Force/time graph acceleration problem

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Homework Statement


[/B]
Figure 5-55 gives, as a function of time t, the force component
Fx that acts on a 3.00 kg ice block that can move only along
the x axis. At t = 0, the block is moving in the positive direction of
the axis, with a speed of 3.0 m/s.What are its (a) speed and (b) direction
of travel at t = 11 s?

problem63.png


Homework Equations


[/B]
##\vec{F}=m \vec{a}##
##a_{avg}= \frac{a_1-a_0}{t_1-t_0}##
##V = V_0 + \vec{a}t##

The Attempt at a Solution



So, if I understand the problem correctly the graph is giving me Fx as a funcion of t, assuming that Fx is the only horizontal force acting on the block a should be given by:

##\vec{F}=m \vec{a}## → ##\vec{a}= \frac {\vec{F}}{m}##

And I think is safe to use the slope formula for the non-horizontal changes in a because they are just a line at a slope, so aavg should be equal to a
My thought process for the problem is that there is not a "smart way" to solve it, we have to just evaluate V from start to finish in order to get V11

So, starting with the problem:

V0 = 3 m/s
m = 3 Kg

a0 = 2/3 m/s2
a2 = 2 m/s2
aavg0/2 = 2/3 m/s2
a2/5 = 2 m/s
aavg5/6 = -2 m/s2
aavg6/9 = -4/3 m/s2
aavg9/11 = 4/3 m/s2

Now that we have the change in acceleration for each part of the graph we can simply calculate the final velocity (I'm using unnecessary parenthesis in the formula just to make more clear the different time frames):

V11 = 3 m/s + (2/3 m/s2 ⋅ 2s) + (2 m/s ⋅ 3s) + (-2 m/s2 ⋅ 1s) + (-4/3 m/s2 ⋅ 3s) + (4/3 m/s2 ⋅ 2s) = 7 m/s

So the answer should be
(a) s11 = 7 m/s
(b) Positive direction


But the answer in the book (Halliday,Resnik, Walker 9th Ed) is 8 m/s, and I really have no idea what could be wrong in my thought process and calculation, any clue?
 

Answers and Replies

  • #2
CWatters
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My thought process for the problem is that there is not a "smart way" to solve it,
Have you studied the impulse-momentum theorem?
 
  • #3
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No, if it helps the problem is in the "Force and motion 1" section, and I assume you are supposed to learn what you suggested later on
 
  • #4
CWatters
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  • #5
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I could do it this way and I will likely try to do so later, but I'm pretty sure the problem is intended to be solvable without that information, I'd assume you learn about that theorem in the kinetic energy section, which comes later on.

Anyway, do you have any idea about solving this problem without that theorem?
I would like to understand the thought process behind solving it with just the knoledge I'm supposed to have at this point more than just solving it easily because it becomes trivial once you have more informations
 
  • #7
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OK I think this is the other way to solve it but it's complicated. I'll have another look when I get back home..
By using calculus it's easy, we know that the rate of change of acceleration is constant (from shape) => let da/dt=j => a=jt (no initial rate of change of acceleration) integrate with respect to time => v=1/2jt2 but a=jt => j=a/t so v=1/2at. But I don't think he is supposed to do this way
With initial rate of change of acceleration v=j0+1/2at
 
  • #8
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By the way, I didn't had time to post this earlier but I understood the problem in my logic, of course it's wrong to use the average formula, because the acceleration is not constant, in this case it's hard for me to find a solution that avoids calculus
 
  • #9
haruspex
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in this case it's hard for me to find a solution that avoids calculus
Yes, it is an integral (of acceleration wrt time, or, equivalently, of force wrt time), but because it's all straight lines you do not need any calculus algebra to do the integrals. Just take the areas.
 
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  • #10
jbriggs444
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Pretty sure you can't use average acceleration to work out the velocity.
Pretty sure you can.
 
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  • #11
CWatters
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Oops you're correct. I've checked and it gives the right answer.
 
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  • #12
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So my thought process was actually correct but I made some misteps in the calculations?

If yes, I can't really come up with the algebraic mistakes I made, and at the moment I am not even sure why my original thoughs are actually ok
 
  • #13
CWatters
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OK so..

V11 = 3 m/s + (2/3 m/s2 ⋅ 2s....etc)
The average acceleration in phase 1 isn't 2/3m/s2. It starts at 2/3m/s2 and ends at 6/3m/s2 so the average is 4/3m/s2

There are also 6 phases as the slope between 5 and 6 isn't the same as 6 and 7 seconds.

The acceleration between 5 and 6 is +ve. The force maybe reducing but it's still +ve until 6 seconds.
 
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  • #14
jbriggs444
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a0 = 2/3 m/s2
a2 = 2 m/s2
aavg0/2 = 2/3 m/s2
What is the average of 2/3 and 2?

aavg5/6 = -2 m/s2
Huh? That acceleration is positive and the average is not two. Did you switch from calculating averages to calculating slopes?
aavg6/9 = -4/3 m/s2
Huh? There is a bend in the acceleration graph between 6 and 9.
aavg9/11 = 4/3 m/s2
The acceleration over this part of the graph is negative, not positive and the average is not 4/3.
Now that we have the change in acceleration for each part of the graph
You have the average rate of change in velocity for each part of the graph. Which has to be multiplied by time. [The equation shows that you knew that. You just didn't say it right]
 
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  • #15
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OK so..



The average acceleration in phase 1 isn't 2/3m/s2. It starts at 2/3m/s2 and ends at 6/3m/s2 so the average is 4/3m/s2

There are also 6 phases as the slope between 5 and 6 isn't the same as 6 and 7 seconds.

The acceleration between 5 and 6 is +ve. The force maybe reducing but it's still +ve until 6 seconds.
I have some questions

You say that a in phase 1 is 4/3m/s2, may I ask why?
The Δa is 4/3m/s2, but it should be considered during the Δt = 2s if I undestand it correctly, so the total Δv during the 2 seconds is 4/3m/s2,
In fact in my calculations I am adding 4/3m/s2 in total for phase 1.
Are you telling me that the Δv is instead twice as much?

Then, I am considering the slope between 5 and 6 in this part of the equation
(-2 m/s2 ⋅ 1s)

But then I am combining 6/7 with 7/9 in this part of the equation
(-4/3 m/s2 ⋅ 3s)
I skipped some steps, but the reasoning behind it was that in the 6/7 phase a is
##\frac{\frac{-4}{3}-0}{1s} = \frac{-4}{3}##
And in the phase 7/9 a is just -4/3, so it was reasonable for me to combine the two phases in one step

And for the last part, I'm not sure if I'm following you, but in the phase 5/6, despite the force being positive, it should decrease the acceleration because less force is being applied right?
 
  • #16
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What is the average of 2/3 and 2?
I'll do it step by step, I've always used this formula

##a_{avg}=\frac{a_1−a_0}{t_1−t_0}##
So, in the first case I get
##a_{avg}=\frac{2−2/3}{2−0}=2/3##

I guess this will answer all of your questions, am I calculating aavg wrongly?
 
  • #17
haruspex
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I've always used this formula
Time to stop, then. Did you really mean that?
 
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  • #18
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Time to stop, then. Did you really mean that?
Well, I want to punch myself for not using Δv now that i think about it.

My god what a stupid mistake...

So, let's say that the problem is just solved if I just use
##a_{avg}= \frac{v_1-v_0}{t_1-t_0}##
 
  • #19
jbriggs444
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So, let's say that the problem is just solved if I just use
##a_{avg}= \frac{v_1-v_0}{t_1-t_0}##
The starting and ending velocities are what you are trying to find. What you are given are the starting and ending accelerations. For an acceleration with a constant rate of change (a straight line on an acceleration-time graph):

[Corrected, thanks @haruspex]
$$a_{avg}= \frac{a_1+a_0}{2}$$
You use final minus initial if you are looking for a rate of change.
You use final plus initial if you are looking for an average.
 
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  • #20
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Oh, I see.
I tend to get tricked by problems involving average values because I actually get to practice it much less than other problems, and this time I made the wrong assumption that my math was right.


Thanks a lot for the help!
 
  • #21
CWatters
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Using average acceleration or impulse-momentum theorem both involve calculating the area under the curve...

By inspection the areas under each phase are... 8, 18, 3, -2, -8, -4 Newton Seconds. Add them up gives 15NS total.

The object has a mass of 3Kg.

15/3 = 5m/s. Plus the 3m/s it starts with = 8m/S
 
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  • #22
haruspex
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The starting and ending velocities are what you are trying to find. What you are given are the starting and ending accelerations. For an acceleration with a constant rate of change (a straight line on an acceleration-time graph):
$$a_{avg}= \frac{a_1+a_0}{t_1-t_0}$$
You use final minus initial if you are looking for a rate of change.
You use final plus initial if you are looking for an average.
Umm..... I think you mean
$$a_{avg}= \frac{a_1+a_0}2$$
 
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