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#### mathmaniac

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- Mar 4, 2013

- 188

a)positive

b)negative

and for non-uniform acceleration?

How would these look like

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- #1

- Mar 4, 2013

- 188

a)positive

b)negative

and for non-uniform acceleration?

How would these look like

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.

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- #3

- Mar 4, 2013

- 188

Sorry,I don't understand a word of you

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.

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

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

- May 31, 2013

- 118

\(\displaystyle a(t)\) 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)$$a(t)=\frac{d(v(t))}{dt}=f$$$$d(v(t))=f.dt$$ $$v(t)=\int f.dt$$ $$v(t)=f.t+constant$$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...

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

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.

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- #6

- Mar 4, 2013

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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?

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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.

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?

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- #8

- Mar 4, 2013

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

So v-u is -ve,so isn't acceleration changing sign?MarkFL said:velocity changes sign

Last edited:

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- Mar 4, 2013

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So why does the object seem to return?

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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.So why does the object seem to return?

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

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- #12

- Mar 4, 2013

- 188

now please go on explaining the case of non-uniform acceleration

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- #14

- Mar 4, 2013

- 188

I was thinking of the situation where no such f(t) is given,that's just stupid,right?

it seems to take any shape

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- #15

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- #16

- Mar 4, 2013

- 188

with the value of a being given by f(t)?

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

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- #17

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.

with the value of a being given by f(t)?

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

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- #18

- Mar 4, 2013

- 188

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

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- #19

- Aug 30, 2012

- 1,157

The position equation is actually a power series:

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

[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

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- #21

- Mar 4, 2013

- 188

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....

- Aug 30, 2012

- 1,157

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.

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....

-Dan

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- #23

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

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....

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

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?