How Do Acceleration and Velocity Graphs Differ?

[IamVector]

Homework Statement

How can we find distance traveled from an acceleration vs velocity graph??

Relevant Equations

can we take small finite values and then limit them or it has something to do with derivatives?

for example take a look on this graph

 

[phyzguy]

An interesting problem. But before we can help, you have to show some effort. What have you tried? What do you know about the relationship between acceleration and velocity? Can you write an equation for how they are related?

 

[FactChecker]

I don't see how it is possible, or even if an acceleration vs velocity plot would help at all.

 

[phyzguy]

I don't see how it is possible, or even if an acceleration vs velocity plot would help at all.

I believe that it is possible, and there is a fairly simple solution. But we need to hear from the OP first.

 

[jack action]

It is done here, so it is definitively possible. If the acceleration is a simple function of velocity, it can also be easily done analytically, knowing that $v=\frac{\partial x}{\partial t}$, $a=\frac{\partial v}{\partial t}$ and $a = f(v)$, where the distance traveled is $\Delta x = \int \partial x = g(v)$.

If you assume that the acceleration is constant, then it is a special case for $f(v)$ where $f(v) = (0)v + constant$. Try to find $g(v)$ for that case and you should recognize a familiar equation of kinematics.

 

[phyzguy]

Well done! you've given the OP the answer with no effort on their part.

 

[FactChecker]

There certainly are special cases where this can be done, but it can not be done in general. Suppose the acceleration is $a(t) = \sin(t)$ and the velocity is $v(t) = -\cos(t)$. For large times, the 'a versus v' graph will just retrace the same line over and over. It would not be possible to tell where it started or stopped. Any proposed solution would need to rule out this type of behavior and can not be a general solution.

 

[IamVector]

Well done! you've given the OP the answer with no effort on their part.

sorry for late reply our time zones might be different ,
back to the problem answer I got is 40(approx)
assumed a/v = constant(c) thus, dv/dx = c so dx= dv/c therefore distance = area under v/c graph.

 

[phyzguy]

There certainly are special cases where this can be done, but it can not be done in general. Suppose the acceleration is $a(t) = \sin(t)$ and the velocity is $v(t) = -\cos(t)$. For large times, the 'a versus v' graph will just retrace the same line over and over. It would not be possible to tell where it started or stopped. Any proposed solution would need to rule out this type of behavior and can not be a general solution.

Why doesn't the following work in general for a function a(v):
\ddot x = a(v) = a( \dot x) \ddot x = \frac {d \dot x}{dt} = \frac {d \dot x}{dx} \frac{dx}{dt} = \dot x \frac{d \dot x}{dx} \dot x \frac{d \dot x}{dx} = a(\dot x) \int \frac {\dot x}{a(\dot x)} d \dot x = \int dx x = \int \frac {v}{a(v)} dv
Of course, you need to know where the function starts and ends, so if the function re-traces itself you need to break it up into pieces, but if you know the path this should work in general.

 

[FactChecker]

@phyzguy , "Why doesn't the following work in general for a function a(v):"

You are considering acceleration as a well-defined function of velocity. That is not true in general. You may be correct that you can often break it up into piecewise well-defined parts, but that is not how I interpret "using the graph". In general, if there is a long time where a(v)==0, so v == constant, you can not calculate the amount of travel while the velocity is that constant. The time variable is lost in an acceleration versus velocity graph.

 

[haruspex]

@phyzguy , "Why doesn't the following work in general for a function a(v):"

You are considering acceleration as a well-defined function of velocity. That is not true in general. You may be correct that you can often break it up into piecewise well-defined parts, but that is not how I interpret "using the graph". In general, if there is a long time where a(v)==0, so v == constant, you can not calculate the amount of travel while the velocity is that constant. The time variable is lost in an acceleration versus velocity graph.

Doesn't the integral work as long as the acceleration is never zero? Maybe also in some cases where it is instantaneously zero. In the present case, we can take v/a at time zero to be the inverse of the initial slope.

 

[FactChecker]

Doesn't the integral work as long as the acceleration is never zero? Maybe also in some cases where it is instantaneously zero. In the present case, we can take v/a at time zero to be the inverse of the initial slope.

Very possibly. I generally agree with your assertion that it can be solved using correct piecewise functions, as long as the case of a(t)==0 for a long time does not occur. But the piecewise function approach is not how I interpret the use of a graph. Given just a graph, a set of (a,v) points that occur somewhere on the path, maybe multiple times, I don't think that even a piecewise function can reliably be recreated.

 

[haruspex]

Very possibly. I generally agree with your assertion that it can be solved using correct piecewise functions, as long as the case of a(t)==0 for a long time does not occur. But the piecewise function approach is not how I interpret the use of a graph. Given just a graph, a set of (a,v) points that occur somewhere on the path, maybe multiple times, I don't think that even a piecewise function can reliably be recreated.

Since the acceleration is never negative, the velocity increases monotonically. It follows that time also progresses left to right in the graph, and acceleration increases monotonically.
That said, the method is still not trivial. One has to draw a second graph by plotting v/a against v and find the area under that.

 

[FactChecker]

Since the acceleration is never negative, the velocity increases monotonically. It follows that time also progresses left to right in the graph, and acceleration increases monotonically.
That said, the method is still not trivial. One has to draw a second graph by plotting v/a against v and find the area under that.

The OP states that the diagram is just one example. It is a specific situation where you can take advantage of the acceleration remaining positive. That can not be done in the general case. The statement of the problem does not specify that the diagram shown is the only case it is asking about.

Any case where the diagram plot includes a point $(v,a) = (c,0)$, where $c \ne 0$ can not be solved. In that case, there is no way to know how long it stayed in that state and how far it moved. Try to determine how far the object moved in the following plot.

 

[jack action]

sorry for late reply our time zones might be different ,
back to the problem answer I got is 40(approx)
assumed a/v = constant(c) thus, dv/dx = c so dx= dv/c therefore distance = area under v/c graph.

You should show your work in more details. You got 40 in what units? between which initial and final points? Shows us how you got this. As you can see, people are not interested otherwise in your problem, and loose themselves in a different conversation.

Assuming that the area under the curve is the distance traveled works only with a velocity vs time graph because $\Delta x = v \Delta t$ where an imaginary rectangle of height $v$ and width $\Delta t$ gives an area $\Delta x$. This is not the case here.

@phyzguy basically did what I wanted you to do on your own in post #9. When I told you to assume the acceleration is constant, it was an easy example for you to find out what final equation you would get, to better understand how it works. Obviously, the acceleration is not constant in your case, it is a function of velocity.

 

[IamVector]

You should show your work in more details. You got 40 in what units? between which initial and final points? Shows us how you got this. As you can see, people are not interested otherwise in your problem, and loose themselves in a different conversation.

Assuming that the area under the curve is the distance traveled works only with a velocity vs time graph because $\Delta x = v \Delta t$ where an imaginary rectangle of height $v$ and width $\Delta t$ gives an area $\Delta x$. This is not the case here.

@phyzguy basically did what I wanted you to do on your own in post #9. When I told you to assume the acceleration is constant, it was an easy example for you to find out what final equation you would get, to better understand how it works. Obviously, the acceleration is not constant in your case, it is a function of velocity.

well you are absolutely a/v can not be a constant we must find a(v) it can be found by tracing the curve but I am unaware of how I checked out on google but can't get it maybe interpolation can be used.

 

[jack action]

This is why you can also use the more crude approach (as done in the link I provided in post #5):
$$\Delta x = \sum \frac{\bar{v}}{\bar{a}} \Delta v$$
Where $\bar{v}$ and $\bar{a}$ are the average velocity & acceleration during interval $\Delta v$ (which you arbitrarily define).

This method is related to the Midpoint Rectangular Approximation Method shown here.

 

[IamVector]

Where did you get this problem? It looks unphysical to me. If you try something like $a = \alpha v^2$, then we have:
$$\frac{dv}{v^2} = \alpha dt$$ $$v = \frac{v_0}{1 - \alpha v_0t}$$
If $v_0 = 0$, then $v(t) = 0$. Otherwise, the velocity blows up at $t = \frac 1 {\alpha v_0}$.

In general, I suspect you'll get physically bizarre solutions if the acceleration increases with velocity.

Note that if $a = \frac{dv}{dt} = f(v)$, then you can integrate to get $v(t)$.

it's from jaan kalda kinematics pdf answer is 39.

 

[PeroK]

well you are absolutely a/v can not be a constant we must find a(v) it can be found by tracing the curve but I am unaware of how I checked out on google but can't get it maybe interpolation can be used.

Where did you get this problem? It looks unphysical to me. If you try something like $a = \alpha v^2$, then we have:
$$\frac{dv}{v^2} = \alpha dt$$ $$v = \frac{v_0}{1 - \alpha v_0t}$$
If $v_0 = 0$, then $v(t) = 0$. Otherwise, the velocity blows up at $t = \frac 1 {\alpha v_0}$.

I suspect that in general you'll get physically bizarre solutions if the acceleration increases with velocity.

Note that if $a = \frac{dv}{dt} = f(v)$, then you can integrate to get $v(t)$.

 

[Vanadium 50]

This thread is a hot mess. I blame the OP, for not posting the complete problem. Then people tried to help and got into the game of "guess what the OP meant". While I understand people are trying to be helpful, they would do the OP a greater service by teaching him/her to follow the rules and use the template.

Note that while he did give us the text, he did not tell us which problem. I can't see any good reason for this.

Here is the text of the problem:

The acceleration of a boat depends on its speed as shown in graph. The boat is given initial speed v0 = 4m/s. What is the total distance traveled until the boat will almost come to rest?

As you can see, there was some critical information that was missing, and because of that people - especially those who tried to solve the problem for the OP - made some incorrect assumptions and jumped to conclusions.

Furthermore, there is about half a page of text following the problem describing how to solve it. I can't see any good reason why the OP didn't mention any of this either.

 

[IamVector]

This thread is a hot mess. I blame the OP, for not posting the complete problem. Then people tried to help and got into the game of "guess what the OP meant". While I understand people are trying to be helpful, they would do the OP a greater service by teaching him/her to follow the rules and use the template.

Note that while he did give us the text, he did not tell us which problem. I can't see any good reason for this.

Here is the text of the problem:
As you can see, there was some critical information that was missing, and because of that people - especially those who tried to solve the problem for the OP - made some incorrect assumptions and jumped to conclusions.

Furthermore, there is about half a page of text following the problem describing how to solve it. I can't see any good reason why the OP didn't mention any of this either.

sorry for inconvenience I am new , will take care of this next time whenever I post any question I apologize for such an ignorant and foolish behavior from my side, I only ask for an apology no explanation in defense.

Moving forward, I solved the problem thankyou to all the helpers for guiding and I appreciate their efforts for correcting me whenever I was wrong.

 

[berkeman]

Moving forward, I solved the problem thankyou to all the helpers for guiding and I appreciate their efforts for correcting me whenever I was wrong.

So this sounds like a good time to close this thread.