# Deriving vector equation for average acceleration

How can I derive the vector equation $$\vec{a}_{avg} = \frac{\vec{v}_2-\vec{v}_1}{t_2-t_1} = \frac{\Delta \vec{v}}{\Delta t}$$ for average acceleration?

I know how to derive the average $$x$$ acceleration, the average $$y$$ acceleration and the average $$z$$ acceleration:

$$a_{x,avg} = {1 \over t_2-t_1} \int_{t_1}^{t_2} a_x(t) dt = {1 \over t_2-t_1} \int_{t_1}^{t_2} {dv_x \over dt} dt = {[v_x(t)]_{t_1}^{t_2} \over t_2-t_1} = {v_x(t_2) - v_x(t_1) \over t_2-t_1}$$

and similarly for the average $$y$$ acceleration $$a_{y,avg}$$ and the average $$z$$ acceleration $$a_{z,avg}$$. However, I don't know how to say that the $$x$$-component of the average acceleration vector $$\vec{a}_{avg}$$ equals the average $$x$$ acceleration, that the $$y$$-component of the average acceleration vector $$\vec{a}_{avg}$$ equals the average $$y$$ acceleration, and that the $$z$$-component of the average acceleration vector $$\vec{a}_{avg}$$ equals the average $$z$$ acceleration. That is, I don't know how to say that:

$$\vec{a}_{avg} = (a_{x,avg},a_{y,avg},a_{z,avg})$$

How can I rigorously show that $$\vec{a}_{avg} = (a_{x,avg},a_{y,avg},a_{z,avg})$$?

## 1 Answer

Since your acceleration vector is $$\vec{a} = \bigg( \frac{d v_x}{dt}, \frac{d v_y}{dt}, \frac{d v_z}{dt} \bigg),$$ averaging over time one gets $$\vec{a}_{avg} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} \vec{a}(t) dt = \bigg( {1 \over t_2-t_1} \int_{t_1}^{t_2} {dv_x \over dt} dt, {1 \over t_2-t_1} \int_{t_1}^{t_2} {dv_y \over dt} dt, {1 \over t_2-t_1} \int_{t_1}^{t_2} {dv_z \over dt} dt \bigg) = \frac{1}{ t_2-t_1 } \big([v_x(t)]_{t_1}^{t_2}, [v_y(t)]_{t_1}^{t_2}, [v_z(t)]_{t_1}^{t_2} \big) = \frac{\Delta \vec{v}}{\Delta t}.$$