How to Find Acceleration from a Velocity-Distance Graph?

In summary, the conversation discusses finding the acceleration of an airplane on a runway at specific points and drawing an A-S graph. The suggested methods involve using the chain rule and solving differential equations, but it is determined that the acceleration is not constant and can be found by using the linear relationship between velocity and distance. The answer in the textbook is a fluke due to the point being chosen in the middle of the interval. The correct approach involves finding the constant slope of the a/v graph and using the linear relationship between velocity and distance to determine the acceleration at specific points. The A-S graph can then be drawn using the computed values of acceleration and distance.
  • #1
takando12
123
5

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?
 
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  • #2
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
takando12 said:
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
Hmmm... I am getting an interesting answer when I try to work out when the plane started at S=0.
 
  • #5
Orodruin said:
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
Alettix said:
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
takando12 said:
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
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
DEvens said:
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
Orodruin said:
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
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
Orodruin said:
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
takando12 said:
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
Orodruin said:
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
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
Orodruin said:
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
takando12 said:
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
Orodruin said:
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
takando12 said:
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
Orodruin said:
There is nothing stupid with spending effort on trying to understand something.
Well said sir.
 

FAQ: How to Find Acceleration from a Velocity-Distance Graph?

1. What is the difference between distance and displacement?

Distance is the total length of the path traveled, while displacement is the straight line distance between the starting and ending points.

2. How are distance and displacement related?

Distance and displacement are related because displacement is a component of the total distance traveled. However, they can have different values depending on the path taken.

3. How is a distance vs displacement graph plotted?

A distance vs displacement graph is plotted by plotting distance on the vertical axis and displacement on the horizontal axis. Each data point represents the distance traveled and the corresponding displacement from the starting point.

4. What does the slope of a distance vs displacement graph represent?

The slope of a distance vs displacement graph represents the average velocity of the object. The steeper the slope, the higher the velocity, and the flatter the slope, the lower the velocity.

5. How can you determine the displacement from a distance vs displacement graph?

The displacement can be determined by finding the x-coordinate of the point where the distance is equal to the total distance traveled. This point represents the displacement from the starting point.

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