of velocity is constant.

4. Can you use the same equations to calculate the displacement of the lift?

Yes, the equations of motion can also be used to calculate the displacement of the lift at a given time. The equation for displacement is s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

5. How can you determine the time taken for the lift to reach a certain velocity?

The time taken for the lift to reach a certain velocity can be calculated by rearranging the equation v = u + at to t = (v - u)/a. Simply plug in the values for v (desired velocity), u (initial velocity), and a (acceleration) to find the time taken.

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