- #1
brotherbobby
- 699
- 163
- 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)##
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.