Position time graphs for different acceleration

[mathmaniac1]

How can I draw the position time graph for uniform acceleration?
a)positive
b)negative
and for non-uniform acceleration?
How would these look like

 

[tkhunny]

Have you considered the standard model?

Constant Acceleration: [tex]a(t) = f[/tex]
Related Velocity: [tex]v(t) = ft + V_{0}[/tex] Where [tex]V_{0}[/tex] is the initial velocity.
Related Position: [tex]x(t) = ½ft^{2} + V_{0}t + H_{0}[/tex] Where [tex]H_{0}[/tex] is the initial location.

For non-constant acceleration, make f ==> f(t). The antiderivatives to v(t) and x(t) may be more challenging, but judicious selection of f(t) can solve that problem.

 

[mathmaniac1]

Have you considered the standard model?

Constant Acceleration: [tex]a(t) = f[/tex]
Related Velocity: [tex]v(t) = ft + V_{0}[/tex] Where [tex]V_{0}[/tex] is the initial velocity.
Related Position: [tex]x(t) = ½ft^{2} + V_{0}t + H_{0}[/tex] Where [tex]H_{0}[/tex] is the initial location.

For non-constant acceleration, make f ==> f(t). The antiderivatives to v(t) and x(t) may be more challenging, but judicious selection of f(t) can solve that problem.

Sorry,I don't understand a word of you

What is a(t),v(t) and f?I think t refers to time.

Can you simplify?Or explain these terms and etc...

 

[mathworker]

Sorry,I don't understand a word of you

What is \(\displaystyle a(t)\),\(\displaystyle v(t) \)and \(\displaystyle f\)?I think t refers to time.

Can you simplify?Or explain these terms and etc...

\(\displaystyle a(t)\) is acceleration as it is assumed to be uniform it is not a function of time so he assumed it to be a constant \(\displaystyle f\) and what is acceleration ? rate of change of velocity which is also time dependent so he assumed it to be v(t)$$a(t)=\frac{d(v(t))}{dt}=f$$$$d(v(t))=f.dt$$ $$v(t)=\int f.dt$$ $$v(t)=f.t+constant$$

 

[MarkFL]

Acceleration is defined to be the time rate of change of velocity:

\(\displaystyle a(t)\equiv\frac{dv}{dt}\)

The average acceleration over some interval is simply the change in velocity divided by the change in time:

\(\displaystyle \overline{a}=\frac{\Delta v}{\Delta t}=\frac{v(t)-v_0}{t-t_0}\)

where \(\displaystyle v\left(t_0 \right)=v_0\)

If we let the beginning of the interval coincide with time $t=0$, and let the acceleration be constant, then we have:

\(\displaystyle a=\frac{v(t)-v_0}{t}\)

and we may arrange this as:

\(\displaystyle v(t)=at+v_0\)

Now, the time rate of change of position is defined to be velocity:

\(\displaystyle v(t)\equiv\frac{dx}{dt}\)

and the average velocity over the interval $[0,t]$ is then:

\(\displaystyle \overline{v(t)}=\frac{x(t)-x_0}{t}\)

Since the velocity function is linear for constant acceleration, we may also write the average velocity as:

\(\displaystyle \overline{v(t)}=\frac{v(t)+v_0}{2}=\frac{at+2v_0}{2}=\frac{1}{2}at+v_0\)

Hence, equating the two expressions for average velocity, we obtain:

\(\displaystyle \frac{x(t)-x_0}{t}=\frac{1}{2}at+v_0\)

\(\displaystyle x(t)-x_0=\frac{1}{2}at^2+v_0t\)

\(\displaystyle x(t)=\frac{1}{2}at^2+v_0t+x_0\)

The case of non-uniform acceleration requires the calculus to determine the position function.

 

[mathmaniac1]

But the parabola stops its journey up after sometime,so what does that imply?
Doesn't it imply that the direction of motion of the object reverses after reaching a specific position,speed?
So the body moved back?wouldn't that mean acceleration being constant changes sign?

 

[MarkFL]

But the parabola stops its journey up after sometime,so what does that imply?
Doesn't it imply that the direction of motion of the object reverses after reaching a specific position,speed?
So the body moved back?wouldn't that mean acceleration being constant changes sign?

No, this means the velocity changed signs, not the acceleration. When the object is moving up, that is against the acceleration, then the object slows, but when the velocity then becomes negative, the acceleration adds to its magnitude, rather than subtracting from it as it did when the object was moving up. The change in the velocity remains the same.

 

[mathmaniac1]

So will an object with some constant+ve acceleration stop and return back all the way?This does not seem practical.

velocity changes sign

So v-u is -ve,so isn't acceleration changing sign?

 

[MarkFL]

If an object is given a positive initial velocity, and the acceleration is positive, then the object will simply speed up (linearly for constant acceleration), and ignoring other forces (such as drag) and special relativity, the velocity will increase without bound.

 

[mathmaniac1]

So why does the object seem to return?

 

[MarkFL]

So why does the object seem to return?

It doesn't return if you throw it down from a cliff. In this case the initial velocity and acceleration are "pointing" in the same direction.

The object only returns if the initial velocity is in the opposite direction as the acceleration.

 

[mathmaniac1]

ok,I get it,the maths puzzled me,I solved the maths
now please go on explaining the case of non-uniform acceleration

 

[MarkFL]

If acceleration varies with time, then you must integrate the acceleration with respect to time to get the velocity, and you must then integrate the velocity to get the position.

 

[mathmaniac1]

oh yeah,you mean integrate the f(t)=a?
I was thinking of the situation where no such f(t) is given,that's just stupid,right?
it seems to take any shape

 

[MarkFL]

Yes, in order to determine the velocity function, we have to know either the acceleration or position functions.

 

[mathmaniac1]

but can't we just use the eq x=ut+1/2at sqr?
with the value of a being given by f(t)?

x=ut+1/2at^2 can be derived for non-uniform acceleration too

 

[MarkFL]

but can't we just use the eq x=ut+1/2at sqr?
with the value of a being given by f(t)?

x=ut+1/2at^2 can be derived for non-uniform acceleration too

That only applies to uniform acceleration. If you review my derivation of this, you will see why. Hint: look at the statement regarding average velocity.

 

[mathmaniac1]

no,my txtbook deroved the same equation taking a=dv/dt,integrating b/w the limits u and v
gives the 1st eq of motion
taking v=dx/dt and integrating and subbing v gives the 2 nd eq as x=ut+1/2at^2
but ther's a difference,a is instantaneous acceleration

 

[MarkFL]

I am talking about the algebra based derivation I posted above. It is much easier though to use the calculus to derive this, and only with $a$ being a constant do you get:

\(\displaystyle x(t)=\frac{a}{2}t^2+v_0t+x_0\)

 

[topsquark]

no,my txtbook deroved the same equation taking a=dv/dt,integrating b/w the limits u and v
gives the 1st eq of motion
taking v=dx/dt and integrating and subbing v gives the 2 nd eq as x=ut+1/2at^2
but ther's a difference,a is instantaneous acceleration

The position equation is actually a power series:
[tex]x = x_0 + v_0t + (1/2)a_0t^2 + (1/6)j_0t^3 + \text{ ... }[/tex]
where j_0 is called the "initial jerk." (x_0 is the initial position, v_0 is the initial speed, etc.) You are using the format of the equation that I have tried very hard to never use in my life because it hides the "initial value" nature of the variables. It leads to confusion.

You can, of course, use your equation but a is no longer an acceleration in the usual sense...just a function of time.

-Dan

 

[mathmaniac1]

There is a Q,derive the eqns of motion using calculus
and my txtbook derived
x=x0+v0t+1/2at sqr
and after all 3 eqns there is a note:
These eqns can be used for non uniformly accelerated motion too.

Now is my textbook wrong?
I think by a he means f(t) for instantaneous acceleration at t.

If needed, I can give the txtbook derivation...

 

[topsquark]

There is a Q,derive the eqns of motion using calculus
and my txtbook derived
x=x0+v0t+1/2at sqr
and after all 3 eqns there is a note:
These eqns can be used for non uniformly accelerated motion too.

Now is my textbook wrong?
I think by a he means f(t) for instantaneous acceleration at t.

If needed, I can give the txtbook derivation...

The derivation is fine, I am sure. But when you are dealing with non-constant acceleration the usual tactic is to simply call x = x(t) and define v, a, etc. by its derivatives. Your form isn't wrong but I've never seen anyone use [tex]x = ut + (1/2)at^2[/tex] for a time varying acceleration.

-Dan

 

[MarkFL]

There is a Q,derive the eqns of motion using calculus
and my txtbook derived
x=x0+v0t+1/2at sqr
and after all 3 eqns there is a note:
These eqns can be used for non uniformly accelerated motion too.

Now is my textbook wrong?
I think by a he means f(t) for instantaneous acceleration at t.

If needed, I can give the txtbook derivation...

If I am reading this correctly, your textbook is essentially positing:

\(\displaystyle \int a(t)\,dt=a(t)\cdot t+c_1\)

and:

\(\displaystyle \int a(t)\cdot t+c_1\,dt=\frac{1}{2}a(t)\cdot t^2+c_1t+c_2\)

Do you see any problem with this?