6
$\begingroup$

Since average velocity is defined as$^1$ $$\vec{\mathbf v}_\mathrm{av}=\frac{\vec{\mathbf x}-\vec{\mathbf x}_0}{t-t_0},$$ where $\vec{\mathbf x}$ denotes position, why is this quantity equal to $$\frac{\vec{\mathbf v}+\vec{\mathbf v}_0}{2},$$ where $\vec{\mathbf v}=\frac{d\vec{\mathbf x}}{dt}$ and $\vec{\mathbf v}_0=\left.\frac{d\vec{\mathbf x}}{dt}\right|_{t=t_0}$, when acceleration is constant?

What in particular about constant acceleration allows average velocity to be equal to the midpoint of velocity?

$^1$: Resnick, Halliday, Krane, Physics (5th ed.), equation 2-7.

$\endgroup$
4
$\begingroup$

Note that $\vec{\mathbf v}_\mathrm{av}$ is defined as the average value of $\vec{\mathbf v}$: $$\vec{\mathbf v}_\mathrm{av}:=\frac{1}{t_1-t_0}\int_{t_0}^{t_1}\vec{\mathbf v}(t)\,\mathrm dt.$$ Since $\vec{\mathbf x}$ is the antiderivative of $\vec{\mathbf v}$, this equals $$\frac{\vec{\mathbf x}(t_1)-\vec{\mathbf x}(t_0)}{t_1-t_0}.$$ However, when acceleration is constant, and thus $\vec{\mathbf v}$ is a line (that is, $\vec{\mathbf v}(t)=\vec{\mathbf a}t+\vec{\mathbf v}_0$), then by plugging into the average value integral, you obtain the equality $$\vec{\mathbf v}_\mathrm{av}=\frac{\vec{\mathbf v}(t_1)+\vec{\mathbf v}(t_0)}{2}.$$

$\endgroup$
6
$\begingroup$

Sticking to one dimension for simplicity, at a constant acceleration, $a$, the distance travelled in a time $t$ is simply:

$$ s = v_0 t + \frac{1}{2}at^2 $$

So the average velocity, $v_{av} = s/t$, is:

$$ v_{av} = v_0 + \frac{1}{2}at $$

But acceleration $\times$ time is just the change in velocity i.e. $at = v - v_0$ so:

$$ v_{av} = v_0 + \frac{1}{2}(v - v_0) = \frac{v_0 + v}{2} $$

$\endgroup$
1
$\begingroup$

I thought I would chime in and give a slightly more fundamental approach in finding the solution. Starting with first principles, we will begin by taking the double integral of a(t), where a(t) = a, and a is a constant. $$ \iint a(t) dt = x(t) = \frac{1}{2}at^2 + C_0t + C_1 $$ Where $$ C_0 = v_i $$ is the initial velocity and $$C_1 = x_i $$ is the initial position we then get position as a function of time: $$ x(t) = \frac{1}{2}at^2 + v_it + x_i $$ Using the definition of average velocity we get $$ v_{avg}=\frac{\Delta x}{\Delta t}=\frac{1}{t_f -t_i}[x(t_f)-x(t_i)] $$ Using some magic this simplifies to $$ v_{avg} = \frac{1}{2}a(t_f+t_i)+v_i $$ We can simplify this further by remembering that $$ v(t)=at+v_i $$ and by rearranging our previous solution we get $$ \frac{1}{2}a(t_f+t_i)+v_i=\frac{[at_f+v_i]+[at_i+v_i]}{2}=\frac{v(t_f)+v(t_i)}{2} $$ Which simplifies to $$ v_{avg} = \frac{v_f+v_i}{2} $$ You can get the same result more directly by using the integral definition for finding the average $$ v_{avg}=\frac{1}{t_f -t_i}\int_{t_i}^{t_f} v(t)dt $$ I chose the long way as it describes the process from a more fundamental perspective.

$\endgroup$

protected by Qmechanic Sep 7 '15 at 5:52

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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