Determining Acceleration Change: How to Analyze a-t Graphs and Calculate Jerk

[UrbanXrisis]

Automotive engineers refer to the time rate of change of acceleration as the "jerk." If an object moves in one dimension such that its jerk "J" is constant. How would I determine expressions for its acceleartion Ax, velocity Vx, and position X, given that its inital acceleration, speed, and position are Axi, Vxi, and Xi, respectively.

How would I show that Ax^2=Axi^2+2J(Vx-Vxi)?

I'm not sure where to start on this one.

 

[Pyrrhus]

You don't know how to integrate, i don't believe you can do this problem.

It's something like

\frac{da_{x}}{dt} = J

a_{x} = \int Jdt

a_{x} = Jt + C
and then you need to find a value for C, and on and on...

or

\int^{a_{x}}_{a_{xo}}da_{x} = \int^t_{0} Jdt

a_{x} ]^{a_{x}}_{a_{xo}}= Jt]^t_{0}

a_{x} - a_{xo}= Jt - 0

Edit: To make it clearer.

 

[robphy]

Automotive engineers refer to the time rate of change of acceleration as the "jerk." If an object moves in one dimension such that its jerk "J" is constant. How would I determine expressions for its acceleartion Ax, velocity Vx, and position X, given that its inital acceleration, speed, and position are Axi, Vxi, and Xi, respectively.

How would I show that Ax^2=Axi^2+2J(Vx-Vxi)?

I'm not sure where to start on this one.

You may recall from the "constant acceleration" case that
v^2=v_0{}^2+2a(x-x_0).
Where did this equation come from? How is it derived?


Now, your problem is to derive
a^2=a_0{}^2+2J(v-v_0)
for the "constant jerk" case.


More hints:

"constant velocity" case: x=x_0+vt
"constant acceleration" case: v=v_0+at and x=x_0+ \ldots
"constant jerk" case: a=\ldots and v=\ldots and x=\ldots

 

[UrbanXrisis]

I'm actually confused by what the question is asking. Can you possibily refrase the question so a high schooler can understand what it's asking? About the "How would I determine expressions for its acceleartion Ax, velocity Vx, and position X, given that its inital acceleration, speed, and position are Axi, Vxi, and Xi, respectively."

What is the question trying to get at? What in the world is a Jerk? Is it when acceleration has a slope on an acc. vs time graph?

Thanks

 

[robphy]

What in the world is a Jerk? Is it when acceleration has a slope on an acc. vs time graph?

Yes! Jerk is the slope of the acc. vs. time graph.

So, you should think back...
when velocity had a constant nonzero slope on a velocity vs time graph, what expressions did you write for the velocity and the position?

Apply the same reasoning your problem.

 

[UrbanXrisis]

I don't think I ever wrote any expression for velocity and position.
It's probably what you wrote...X=Xo+vt

I'm not sure how to apply this to my problem since I don't know what the question is asking. Probably a rephrase of the question would be most helpful.

Thanks

 

[Pyrrhus]

Urban, i don't know how you can do this problem without Riemann's Integrals, but Rophy's idea is quite good, the kinematic equations you know for constant acceleration happens when dv/dt = a, where a is a constant. Now for this problem it will be da/dt = j, where j is a constant. Imagine how can you use the kinematic equations you know for this case.

 

[Pyrrhus]

Hey Urban, i was thinking about a good resource for helping you learn integrals.

It's a Calculus E-Book. [It's on PF's links, Mathematics Resources]

https://www.physicsforums.com/local_links.php?action=jump&id=18 [size: 6100 kb]

 

[UrbanXrisis]

Thanks a lot. Even if I did learn to do integral, how would this help me in this problem? What does getting the area under a position v time, velocity v time, and acceleration v time, graph tell me?

 

[Pyrrhus]

dx = vdt (Velocity vs time graph area it will give displacement)
dv = adt (Acceleration vs time graph area it will give velocity)

 

[UrbanXrisis]

sooo... da/dt = j ... da=jdt will give me acceleration?

When you say "velocity" is it average velocity?

 

[Pyrrhus]

In a Jerk vs Times graph yes

Yes average velocity.

 

[UrbanXrisis]

When the question asks "How would I determine expressions for its acceleartion Ax, velocity Vx, and position X" is it asking for me to derive a formula with variables Axi, Vxi, and Xi...so it would be X=blah*J^2+blah*J+blah...and Vx would be the dirivative of X...so on?

Also, how must I use Integral to obtain this fomula for Jerk?

 

[Pyrrhus]

When the question asks "How would I determine expressions for its acceleartion Ax, velocity Vx, and position X" is it asking for me to derive a formula with variables Axi, Vxi, and Xi...so it would be X=blah*J^2+blah*J+blah...and Vx would be the dirivative of X...so on?

Also, how must I use Integral to obtain this fomula for Jerk?

Look at my first post here.

 

[UrbanXrisis]

where did the variable C come from?

 

[UrbanXrisis]

Also, when you write "dx/dt = v" are you saying the integral of the relationship between displacement and time is the average velocity?

 

[Pyrrhus]

where did the variable C come from?

Well if you have y = x^2 + 4 or y = x^2 o y= x^2 + 53, the derivatives will be the same, 2x, so you put a C when you integrate representing that unknown constant.

 

[Pyrrhus]

Also, when you write "dx/dt = v" are you saying the integral of the relationship between displacement and time is the average velocity?

\frac{dx}{dt} is Instantenous Velocity while \frac{\Delta x}{\Delta t} is average velocity, so if you integrate and get the whole area down a curve of Acceleration vs Time, it will be the average velocity, it depends on what you're getting.

 

[UrbanXrisis]

therefore, Instantenous Acceleration is the Jerk.

to determine expressions for its acceleartion Ax, velocity Vx, and position X, all I have to do is find the integral of the Jerk for Ax, the integral of Ax for Vx, then the integral of Vx for X.

 

[Pyrrhus]

Exactly, all depending of course on time differential.

 

[UrbanXrisis]

ahhh, I see where this is heading. It'll take me some time to learn how to do integral but once I have that down, I'll attempt this problem again. There is also a thread that I started that has not been answered in a while in which I'm confused once again...
https://www.physicsforums.com/showthread.php?t=44283

Thanks for all the help

 

[james1232]

okay guys I am a grade 11 student and i got a question that i don't understand, here it is. 32) a) derive an equation for the change in acceleration by using a graph analysis technique on a a-t (acceleration vs time) graph. (note. The name given tothe change in acceleration per unit time is called a jerk)

b) write an equation for the area under the graph . what does this represent.


so I thought that the equation to calculate acceleration in a = v/delta t but i don't have any clue what the equation is for the change in acceleration, i figure maybe if they were two straight lines you could just do a2 - a1 = change in acceleration, but on a a-t graph how would that work, because it would have to be two straight lines, not a diagonal line... if anyone has any ideas that would be great.