Maximum Velocity= 4.8 m/s2Deceleration= -1.2m/s2

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Homework Statement


a lift accelerates from rest at 1.2m/s2 for 4 seconds..it travels at a constant velocity for 8 seconds and comes to rest in 3 seconds.calculate 1)the maximum velocity 2)deceleration in the final 3 seconds

Homework Equations


calculate 1)the maximum velocity 2)deceleration in the final 3 seconds

The Attempt at a Solution


v=u+at 0+1.2(4) =4.8 m/s2
 
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alsy said:

Homework Statement


a lift accelerates from rest at 1.2m/s2 for 4 seconds..it travels at a constant velocity for 8 seconds and comes to rest in 3 seconds.calculate 1)the maximum velocity 2)deceleration in the final 3 seconds

Homework Equations


calculate 1)the maximum velocity 2)deceleration in the final 3 seconds

The Attempt at a Solution


v=u+at 0+1.2(4) =4.8 m/s2
If you multiply acceleration (m/s2) by time (s), what are the resulting units?

What about the answer to part 2 of this question?
 
alsy said:

Homework Equations


calculate 1)the maximum velocity 2)deceleration in the final 3 seconds
Those are not relevant equations. They are part of the problem statement.

List the equations that you know which might be relevant to the problem you are solving.
 
v = u +at

I have no deceleration formula
 
ok
deceleration v = u +at
 
v=u+at 0+1.2(4) =4.8 m/s
 
v=u+at

8+(1.2)(4)=12.8 m/s
 
alsy said:
v=u+at

8+(1.2)(4)=12.8 m/s
You're very cryptic in describing what these calculations are supposed to be.

alsy said:

Homework Statement


a lift accelerates from rest at 1.2m/s2 for 4 seconds..it travels at a constant velocity for 8 seconds and comes to rest in 3 seconds.calculate 1)the maximum velocity 2)deceleration in the final 3 seconds

calculate 1)the maximum velocity 2)deceleration in the final 3 seconds

Remember, the elevator accelerates only for the first 4 seconds. After the initial acceleration, the elevator travels at constant velocity for 8 seconds. Does this calculation show that the elevator is traveling at constant velocity?

If velocity is constant, what is the acceleration?
 
What was that in response to? The last question asked was "The elevator takes 3 seconds to come to a stop. What's the deceleration?" Your answer "12.8 m/s" is not an acceleration. It appears you are answering "If a object has initial speed 8 m/s and accelerates at 1.2 m per second per second for 4 seconds, what is its final speed" but I don't see that question anywhere! You have repeatedly written "v= u+ at". Do you understand what "v", "u", "a", and "t" represent?
 
alsy said:
4+(1.2)(4)=8.8 m/s
I'm afraid you've lost the plot and are just typing in random equations here, with little or no understanding of what you are doing.

It's OK to, you know, type in actual words in your reply, to say "I don't understand" or "I need help with this".
 
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v=u+at

I have calculated part one as v=u+at 0+1.2(4) =4.8 m/s

the part 2 deceleration I'm using same formula n

v=u+at and as you say using constant velocity of 4

4+(1.2)(4)=8.8 m/s

that's what I'm getting
 
Once again, what is that in response to and what are you trying to calculate here? It seems that the only question (of the two you originally asked) was "what is the acceleration and, once again, "8.8 m/s" is NOT an acceleration.

Now you appear to be answering "if an object has initial speed 4 m/s and accelerates at 1.2 m per sec per sec for 4 seconds, what is its final speed", but, again, I see no where that question was asked!

Please, when you post a calculation like this, tell us what you are trying to calculate so we can tell you whether it is correct or not!
 
alsy said:
v=u+at

I have calculated part one as v=u+at 0+1.2(4) =4.8 m/s

the part 2 deceleration I'm using same formula n

v=u+at and as you say using constant velocity of 4

4+(1.2)(4)=8.8 m/s

that's what I'm getting
Perhaps you are simply misreading the question.
"u" is the initial speed, "a" is the acceleration (or deceleration if negative), and t is the time the acceleration lasts. In the first part you had u= 0 m/s, a= 1.2 m per sec per sec and t= 8 sec so v= 0 m/s+ (1.2 m/s^2)(4)s= 0+ 4.8 = 4.8 m/s. Yes, that is the correct answer. Now you are just doing the same changing u to 4 m/s?

If you are now doing part b, you are told that it take 3 sec to go back to 0 and are asked to find the acceleration. So now v= 0 m/s, u= 4.8 m/s, and t= 3. So "v= u+ at" becomes 0= 4.8+ a(4). Solve that equation for a.
 
alsy said:
v=u+at

I have calculated part one as v=u+at 0+1.2(4) =4.8 m/s

the part 2 deceleration I'm using same formula n

v=u+at and as you say using constant velocity of 4

Why do you say the constant velocity is 4? You just calculated 4.8 m/s in response to the first question.

4+(1.2)(4)=8.8 m/s

that's what I'm getting

The elevator stops in 3 seconds. You have to use 3 seconds when calculating the deceleration.
 
0= 4.8+ a(4)
0= 4.8 +4A
a=-1.2 m/s
 
0= 4.8+ a(3)
0=4.8 +3a
a=-1.6 m/s
 
alsy said:
v=u+at

I have calculated part one as v=u+at 0+1.2(4) =4.8 m/s
Can you explain in words why you set ##u=0##, ##a = 1.2##, and ##t = 4##?

the part 2 deceleration I'm using same formula n

v=u+at and as you say using constant velocity of 4
4 what? Where did this 4 come from?

4+(1.2)(4)=8.8 m/s
Again, can you explain in words why you set ##u=4##, ##a = 1.2##, and ##t=4## here? You need to understand that the variables and equations have meanings, and you can't just plug in a number because it appears somewhere in the problem and has the right units.

The problem statement asks you to find the "deceleration in the final 3 seconds." Don't you think the information that the deceleration took 3 seconds should appear somewhere in your solution? If you're looking for deceleration, what are you solving for — a position, a velocity, or an acceleration?

that's what I'm getting
 
u=initial velocity
a=acceleration
t=time
8-4=4 constant velocity
 
its correct now

a=1.6 m/s
 

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