Distance vs displacement graph

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Homework Statement


The V-S graph ( x axis as S and y axis as V)of a airplane on a runway is as follows: for 0-100 meters the velocity increases steadily in a straight line from 0-40 m/s. For the remaining 100 -200 meters, the velocity increases steadily from 40 -50m/s in a straight line. The slope is more in the first half. I need to find the acceleration of the plane at S=50m and S=150 m. I also need to draw an A-S graph.

Homework Equations




The Attempt at a Solution


I tried to find out what the slope dv/ds was . Using the chain rule. dv/dt * dt/ds= a/v and that's 1/t. Didn't do anything else productive. Any help?
 

Answers and Replies

  • #2
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Hi!

I think using the chain rule is a good start, but I would do it this way: dv/ds * ds/dt = dv/dt
Now, from the graph we know that dv/ds = K (a constant).
Rewriting dv/dt = d^2s/dt^2, this should give us a solvable differential equation I think. :)
 
  • #3
Orodruin
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that's 1/t
This assumes constant acceleration from ##v = 0## at ##t = 0##, which is not the case in the second part. However, your start is good. You now have
$$
\frac{dv}{ds} = \frac av.
$$
From your information, you should be able to find both ##dv/ds## and ##v## at any given point and from there be able to solve for ##a##.
 
  • #4
DEvens
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Hmmm.... I am getting an interesting answer when I try to work out when the plane started at S=0.
 
  • #5
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This assumes constant acceleration from v=0v = 0 at t=0t = 0, which is not the case in the second part.
Can we split in into two parts with two different constant accelerations and solve it?
1) At 50 m (0-100 interval)
assuming const. acceleration a , v2-u2=2as
a = 1600/100*2
a = 8m/s2
2) 100m (100-200 interval)
const.acceleration a1 v2-u2=2a1s
2500-1600= 2*100*a1
a1=4.5m/s2
this matches the answers in the back of my textbook( ecstasy...) but i don't understand exactly. Why is the acceleration constant (in two separate parts) because dv/ds=1/t? please bear with me, I am completely new to calculus and using it in physics.
 
  • #6
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Rewriting dv/dt = d^2s/dt^2, this should give us a solvable differential equation I think. :)
sorry, i'm kinda new to calculus. I get everything you said , but I don't know how to proceed from d^2s/dt^2. A little help please?
 
  • #7
DEvens
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Can we split in into two parts with two different constant accelerations and solve it?
[snips]
No, the acceleration is not constant.

I will give you a hint. What is the derivative with respect to ##t## of ##e^{At}## ? So the problem has (for the first part) ##V = A S##.
 
  • #8
Orodruin
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No, the acceleration is not constant in either part. You are overcomplicating things. You already have v as a function of s as well as an expression for the acceleration which only involves this function. There is no need to involve time or any other assumptions.
 
  • #9
Orodruin
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No, the acceleration is not constant.

I will give you a hint. What is the derivative with respect to ##t## of ##e^{At}## ? So the problem has (for the first part) ##V = A S##.
In my opinion, involving the time in the solution (other than in applying the chain rule as above) is only an unneseccary complication. There is no need to find the displacement or velocity as a function of time.
 
  • #10
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You already have v as a function of s as well as an expression for the acceleration which only involves this function.
I really do not understand what to do after that. at s=50m , v=20m ? then?
But why is the answer turning out to be right with my previous post ? i checked the back again . Is it just fluke?
 
  • #11
Orodruin
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Yes, it is a fluke that results from the point being chosen in the middle of the interval. The correct approach should give you the same end result. A pretty non-pedagogical task, which gives the same results with the wrong approach ...
 
  • #12
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Yes, it is a fluke that results from the point being chosen in the middle of the interval. The correct approach should give you the same end result. A pretty non-pedagogical task, which gives the same results with the wrong approach ...
Oh. It turns out to be the average acceleration which is in the middle?
Coming to the right method. I need help please, i don't know how to proceed after dv/ds= a/v. At s= 50 m , I suppose v= 20m/s. what then?
 
  • #13
Orodruin
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Oh. It turns out to be the average acceleration which is in the middle?
Coming to the right method. I need help please, i don't know how to proceed after dv/ds= a/v. At s= 50 m , I suppose v= 20m/s. what then?
Then you need to compute dv/ds. This should be easy, since v is a linear function of s.
 
  • #14
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Then you need to compute dv/ds. This should be easy, since v is a linear function of s.
okay, dv/ds for the first half is 40/100=2/5
a/v=2/5 and v=20.Hence v=8m/s2. That right?
Just a few things cleared up:
1) Does the fluke work because it's average acceleration?
2) And i need to draw the a-s graph as well. Must i just use the values of a and s that i computed and draw it ?Or is there a more formal method involving calculus?
 
  • #15
Orodruin
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Well, you have already concluded that dv/ds is constant over each part and you know that v is a linear function of s. What does this mean for a?
 
  • #16
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Well, you have already concluded that dv/ds is constant over each part and you know that v is a linear function of s. What does this mean for a?
Well....is the acceleration is increasing in the first half as the a/v slope is constant and the velocity is increasing ?and then the a/v slope gets smaller so, the acceleration decreases?
 
  • #17
Orodruin
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Well....is the acceleration is increasing in the first half as the a/v slope is constant and the velocity is increasing ?and then the a/v slope gets smaller so, the acceleration decreases?
The easiest way to understand it is to just multiply both sides by v to obtain ##a = v (dv/ds)##. In both segments, ##dv/ds## is constant and ##v## is a linear function. You are correct that the acceleration decreases in the transition between the two sections, but then it increases again as ##dv/ds## is again constant.
 
  • #18
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The easiest way to understand it is to just multiply both sides by v to obtain ##a = v (dv/ds)##. In both segments, ##dv/ds## is constant and ##v## is a linear function. You are correct that the acceleration decreases in the transition between the two sections, but then it increases again as ##dv/ds## is again constant.
Sorry for the late reply. Yes it makes sense for it to increase again. Well that's about it........thank you for all the help and for bearing with my infinite stupidity!!
Merci monsieur Orodruin. Good day.
 
  • #19
Orodruin
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Sorry for the late reply. Yes it makes sense for it to increase again. Well that's about it........thank you for all the help and for bearing with my infinite stupidity!!
Merci monsieur Orodruin. Good day.
There is nothing stupid with spending effort on trying to understand something.
 
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  • #20
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There is nothing stupid with spending effort on trying to understand something.
Well said sir.
 

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