Average velocity for two different types of motion along same path

[brotherbobby]

Homework Statement

A particle travels a distance $L$ in two ways.

(i) First it travels $\textbf{half the distance}$ at $v_1$ and the other half at $v_2$. What is the average velocity $\bar v$?

(ii) Suppose it travels $\textbf{half the time}$ at $v_1$ and the other half of time at $v_2$. What is $\bar v$?

(iii) Take the difference and show that one average is definitely bigger. (Assume all v’s are positive.) Explain.

Relevant Equations

Average velocity $\bar v=\dfrac{\text{Total displacement made}}{\text{Total time taken}}=\dfrac{\Delta x}{\Delta t}$.

Statement : Let me copy and paste the problem as it appears in the text.

Attempt : I could solve (i) and (ii). Part of (iii) also. It is in the explaining bit where I am stuck, conceptually.
I copy and paste my solutions using $\text{Autodesk Sketchbook}^{\circledR}$ hoping am not violating anything.

My answers are right, but I could not explain (iii). Why must $\bar{v_2}>\bar{v_1}\,?$
Is there a conceptual reason?

Intuitively, I have a doubt that actually comes earlier still. Why should the two average velocities be different at all (let alone argue why one more than another)? Is there something inherent in the definition of average velocity that makes them different?

A help would be welcome on these conceptual points.

 

[PeroK]

The averages can be equal if the speeds are equal. But, given the speeds $v_1$ and $v_2$ are different:

If you travel half the distance at a lower speed, then you travel at this speed for a longer time. This means that the time average is nearer the lower speed.

 

[brotherbobby]

The averages can be equal if the speeds are equal. But, given the speeds $v_1$ and $v_2$ are different:

If you travel half the distance at a lower speed, then you travel at this speed for a longer time. This means that the time average is nearer the lower speed.

Thank you, that explains it.
I was thinking along the lines of calculus. If I have a distance $L$ to cover with two different speeds, it seems that the shortest time in which to do so is to halve the time of travel between them. Is that true?

 

[PeroK]

Thank you, that explains it.
I was thinking along the lines of calculus. If I have a distance $L$ to cover with two different speeds, it seems that the shortest time in which to do so is to halve the time of travel between them. Is that true?

No. You travel as much distance as possible at the greater speed.

 

[robphy]

Here's a velocity-vs-time graph to visualize the calculation
and to support the interpretation by @PeroK .
Using nice numbers,
let $L=10\rm\ m$ and $v_1=1\rm\ m/s$ and $v_2=4\rm\ m/s$.

In both cases, the total area under velocity-curve represents $L$.

• The red rectangles represent the motion for equal displacements at each velocity.
  So, the red-rectangles have equal area,
  with height equal to the velocity and thus width equal to the time-spent.
  The rectangle for the average-velocity also has an equal area (the total displacement $L$),
  whose height ($v_{i,avg}$) is determined by the base (the total time-spent $\Delta t_i$).

• The green rectangles represent the motion for equal durations at each velocity.
  So, the green-rectangles have equal bases,
  with height equal to the velocity and thus area equal to the displacement traveled.
  The rectangle for the average-velocity also has an equal area (the total displacement $L$),
  whose height ($v_{ii,avg}$) is determined by the base (the total time-spent $\Delta t_{ii}$).

 

[Lnewqban]

I was thinking along the lines of calculus. If I have a distance $L$ to cover with two different speeds, it seems that the shortest time in which to do so is to halve the time of travel between them. Is that true?

 

[robphy]

@Lnewqban
Note that $\displaystyle v_{avg}\equiv\frac{\int v\ dt}{\int dt}$, which is generally not-equal to $\frac{1}{2}(v_1+v_2)$. (See the calculation in the OP.)

In my opinion,
it is better to specify a slope on a position-vs-time diagram instead of an angle.
[The spacetime geometer in me is cringing at the sight of that angle.]

 

[Lnewqban]

@robphy
Points taken.
Post has been edited.
Thank you.

 

[Delta2]

You consider the average velocity over time in all three cases. If (i) was considering the average velocity over distance then the answer would be for (i) $\frac{v_1+v_2}{2}$ too but then in (iii) we would compare apples to oranges but that's another story ...

The average of a function $f(x)$ over $x\in[x_1,x_2]$ is by definition $$\frac{\int_{x_1}^{x_2} f(x) dx}{x_2-x_1}$$.

Applying this definition for $f(x)=v(x)$ ,v the velocity and $x=t$ (the time) we get for the average of velocity over time as equal to $$V=\frac{\int_{t_1}^{t_2} v(t) dt}{t_2-t_1}$$

And for $f(x)=v(x)$ $x=s$ the distance covered we get V' the average of velocity over distance as equal $$V'=\frac{\int_{s_1}^{s_2} v(s) ds}{s_2-s_1}$$ where $$s_1=s(t_1) and s_2=s(t_2)$$

BUT because $$ds=v(t)dt$$ we can (integration by substitution is involved here) write $V'$ as $$V'=\frac{\int_{t_1}^{t_2} v^2(t) dt}{s(t_2)-s(t_1)}$$ so we can see that V' is in general different than V.

Worth noting once case where they are equal (they might be other cases too) is when $v(t)=v(s)=c$ where c a constant.

 

[brotherbobby]

Sorry for coming in late. I am the OP for the problem. The problem is clearly less trivial than it looks, if one wants to understand it conceptually. The mathematics is however straightforward.
Let's have the statement of the problem again, slightly changed to help understanding.

Statement : A particle travels a distance $L$ in two ways. First it covers half the distance at some speed $v_1$ and the other half with speed $v_2$. Then it travels half the time with $v_1$ and the other half with $v_2$. Calculate the average speeds (say $v_L$ and $v_T$ respetively) and argue which one is greater and why.

Attempt : I have already solved it (see post# 1). I also showed that $\boxed{v_L<v_T}$ and the struggle is to understand why. Let us assume $v_1<v_2$.
I suppose the answer lies in the silly (and embarrassing) error I made in post #3 above which @PeroK explained.

Nothing is kept equal in the point above. So it is different from the question. But it makes the important case that total time will be less if a greater distance was travelled at the higher speed. Let's call this concept as $\boldsymbol C$ and I'd refer to it later.

In this problem, since the total distance travelled remains the same, it is the times we should compare - I capture the point in the writings scibbled below.

Clearly, since we know the answer, if we call the first time of travel as $t_L$ and the second as $t_T$, we have $t_L>t_T$. But how is it so, conceptually?
We focus on the times.

For the second case, equal times, we have $L_1<L_2$.Hence, as per concept $\boldsymbol C$ above, when the object moves with equal times, it also covers a lower distance with the lower speed and therefore ends up taking less time on the whole.

But when it moves covering equal distances, the distance covered with the lower velocity is as much as that with the higher. Using concept $\boldsymbol C$ again, it does not change its distance accordingly, and ends up taking more time on the whole.

I believe this is the point that @PeroK made in post #2, assuming my reasoning is correct.

 

[PeroK]

Here's an interesting related problem. Suppose we have two batters (could be cricket or baseball, depending on your taste). Suppose A has a better batting average than B in the first half of the season; and, a better batting average than B in the second half of the season. Prove that A must have a better batting average than B across the whole season.

 

[Delta2]

Here's an interesting related problem. Suppose we have two batters (could be cricket or baseball, depending on your taste). Suppose A has a better batting average than B in the first half of the season; and, a better batting average than B in the second half of the season. Prove that A must have a better batting average than B across the whole season.

Isn't this, at least in the intuitive-qualitative level , kind of obvious or did I completely misunderstood the statement of this?

 

[PeroK]

Isn't this, at least in the intuitive-qualitative level , kind of obvious or did I completely misunderstood the statement of this?

Then it should be easy to prove!

 

[Delta2]

Ye ok we both know that in math there are quite a few things that are straightforward in the intuitive-qualitative level of logic, but in the formalistic-rigorous level of logic they become not so straightforward.
And vice versa!

 

[PeroK]

If you can't prove it, perhaps it's not true? You could look for a counterexample.

 

[Delta2]

I think I can prove it , I just wanted to say that somethings that look easy to prove are not so easy after all.

 

[brotherbobby]

Here's an interesting related problem. Suppose we have two batters (could be cricket or baseball, depending on your taste). Suppose A has a better batting average than B in the first half of the season; and, a better batting average than B in the second half of the season. Prove that A must have a better batting average than B across the whole season.

I'm the OP and I'd respond to your task above in a moment. But for now, the author has given his answers to the problem I raised in this thread. Most are fine, except for a bit of thinking he's raised towards the end.
I'd paste the problem first and the answers next.

Attempt : So let's see. In the extreme case when $v_1=0$, what happens to the average speed when the particle travels dividing the total distance into two halves. With $v_1=0$, the time for the first half $t_1=\infty$. The total time $t_1+t_2 = \infty$ which makes the average speed $\boxed{\bar v_L = 0}$. This can also be seen from the formula for average speed which we found. $\bar v_L = \frac{2v_1v_2}{v_1+v_2}=0\;\text{for}\;v_1=0$.

For the second case, where the particle travels dividing the total time into two halves, if the first speed $v_1=0$, the distance covered with that speed $L_1=0$. The whole of the remaining distance would have to be covered therefore with the other speed $v_2$. That would take a time of $t_2=\dfrac{L}{v_2}=t_1=\;\text{(some)}\; t$, since both times are the same. Hence, the average speed in this case $\bar v_T = \frac{L}{t_1+t_2}=\frac{v_2t}{t+t}=\boxed{\frac{v_2}{2}}$. This agrees with the answer : $\bar v_T = \frac{v_1+v_2}{2}\Rightarrow \boxed{\bar v_T = \dfrac{v_2}{2}}\;\text{for}\; v_1=0$.

I suppose the author is trying to show, as a form of check, that the first average speed is lower. It is is because of the lower speeds of the two. If the first speed $v_1\rightarrow 0$, the first average speed $v_L\rightarrow 0$.

I hope my reasoning is correct.

 

[brotherbobby]

Here's an interesting related problem. Suppose we have two batters (could be cricket or baseball, depending on your taste). Suppose A has a better batting average than B in the first half of the season; and, a better batting average than B in the second half of the season. Prove that A must have a better batting average than B across the whole season.

Let A have a batting average of $a$ in the first season and $b$ the second season. Assume he has played $m$ matches the first season and $n$ the second season.
Let B have a batting average of $c$ in the first season and $d$ the second season. Assume he has played $p$ matches the first season and $q$ the second season.
We are given
\begin{equation}
\dfrac{a}{m}+\dfrac{b}{n}>\dfrac{c}{p}+\dfrac{d}{q}\qquad\text{(given)}
\end{equation}
We have to show that
\begin{equation}
\color{blue}{\dfrac{a+b}{m+n}>\dfrac{c+d}{p+q}\qquad\text{(to prove)}}
\end{equation}
Subtracting the L.H.S. of the two equations (1) and (2) , I write out the following using $\text{Autodesk Sketchbook}^{\circledR}$

Likewise, we should have for the R.H.S. of the two equations (1) and (2) :

Let's call $\dfrac{a}{m}+\dfrac{b}{n}>\dfrac{a+b}{m+n}\qquad (3)$

and $\dfrac{c}{p}+\dfrac{d}{q}>\dfrac{c+d}{p+q}\qquad (4)$

By how much are these greater? From the picture above, we have $\frac{a}{m}+\frac{b}{n}-\frac{a+b}{m+n}=\frac{m^2b+n^2a}{mn(m+n)}$.

Likewise, $\frac{c}{p}+\frac{d}{q}-\frac{c+d}{p+q}=\frac{p^2d+q^2c}{pq(p+q)}$

These are the amounts we are "taking" away from the originals. If we can show that we are taking away less from the first expression than from the second, then we are done, given the original inequality. I put the reasoning to the right.

What is $\dfrac{m^2b+n^2a}{mn(m+n)}-\dfrac{p^2d+q^2c}{pq(p+q)}?$

This is where am stuck. Is the last expression < 0?

Are the terms marked in red less than zero? The values $a,\dots,d, m,\dots,q$ are chosen arbitrarily with the only constraint : $\dfrac{a}{m}+\dfrac{b}{n}>\dfrac{c}{p}+\dfrac{d}{q}$.

I don't see how the latter can be used to prove that the all those terms come out negative.

Request : I am open to any suggestion to solve this problem, which only appears easy intuitive. A thanks to @PeroK for suggesting it.

 

[PeroK]

If you get stuck trying to prove something you should look for a counterexample.

 

[brotherbobby]

The problem you (@PeroK) stated in post # 11 looks obvious to be generally true. So no counterexample to it should exist.
But I cannot prove it mathematically.
Any hint?

 

[PeroK]

The problem you (@PeroK) stated in post # 11 looks obvious to be generally true. So no counterexample to it should exist.
But I cannot prove it mathematically.
Any hint?

It's not true!

 

[PeroK]

Or, for player A:
$$(1\times 50) + (9 \times 20) = 230/10$$Player B:
$$(9 \times 40) + (1\times 10)= 370/10$$

 

[brotherbobby]

I suppose the trick to make it work is to give B more games with the better average and A less games with the better average and vice versa.