How does the result of derivative become different from average ratio calculation?

Question

Lets give an example. Velocity, $v=ds/dt$. If we know the value of $s$ (displacement) and $t$ (time), we can instantly find the value of $v$. But then this $v$ will be the average velocity. Now consider, $$s= ut+.5at^2,$$ which is equal to 100 meter. If a body travels 100 meter distance in 10 seconds, then we can say, $$v=s/t= (ut+.5at^2)/t = u+.5at.$$ And we know for sure this is average velocity. But when we consider $ds/dt$, the formula becomes $u+at$.

Here, u+at and u+.5at can never be equal, meaning average velocity in this case is not equal to actual velocity. But we know the answer for the given circumstance will be 10 meter per second velocity. So how this is incorrect? What did I miss? Can someone explain intuitively or logically and in details without mentioning formal definitions and jargon terms?

Answer 1

You compare apples to the oranges. Average speed by definition is : $$\tag 1 \overline v = \frac {u+v}{2} $$

Final speed of the object after accelerating it for $t$ seconds is :

$$\tag 2 v = u+at $$

Substituting (2) into (1) :

$$ \tag 3 \overline v = \frac {2u+at}{2} = u + \frac 12 at $$

We get an average speed when object accelerates uniformly over whole time period $t$.

So the problem is that you are mixing different concepts here. (2) kinematical equation shows object speed increase after arbitrary time lapse. While equation (3) shows average object speed which will be in case when object is accelerating and it has elapsed half-period of total time covering full object trajectory. Hence to some sense 3-rd equation is specific instance of more general 2-cond equation, which doesn't care about how you interpret time elapsed for an object.

Btw, $v(t)=ds/dt$ is not average speed, but rather instant speed at some particular time moment $t$, because of infinitesimal quantities involved.

Answer 2

The problem you are having is that you are trying to define "average velocity" in more than one way, but those different definitions are not actually compatible.

For example, you can claim "a tiger is a big cat with sharp teeth and black stripes" and also claim "a tiger is also a grey mammal with a long nose," but then it will be confusing when one "tiger" eats plants and another "tiger" eats meat.

In your first paragraph you say "$v = \frac{ds}{dt}$" is average velocity. Then a few sentences later you say "$v = s/t$" is average velocity. These are two different possible definitions. Do they match? You see for yourself that they do not match.

And why should they match? Here are three other possible definitions of "average velocity," none of which give the same result

1. $v = \frac{v(0)+v(t)}{2}$
2. $v = \frac{v(0)+v(t/2)+v(t)}{3}$
3. $v = \frac{s}{2t}$

So how do we decide which one should be the real definition of "average velocity"? We choose the definition that is the most useful in the most cases. Historically, the definition that has turned out to be the most useful is the definition that satisfies two rules:

1. When the velocity is constant the average velocity should match the constant velocity.
2. If you know the average velocity over a time period $t$ you should be able to calculate the displacement with the simple equation $s = v_{avg}t$

Now once we've decided on that definition of average velocity we need to stick with it, and we shouldn't be surprised when other calculations don't give the same result. We also shouldn't be surprised if there are other useful things to know about velocity. "Average velocity" tells us a specific thing about velocity, but it doesn't tell us everything and so it's helpful to have other concepts like "instantaneous velocity" to give us information that "average velocity" doesn't.

Answer 3

The time derivative of $\textit s$ is merely giving the slope of the function $\textit s(t)$ at any given value of $\textit t$, or, what the speed will be at any given time. Taking the same function, and dividing it by some value of $\textit t$, is essentially using the distance formula which $\textit will$ give the $\textit average$ speed over the specified time $\textit interval$.

Answer 4

Suppose your velocity on a succession of days is $0$, then $1$, then $2$, then $3$, then $4$, then $5$, then $6$. On the last day, your average velocity for the trip is the average of those numbers, which is $3$. But your current velocity is still 6.

Moral: There's no reason to think that your average past velocity (which is what you're computing with $s/t$) should equal your current velocity (which is what you're computing with $ds/dt$).

Answer 5

There is no problem. In your example of $100 \,\mathrm m$ in $10\,\mathrm s$, you have implicitly chosen $a=0$. Plug that into either of your formulas and you will see them giving the same result.

If $a$ is not equal to zero, $a\neq0$, then the two speed values you find do not represent the same property. The former is indeed the average speed, but the latter is instantaneous speed. The speed changes constantly (since $a$ is non-zero), so the instantaneous speed is different at any new moment. If you plug a value of $t$ into your two expressions, then the former will give you the average speed from the beginning and until this value of $t$ seconds, whereas the latter will give you the instantaneous speed after $t$ seconds.

Answer 6

If you ever run 100m you would know that your speed at beginning und end are different. So even if you make it in 10s, your speed at the beginning and the one at the end are not 10m/s. If you know exactly the distance s depending on t so s(t) , ds/dt gives you the speed at any time, so for example at the moment t=3s this may be greater or smaller than the average speed, depending on s(t) If you have $s(t)=u*t+0.5t^2$ with positive a than your sped at beginning is $v_a=u$ and at the end $v_e=u+0,5a\cdot100s^2$ the average would be $\frac{v_a+v_e}{2}$

Answer 7

The average velocity between a given time $t$ and another $t+\Delta t$ is:

$$\bar v = \frac{s(t+\Delta t) - s(t)}{\Delta t} = \frac{v_0(t+\Delta t)+\frac{1}{2}a(t+\Delta t)^2 - (v_0t + \frac{1}{2}at^2)}{\Delta t}$$ The result after simplifications is: $$\bar v = \frac{v_0\Delta t+at\Delta t+a\frac{\Delta t^2}{2}}{\Delta t} = v_0+at+a\frac{\Delta t}{2}$$

If the time interval is very short, we can neglect $\Delta t$ with respect to the other terms, approaching the instantaneous velocity: $$v = v_0+at$$

On the other hand, if the initial time is zero $(t=0)$ the expression becomes for any $\Delta t$ (small or large):

$$\bar v = v_0+a\frac{\Delta t}{2}$$

Note that the result is consistent because when both conditions are met ($t=0$ and $\Delta t$ very small) we have: $v = v_0$