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.


2 Answers 2


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:

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


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 $$\vec{\mathbf v}_\mathrm{av}=\frac{\vec{\mathbf v}(t_1)+\vec{\mathbf v}(t_0)}{2}.$$


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