0
$\begingroup$

I just had a quick question on something I learned in physics the other day. We were taught that average velocity, $$\bar{v} = \frac{(x_f - x_i)}{(t_f - t_i)}.$$ I was wondering if this is only true if the displacement-time graph is linear or constant. Is this the case for similar equations (ex. average acceleration, average jerk, etc...)?

Similarly, is the equation average $$\bar{a} = \frac{(a_f + a_i)}{2}$$ true for all $a$, where the $a$-time graph is linear or constant?

$a$ is a placeholder for velocity, acceleration, jerk, etc.

$\endgroup$
2
$\begingroup$

The average is applicable no matter what the displacement-time graph is like.

Let me first introduce the shorthand $\Delta x = x_f - x_i$, and likewise $\Delta t = t_f -t_i$. Then the average velocity is $\bar{v}=\frac{\Delta x}{\Delta t}$. The average velocity can be thought of as "What velocity must I move at, for a total time $\Delta t$, to have moved a total amount $\Delta x$?"

No matter how you're actually moving, you could be accelerating around like a maniac, you'll still travel an amount $\Delta x$, in a time $\Delta t$, and so there's always an equivalent constant velocity you could travel at for a time $\Delta t$, to end up travelling the exact same amount, $\Delta x$.

What you may have picked up on though, is that for the case where your displacement graph is a straight line, then your actual velocity will be the same as the average velocity.


EDIT: to address comment

Consider a displacement time graph of x^2 that is defined for 0<=t<=2. If u use the average velocity formula, you have 4-0/2-0 = 2. But if you take the tangent at each point and average all values you get what I call the "true" average velocity. What I meant in my previous comment was that the true average velocity = the calculated ( straight line) iff the displacement-time graph is linear. Is this true?

First, when you say a graph of $x^2$, I assume you mean a graph of $x=t^2$.

Let's be more general, let's say there's a position as a function of time, $x(t)$. The tangeant of the position is the velocity $v(t)$. If we want to find the average of this tangeant across the entire time, the correct way to do this is with integration. That gives you

$$\frac{\int^{t_f}_{t_i} v(t) \text{d}t}{\int^{t_f}_{t_i} \text{d}t}= \frac{[x(t)]^{t_f}_{t_i}}{[t]^{t_f}_{t_i}} = \frac{x_f - x_i}{t_f - t_i}.$$

That is it gives you the same equation, for all cases. So the answer to your question is "no".

I'm not sure if you've come across calculus yet, but the integral is doing the same procedure you outline. At each point of the curve, it adds up all the velocities, then dividing on the bottom by $t$ gives you the average.

$\endgroup$
  • $\begingroup$ Right. So correct me if I'm wrong: Average velocity assumes that the particle is taking a straight line to its endpoint. However, I must make another thing clear. If this is the case, then that means the calculated average velocity =/= the true average velocity for all graphs that are not linear. $\endgroup$ – Typical Highschooler Aug 22 '17 at 1:43
  • $\begingroup$ Yes average velocity is the straight line velocity that would take you $\Delta \vec{r}$ in time $\Delta t$. When you say the "true average velocity", what do you mean by that term? $\endgroup$ – CDCM Aug 22 '17 at 10:00
  • $\begingroup$ Consider a displacement time graph of x^2 that is defined for 0<=t<=2. If u use the average velocity formula, you have 4-0/2-0 = 2. But if you take the tangent at each point and average all values you get what I call the "true" average velocity. What I meant in my previous comment was that the true average velocity = the calculated ( straight line) iff the displacement-time graph is linear. Is this true? $\endgroup$ – Typical Highschooler Aug 22 '17 at 14:33
  • $\begingroup$ @TypicalHighschooler I've updated my answer in response to your comment. $\endgroup$ – CDCM Aug 22 '17 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.