Contradiction in the equations of motion ?

[modulus]

Contradiction in the equations of motion...??!

I' ve found a strange contradiction between the fist and second equations of motion. First we start with the second equation:

s = ut + 1/2at²

We factor 't' out, whoch gives us:

s = t(u + 1/2at)

We divide both sides by 't', and on the LHS we get 's/t' which should be equal to 'v', i.e., the velocity:

v = u + 1/2at

But, the first equation goes as the following:

v = u + at

And that contradicts our previous result...but...how??!

 

[espen180]

Dividing s by t to get v only works when v is constant. This is not the case when a is nonzero.

In order to get v, you must differentiate with respect to t, which will give you v=u+at

 

[Cthugha]

Think about what your s and v actually are.
s is the traveled distance at time t and v is the momentary velocity at time t.
If you now divide s by t you will get the average velocity for the timespan 0 to t, which is of course different from the momentary velocity at time t.

 

[modulus]

All right, I understand the thing about not being able to convert 's/t' directly into 'v', because it is not constant. So am I supposed to do something like this:

d(ut + 1/2at²) / dt = v
= d(ut)/dt + d(1/2at²)/dt = v
= u*dt/dt + 1/2a*d(t²)/dt = v
= u + 1/2a*2t = v
= v = u + at

OK, so I get that, but let's suppose I want to find the average velocity, then can I use the formula I derived (v = u + 1/2at), will it be correct?
Let's suppose I want to find the rate of change of the average velocity, can I do this:

d(u + 1/2at)/dt = rate of change of average velocity
= d(u)/dt + d(1/2at)/dt
= 0 + 1/2a*dt/dt
= 1/2a*1
= rate of change of average velocity = 1/2a

So... is that correct?

 

[Sidnv]

I don't think you can differentiate average velocity because in this case the limit dt -> 0 will not be valid.

 

[gmax137]

Just checking here: you do realize that the equations and derivatives you're doing assume constant acceleration, right? In other words, your "a" is a constant value and d(a)/dt = 0. That's why you get the 1/2 factor in your average velocity: if "a" is constant , then average "v" is just 1/2(v0 + v(t)). Pick some value for "a" (such as 32 ft/sec^2 or 9.8 m/sec^2) and then plot "a", "v" and "s" from 0 to say 10 seconds. If you look at the graphical results what's going on may be more obvious.

 

[modulus]

Just checking here: you do realize that the equations and derivatives you're doing assume constant acceleration, right? In other words, your "a" is a constant value and d(a)/dt = 0. That's why you get the 1/2 factor in your average velocity: if "a" is constant , then average "v" is just 1/2(v0 + v(t)). Pick some value for "a" (such as 32 ft/sec^2 or 9.8 m/sec^2) and then plot "a", "v" and "s" from 0 to say 10 seconds. If you look at the graphical results what's going on may be more obvious.

I understand that the acceleration is constant, but am I doing anything wrong with that while figuring out the derivatives?

Also, if the average velocity is (v + u)/2, then how is the rate of change of the average velocity (which should be equal to (v + u)/2t, according to the equation stated earlier) equal to half of the acceleration. Because, if it will be equal to half of the acceleration, then the rate of change of average velocity should be equal to (v - u)/2t

Also, I still don't understand why I cannot differentiate the average velocity with respect to time (as 'Sidnv' said).

When you tell me to make a graph of "v", is '"v"' the final velocity or the average velocity?

 

[vanesch]

Also, I still don't understand why I cannot differentiate the average velocity with respect to time (as 'Sidnv' said).

You can of course differentiate the average velocity wrt. time, but you get something else than the instantaneous acceleration. Differentiating the average velocity wrt. time gives you the *rate of change of the average velocity*. Taking the derivative of instantaneous velocity gives *you the rate of change of instantaneous velocity* which equals the second derivative of position wrt. time.

Look at it in a purely mathematical form without thinking of the physics. Consider a function s(t). (here, it will be "position as a function of time", but let's think of it as an arbitrary function s of a variable t).

On one hand, you calculate:

\frac{d}{dt} \left( \frac{s(t)}{t} \right)

On the other hand you calculate:
\frac{d^2}{dt^2} \left( s(t) \right)

That's not the same.

 

[modulus]

Taking the derivative of instantaneous velocity gives *you the rate of change of instantaneous velocity* which equals the second derivative of position wrt. time

What is the 'second derivative of position wrt. time? Shouldn't the rate of change of instntaneous velocity imply be instantaneous acceleration?

On one hand, you calculate:



On the other hand you calculate:

I believe we'll be using the second one to calculate the innstantaneous acceleration. And, the first one is the right approach for calculating 'the rate of change of average velocity', and that's what I did (right?)...so, basically I think this impllies both of my results are correct:

rate of change of average velocity = 1/2a

v(avg) = u + 1/2at

 

[vanesch]

What is the 'second derivative of position wrt. time? Shouldn't the rate of change of instntaneous velocity imply be instantaneous acceleration?

Velocity (instantaneous velocity) is the (first) derivative of position wrt. time. Acceleration (instantaneous acceleration) is the (first) derivative of instantaneous velocity wrt. time, and hence the second derivative of position wrt. time.

For uniform acceleration, the position as a function of time is given by:
s(t) = s0 + v0.t + a/2 t^2

The first derivative of this function, wrt t, is: v(t) = v0 + a/2 *2t = v0 + a.t
This is the instantaneous velocity

The second derivative is a(t) = a, a constant. It is the instantaneous acceleration, and in this case, it is a constant (it was the defining property of this specific motion).

I believe we'll be using the second one to calculate the innstantaneous acceleration. And, the first one is the right approach for calculating 'the rate of change of average velocity', and that's what I did (right?)...so, basically I think this impllies both of my results are correct:

Yes.

However, the "rate of change of average velocity" is not the instantaneous acceleration. This is because you've replaced a derivative with respect to t (instantaneous velocity) by a division by t (average velocity).

 

[modulus]

Thank a lot!
You guys cleared up a lot for me.