Problem on Uniform acceleration

[taichi2910]

Homework Statement

Please can somebody help me,I have got this problem from my college but i have NO clue how to solve the problem, this is how it goes:

A lift rises 50m in 15 seconds . During the first quarttance distance the velocity is uniformly accelerating and during the last tenth portion it is uniformly retarded. if the velocity is constant durning the centre portion of travel calculate:

-1) the velocity in the centre portion of travel
-2) the acceleration and the retardation values
-3) why has question 2 got both positive and negative values



please help me in any way you can by either solving the problems and explanation of solving
Any help would be highly appreciated and thank you in advance.

 

[drizzle]

this should be posted in the Introductory Physics

start with solving 1, find the speed which is constant, first you do need to subtract [¼(15 sec) + 1/10(15 sec)] from 15, as for the last question why would the retardation be negative you think? If it isn’t, then the elevator will crash up there!

 

[rl.bhat]

this should be posted in the Introductory Physics

start with solving 1, find the speed which is constant, first you do need to subtract [¼(15 sec) + 1/10(15 sec)] from 15, as for the last question why would the retardation be negative you think? If it isn’t, then the elevator will crash up there!

In the problem 1/4^th and 1/10^th of the distance is mentioned not the time.
Now the lift starts from rest and acquires a velocity v in quarter distance in time t₁. The average velocity is v/2 and distance moved is d₁. Then d₁= 1/2*vt₁. It moves some distance d₂ with uniform velocity during time t₂ which is equal to vt2. And the remaining distance d₃ during time t₃ which is equal to 1/2*v*t₃. Add these three and solve for v.

 

[drizzle]

In the problem 1/4^th and 1/10^th of the distance is mentioned not the time.
...

gheeeeeeeeeeeeee you're right how did I miss it, still sleepy I think


note: the third part of your solution should be negative, that is -1/2*v*t₃

 

[drizzle]

for a better understanding:

in the quarttance distance part: d₁=[(v₁-v₀)/2]*t₁ ; v₀=0

in mid-distance: d₂=[v₁]*t₂ ; the velocity in midway v₁ is constant

in the tenth portion part: d₃=[(v₃-v₁)/2]*t₁ ; v₃=0, the lift stops

retyping those as follows

d₁=[v₁ /2]*t₁ ==> d₁=1/4[50 m]

d₂=[v₁]*t₂ ==> d₂= 50-[d₁+d₃]

d₃=[-v₁/2]*t₃ ==> d₃=1/10[50 m]

v₁ here is the velocity in the centre portion of travel

what a lucky guy you are taichi2910 you got an answer :tongue:

 

[rl.bhat]

gheeeeeeeeeeeeee your right how did I missed it, still sleepy I think


note: the third part of your solution should be negative, that is -1/2*v*t₃

The average velocity is positive.

 

[drizzle]

The average velocity is positive.

the last velocity value SHOULD be negative so you can calculate the retardation and get a negative value

 

[rl.bhat]

the last velocity value SHOULD be negative so you can calculate the retardation and get a negative value

In the equation v = v_o - at, v_o need not be negative. In the problem the direction of the velocity remains the same from beginning to end.

 

[drizzle]

In the equation v = v_o - at, v_o need not be negative. In the problem the direction of the velocity remains the same from beginning to end.

where did you get the minus from?! in the known kinematic equation it's plus, that is;

v = vo + at

 

[rl.bhat]

where did you get the minus from?! in the known kinematic equation it's plus, that is;

v = vo + at

In the given problem, in the last portion of the motion, the lift is moving with the retardation. The above equation is for that portion of the motion. In your post you have mentioned that the velocity in that portion SHOULD be negative.

 

[drizzle]

In the given problem, in the last portion of the motion, the lift is moving with the retardation. The above equation is for that portion of the motion. In your post you have mentioned that the velocity in that portion SHOULD be negative.

hey rl.bhat, first you prove it mathematically throughout your conclusions in analyzing the motion of the lift, when the (minus) sign appears then you have a physical explanation to it, which is the lift is stopping after 15 secs and it retardates.