In the undergraduate text of physics, the equation for velocity and displacement at constant acceleration are given in scalar form. For example, my text reads

$$v^2 = v_0^2 + 2a(x-x_0)$$

But today, I am reading another text which gives the vector form for displacement. I am trying to write above equation in vector form also but how do I deal with that square in vector? I am tackling that square with dot product of two vector such that $v^2 \to \vec{v}\cdot\vec{v}$, is that correct? So what about the multiplication of acceleration and the displacement? dot product again?

This equation looks a bit confusing to me. Let's say I throw a stone upright with initial speed $v_0$ and it reaches the highest point sometimes later (so $v=0$), and the coordinate is given with y axis down and x axis horizontal. Note that the displacement is negative since the coordinate's yaxis is downward, also the gravity is along the positive y axis so the equation becomes $$ 0 = v_0^2 + 2g(-x) $$ which gives $$ x = \frac{v_0^2}{2g} $$ is that right?


1 Answer 1


Here's a breakdown of the relevant vector algebra.

$ \frac{d\vec x}{dt} = \vec v, \qquad \frac{d\vec v}{dt} = \vec a $

$ \frac{d}{dt} v^2 = \frac{d}{dt} ( \vec v \cdot \vec v) = \frac{d\vec v}{dt} \cdot \vec v + \vec v \cdot \frac{d\vec v}{dt} = 2 \vec a \cdot \vec v $

$ \int_{t_0}^t dt' \frac{d}{dt'} v^2 = v^2 - v_0^2 $

$ \int_{t_0}^t 2 \vec a \cdot \vec v dt' = 2 \int_{{\vec x}_0}^{\vec x} \vec a \cdot d \vec x' = 2 \vec a \cdot \int_{{\vec x}_0}^{\vec x} d \vec x' = 2 \vec a \cdot (\vec x - {\vec x}_0 ) $

Note that in the last line the constant acceleration vector assumption is necessary for pulling the dot product outside of the integral.

Finally, if the acceleration vector is in the same direction as the displacement then it reduces to the familiar scalar equation:

$ v^2 - v_0^2 = 2 \vec a \cdot (\vec x - {\vec x}_0 ) = 2a\| \vec x - {\vec x}_0 \| $.


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