11.2: Constant Acceleration

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As was done in one-dimensional kinematics, we may derive a set of equations for the motion of a particle under a constant acceleration. In two or three dimensions, though, it's a constant acceleration vector a. If the acceleration vector a is constant, we can bring it outside the integral sign of Eq. (11.1.4) just as we do with constant scalars. We get

\[\mathbf{v}(t)=\int \mathbf{a} d t=\mathbf{a} \int d t\]

or

\[\mathbf{v}(t)=\mathbf{a} t+\mathbf{C}\]

where \(\mathbf{C}\) is the constant of integration. By setting \(t=0\), we can see that physically, just as in one-dimensional kinematics, \(\mathbf{C}=\mathbf{v}_{0}=\mathbf{v}(0)\) represents the velocity vector at time \(t=0\), so

\[\mathbf{v}(t)=\mathbf{a} t+\mathbf{v}_{0}\]

Substituting this result into Eq. (11.1.3), we have

\[
\begin{align}
\mathbf{r}(t) & =\int\left(\mathbf{a} t+\mathbf{v}_{0}\right) d t \\[6pt]
& =\int \mathbf{a} t d t+\int \mathbf{v}_{0} d t \\[6pt]
& =\mathbf{a} \int t d t+\mathbf{v}_{0} \int d t
\end{align}
\]

or

\[\mathbf{r}(t)=\frac{1}{2} \mathbf{a} t^{2}+\mathbf{v}_{0} t+\mathbf{r}_{0}\]

where \(\mathbf{r}_{0}=\mathbf{r}(0)\) is the position vector at time \(t=0\).

The remaining constant-acceleration formula is a formula for \(\mathbf{v}(\mathbf{r})\), in which we eliminate time \(t\) to get an expression for velocity in terms of position. We did this in one dimension by solving the equation for \(v(t)\) for \(t\), then substituting into the equation for \(x(t)\) and solving for \(v\). Unfortunately, that technique won't work with vectors, because it would require dividing by a vector, which is not defined. Instead, being guided by the knowledge that the vector formula must reduce to the known scalar formula when the vectors are one-dimensional, we proceed as follows. Start with Eq. \(\PageIndex{7}\) for \(\mathbf{r}(t)\) for constant acceleration:

\[
\begin{align}
\mathbf{r}(t) & =\frac{1}{2} \mathbf{a} t^{2}+\mathbf{v}_{0} t+\mathbf{r}_{0} \\[6pt]
\mathbf{r}-\mathbf{r}_{0} & =\frac{1}{2} \mathbf{a} t^{2}+\mathbf{v}_{0} t
\end{align}
\]

Now take the dot product of both sides with the acceleration a:

\[
\begin{align}
\mathbf{a} \cdot\left(\mathbf{r}-\mathbf{r}_{0}\right) & =\mathbf{a} \cdot\left(\frac{1}{2} \mathbf{a} t^{2}+\mathbf{v}_{0} t\right) \\[6pt]
& =\frac{1}{2} a^{2} t^{2}+\mathbf{a} \cdot \mathbf{v}_{0} t,
\end{align}
\]

and multiply both sides by 2 :

\[2 \mathbf{a} \cdot\left(\mathbf{r}-\mathbf{r}_{0}\right)=a^{2} t^{2}+2 \mathbf{a} \cdot \mathbf{v}_{0} t \]

The left-hand side looks similar to the second term on the right-hand side of the one-dimensional Eq.\(\PageIndex{7}\), but we still need to eliminate \(t\) on the right-hand side. To do that, let's start by working on the first term on the right-hand side of Eq. \(\PageIndex{12}\). Starting with Eq. \(\PageIndex{3}\), we have

\[
\begin{align}
\mathbf{v}(t) & =\mathbf{a} t+\mathbf{v}_{0} \\[6pt]
\mathbf{v}-\mathbf{v}_{0} & =\mathbf{a} t
\end{align}
\]

Now take the dot product of the left-hand side of Eq. \(\PageIndex{10}\) with itself, and dot the right-hand side with itself:

\[
\begin{align}
\left(\mathbf{v}-\mathbf{v}_{0}\right) \cdot\left(\mathbf{v}-\mathbf{v}_{0}\right) & =(\mathbf{a} t) \cdot(\mathbf{a} t) \\[6pt]
v^{2}-2 \mathbf{v} \cdot \mathbf{v}_{0}+v_{0}^{2} & =a^{2} t^{2}
\end{align}
\]

Next, let's work on the second term on the right-hand side of Eq. \(\PageIndex{12}\). To do this, let's take the dot product of both sides of Eq. \(\PageIndex{14}\) with \(\mathbf{v}_{0}\) :

\[
\begin{align}
\mathbf{v}_{0} \cdot\left(\mathbf{v}-\mathbf{v}_{0}\right) & =\mathbf{v}_{0} \cdot \mathbf{a} t \\[6pt]
2 \mathbf{v} \cdot \mathbf{v}_{0}-2 v_{0}^{2} & =2 \mathbf{a} \cdot \mathbf{v}_{0} t
\end{align}
\]

Now we have all the pieces we need to eliminate \(t\). In Eq. \(\PageIndex{12}\), we use Eq. \(\PageIndex{16}\) to replace \(a^{2} t^{2}\), and we use Eq. \(\PageIndex{18}\) to replace \(2 \mathbf{a} \cdot \mathbf{v}_{0} t\) :

\[
\begin{align}
2 \mathbf{a} \cdot\left(\mathbf{r}-\mathbf{r}_{0}\right) & =\left(v^{2}-2 \mathbf{v} \cdot \mathbf{v}_{0}+v_{0}^{2}\right)+\left(2 \mathbf{v} \cdot \mathbf{v}_{0}-2 v_{0}^{2}\right) \\[6pt]
& =v^{2}-v_{0}^{2}
\end{align}
\]

or

\[v^{2}=v_{0}^{2}+2 \mathbf{a} \cdot\left(\mathbf{r}-\mathbf{r}_{0}\right)\]