Homework Help: Velocity Query

1. Nov 7, 2015

alsy

1. The problem statement, all variables and given/known data
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

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

3. The attempt at a solution
v=u+at 0+1.2(4) =4.8 m/s2

2. Nov 7, 2015

SteamKing

Staff Emeritus
If you multiply acceleration (m/s2) by time (s), what are the resulting units?

3. Nov 7, 2015

Staff: Mentor

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.

4. Nov 7, 2015

alsy

v = u +at

I have no deceleration formula

5. Nov 7, 2015

alsy

units would be m/s

6. Nov 7, 2015

SteamKing

Staff Emeritus
You're staring at it and you don't realize it.

Think about the information you are given and what you can calculate.

7. Nov 7, 2015

alsy

ok
deceleration v = u +at

8. Nov 7, 2015

SteamKing

Staff Emeritus
And ...?

9. Nov 7, 2015

alsy

v=u+at 0+1.2(4) =4.8 m/s

10. Nov 7, 2015

SteamKing

Staff Emeritus
Yes, that's the velocity that the elevator reaches after accelerating.

The elevator takes 3 seconds to come to a stop. What's the deceleration?

11. Nov 7, 2015

alsy

v=u+at

8+(1.2)(4)=12.8 m/s

12. Nov 7, 2015

SteamKing

Staff Emeritus
You're very cryptic in describing what these calculations are supposed to be.

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?

13. Nov 7, 2015

HallsofIvy

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?

14. Nov 7, 2015

alsy

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

15. Nov 7, 2015

SteamKing

Staff Emeritus
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".

16. Nov 7, 2015

alsy

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

17. Nov 7, 2015

HallsofIvy

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!

18. Nov 7, 2015

HallsofIvy

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.

19. Nov 7, 2015

SteamKing

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

The elevator stops in 3 seconds. You have to use 3 seconds when calculating the deceleration.

20. Nov 7, 2015

alsy

0= 4.8+ a(4)
0= 4.8 +4A
a=-1.2 m/s

21. Nov 7, 2015

SteamKing

Staff Emeritus
Note: 3 seconds ≠ 4 seconds. You're just repeating the same calculation over and over.

22. Nov 7, 2015

alsy

0= 4.8+ a(3)
0=4.8 +3a
a=-1.6 m/s

23. Nov 7, 2015

vela

Staff Emeritus
Can you explain in words why you set $u=0$, $a = 1.2$, and $t = 4$?

4 what? Where did this 4 come from?

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?

24. Nov 7, 2015

alsy

u=initial velocity
a=acceleration
t=time
8-4=4 constant velocity

25. Nov 7, 2015

HallsofIvy

No, 0= 4.8+ a(3). The deceleration back to 0 speed took 3 seconds, not 4

Solve 0= 4.8+ 3a
(And do NOT switch from "a" to "A".)