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We know average velocity in its strictest sense, means total displacement over total time taken: $$\frac{X_f-X_i}{T_f-T_i}$$

There's a special case, when a body is moving in a straight line with a constant acceleration. Of course since its acceleration is constant, it has to be a rectilinear motion. In this case average velocity (over a time interval) is simply $$V_{avg}=\frac{V_1+V_2}{2} \tag 1$$

It can be proved easily, using the equations of motion. The important point is, this formula works only when a body is moving with a "constant acceleration". As per my teachers and the books I have.

The problem is, why I'm actually putting this post up, there's another case. If the body moves with a velocity $V_1$ for a time interval $t$, and then it moves with a velocity $V_2$ for the same amount of time $t$. In other words, the body traveling with a velocity $V_1$, takes time $t$ to go from a point $A$ to another point $B$, and then it goes from point $B$ to another point $C$ and again, it takes time $t$, traveling with a velocity $V_2$.

In this case, when time intervals are equal, we calculate average velocity (or average speed) by taking the arithmetic mean of individual velocities, and this formula (as per my teachers and the books I have). So, in this case : $$V_{avg}=\frac{V_1+V_2}{2}\tag 2$$

But equation (1) and equation (2) are completely identical. Isn't that strange? And odd? Because equation (1) should be valid 'only' when the acceleration is constant. But in the second case, acceleration of the body is not constant during the course of its motion. It's acceleration is zero as it goes from $A$ to $B$, then its acceleration changes as it velocity changes from $V_1$ to $V_2$, and then its acceleration is constant from $B$ to $C$. Even though its acceleration is non-constant, the formula for finding out its average velocity is exactly the same, as in the first case.

Please explain what's going on here. Because the formula $V_{avg}=\frac{V_1+V_2}{2}$ should be valid if and only if acceleration is uniform. (As per my books and my teachers). Thanks

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    $\begingroup$ Why do you think the "only if" part in the last sentence should hold? $\endgroup$ – Jasper Oct 2 '18 at 12:21
  • $\begingroup$ Here's a hint for your question. Average velocity=(total distance travelled)/(total time taken) this is the correct defination of average velocity $\endgroup$ – Sourabh Oct 2 '18 at 13:56
  • $\begingroup$ Let's look at the fact that your velocity jumped from $v_1$ to $v_2$ instantly. Isn't that strange? $\endgroup$ – harshit54 Oct 2 '18 at 14:45
  • $\begingroup$ @Harshit Yeah I find that strange. It was actually a question from a book, I didn't make it up. And yes, when I read the question, I did find it a bit strange. I mean, velocity instantly jumps from V1 to V2 at B. It implies that acceleration must have been infinitely large, at B, which doesn't sound very practical. But, as I said, it's a question from a book. I actually wanted to ask how it was possible. But that's against the rules, I guess we can't ask two different question in the same post. But since you've mentioned it, can you explain if it's possible? And how? Thank you $\endgroup$ – π times e Oct 2 '18 at 16:11
  • $\begingroup$ Can you please write the exact statement of the question. $\endgroup$ – harshit54 Oct 2 '18 at 16:12
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You are mixing things. Those formulae are not the same; the problem is that you are using the same symbols, but the meaning is different.

For the rectilinear uniformly accelerated movement, you use

$$v_{average}=\dfrac{v_{initial}+v_{final}}{2}$$

Whereas ni the other case, you use

$$v_{average}=\dfrac{v_{1}+v_{2}}{2}$$ Meaning "velocity in the first part" and "velocity in the second part", but those velocities are constant.

So, in the first case, $v$ is continuously changing (constant acceleration), while in the second one, those are two constant velocities. First one value for some interval, then another value along the rest of the movement. They are different things.


Of course there's something else that tells you why they look so similar (not being the same). Take the definition of a mean value. In this case, we will talk about velocity but it works for any magnitude.

The movement lasts a total time $T$, and it is divided into 2 "parts", in which velocity is different. Let's say

1→$v_1, t_1$

2→$v_2, t_2$,

Meaning that "part one" lasts $t_1$ seconds and velocity is $v_1$ in that part. Same for the 2nd one. Obviously, $T=t_1+t_2$ because it is the total time.

Then, the average velocity is

$$ v_{average}=\dfrac{v_1t_1+v_2t_2}{T}$$

(Please, don't say "as my book and teachers say", you need to understand where this formula comes from. If you don't, let me know in the comments and I'll add it here).

And, of course, you can rewrite it as

$$ v_{average}=v_1 \frac{t_1}{T} + v_2 \frac{t_2}{T}$$.

So, no matter what the values are, you have $ v_{average}=\dfrac{v_1+v_2}{2}$, as long as, and only if

  • Those velocities are constant along their intervals.
  • All the intervals have the same duration.

That is,

If $t_1=t_2$, you can use that formula.

But this formula can be generalized for any number of intervals

$$ v_{average}=\dfrac{v_1t_1+v_2t_2+v_3t_3+...}{T}$$

Even in the case where the intervals are so small that velocity is contiousuly changing. This is a definition which still holds. So, yyou can write

$$ v_{average}=\dfrac{\int_{ini}^{final} v dt}{T} $$

So what happens if velocity varies linearly, as in the uniformly accelerated straight motion? Then $v(t)=v_0+at$, so

$$ v_{average}=\dfrac{\int_{ini}^{final} [v_0+at] dt}{T}= $$

$$ = \frac{v_0T + aT^2/2}{T}= v_0+\frac{(v_F-v_0)}{2}=\frac{v_0+v_F}{2}$$

A formula with the same form, but the meaning is different, because now they're not 2 intervals, but a continuously changing velocity.

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  • $\begingroup$ Thanks for explaining it so thoroughly. Just to clarify, what I meant to say was, in the second case, the body's initial velocity was V1 (even though it didn't change until it reached B), and it's final velocity was V2. I was just referring to it's initial and final velocity, which is V1 and V2 respectively. And we end up taking the arithmetic mean of the two values. That's what I meant. $\endgroup$ – π times e Oct 2 '18 at 16:21
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Because the formula Vavg = (V1 + V2)/2 should be valid if and only if acceleration is uniform. (As per my books and my teachers).

Obviously, as you've shown yourself, this is not the case.

If in doubt regarding any special case, you can go back to the definition of the average velocity and you'll get the correct results, including the results for your two special cases.

What the book and your teacher probably meant to say was that, when the acceleration is constant, this formula can be applied and when not - it, generally, shouldn't. That does not preclude, though, that, in some other special cases, the result could be the same.

For instance, we can state that the result would be the same, if the speed curve $v(t)$ is symmetric relative to the mid-point between points $(v_1,t_1)$ and $(v_2,t_2)$. This covers both of your special cases.

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  • $\begingroup$ Makes sense now. Maybe, that's what my teachers meant. Thank you! $\endgroup$ – π times e Oct 2 '18 at 16:14
  • $\begingroup$ @πtimese Glad to help. $\endgroup$ – V.F. Oct 2 '18 at 16:53
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It's incorrect to use the word "only". (1) is always valid if acceleration is constant, but that doesn't exclude the other situation. It still can be valid in certain special situations.

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