# Why is average velocity the midpoint of initial and final velocity under constant acceleration?

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.

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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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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:

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

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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.

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## protected by Qmechanic♦Sep 7 '15 at 5:52

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