Average velocity ##\bar{v}## for a uniformly accelerating particle

[brotherbobby]

Homework Statement

Show that if a particle is undergoing uniform acceleration, its average velocity over a time $t$ may be written as $\bar{v} = \dfrac{v_f+v_i}{2}$ or as $\dfrac{x_f-x_i}{t}$, where $f$ and $i$ refer to initial and final values.

Relevant Equations

1. Average velocity is defined to be $\quad\bar{v} \overset{\text{def.}}{=}\dfrac{\Delta x}{\Delta t}$.
2. For uniformly accelerated motion, final velocity $\quad v_f=v_1+a_0(t_f-t_i)$

Statement of the problem : Let me copy and paste the problem from the text. Please note it's part (i) that I am seeking to answer. I know the answer to part (ii).

Solution : To show where my error appears, I take the time(s) to be $t_f$ and $t_i$, whereby the given time interval $t = t_f-t_i$. The average velocity $\small{\quad\bar{v}\overset{\text{def.}}{=}\dfrac{\Delta x}{\Delta t}= \dfrac{\int\limits_{t_i}^{t_f}v(t)dt}{t_f-t_i}=\dfrac{\int\limits_{t_i}^{t_f}(v_i+a_0t)dt}{t_f-t_i}=v_i+\dfrac{a_0}{t_f-t_i}\left[ \dfrac{t^2}{2} \right]_{t_i}^{t_f}=v_i+\dfrac{a_0}{2(t_f-t_i)}\left( t_f^2 - t_i^2\right) = v_i+\dfrac{a_0}{2}\left( t_f+t_i \right)=\dfrac{v_i+\{v_i+a_0\left( t_f+t_i\right)\}}{2}}$
The last term is not equal to the final velocity $v_f$ : $\quad\dfrac{v_i+\overbrace{\{v_i+a_0\left( t_f+t_i\right)\}}^{\ne v_f}}{2}\ne\dfrac{v_i+v_f}{2}$.
We note that the final velocity $v_f=v_i+a_0(t_f-t_i)$. For me, the minus sign $-$ is replaced with a plus$+$ sign.
This is where I am stuck. Request : A hint would be welcome.

 

[PeroK]

We note that the final velocity $v_f=v_i+a_0(t_f-t_i)$. For me, the minus sign $-$ is replaced with a plus$+$ sign.

You'll have to explain why you made that a plus sign!

 

[brotherbobby]

You'll have to explain why you made that a plus sign!

From my calculations above. The two terms which have "$-$" signs cancel, leaving the term $t_f+t_i$.

 

[PeroK]

View attachment 337949
From my calculations above. The two terms which have "$-$" signs cancel, leaving the term $t_f+t_i$.

You have integrated wrongly. You should find that the integral of velocity for constant acceleration is
$$x_f -x_i = v_i(t_f-t_i) +\frac 1 2 a(t_f^2 -t_i^2)$$
Ps sorry, I meant:
$$x_f -x_i = v_i(t_f-t_i) +\frac 1 2 a(t_f-t_i)^2$$

 

[brotherbobby]

I don't follow you. What you wrote is exactly what I did.

$x_f -x_i = v_i(t_f-t_i) +\frac 1 2 a(t_f^2 -t_i^2)$

If you divide both sides of the equation by $t_f-t_i$, you get $\bar v= v_i+\dfrac{1}{2}a_0(t_f+t_i)$, which is what I got.

You have integrated wrongly

Can you show me where in my workings in post #1 above?

 

[erobz]

Your integration is fine. What you have inadvertently done is used the wrong function for $v(t)$. I ask you to evaluate $v(t)$ at $t=t_o$ and tell us what you get.

 

[brotherbobby]

Your integration is fine. What you have inadvertently done is used the wrong function for $v(t)$. I ask you to evaluate $v(t)$ at $t=t_o$ and tell us what you get.

I got you. We should have $v(t) = v_i+a_0(t-t_i)$. Hence when $t=t_i$, $v(t_i)=v_i$.

 

[brotherbobby]

I show you the calculations below in rough. Thanks. I have solved the problem.

 

[erobz]

It will save you significant algebra if you do a $u$ substitution with $u = t - t_o$ to evaluate the integral.

$du = dt$

$u_f = t_f - t_o$

$u_o = t_o -t_o = 0$

$$ \bar{v} = \frac{ \int_{0}^{u_f} \left( v_o + a_o u \right) du}{u_f} $$

$$ \implies \bar{v} = \frac{v_o u_f + \frac{1}{2} a_o u_f^2 }{u_f} $$

$$ \implies \bar{v} = v_o + \frac{1}{2} a_o u_f = v_o + \frac{1}{2} a_o ( t_f - t_o ) = \frac{v_o+v_f}{2} $$