Fix typos DescribingMotionIn1D

tex/DescribingMotionIn1D.tex

@@ -542,7 +542,7 @@ \subsection{Solutions}
 \end{align*}
 We want the left hand side to be $2a(x-x_0)$, so we multiply each term by $2a$:
 \begin{align*}
-2a(x-x_0)x&=(2a)v_0\left( \frac{v-v_0}{a}\right) +(2a)\frac{1}{2}a\left( \frac{v-v_0}{a}\right) ^2\\
+2a(x-x_0)&=(2a)v_0\left( \frac{v-v_0}{a}\right) +(2a)\frac{1}{2}a\left( \frac{v-v_0}{a}\right) ^2\\
 2a(x-x_0)&=(2v_0)a\left(\frac{v-v_0}{a}\right)+a^2\left( \frac{v-v_0}{a}\right) ^2\\
 2a(x-x_0)&=2v_0(v-v_0)+(v-v_0)^2
 \end{align*}
@@ -558,7 +558,7 @@ \subsection{Solutions}
 2a(x-x_0)&=v^2-v_0^2\\
 \therefore v^2-v_0^2&=2a(x-x_0)\\
 \end{align*}
-If you choose a coordinate system such that $x_0$, this equation becomes $v^2-v_0^2=2ax$.
+If you choose a coordinate system such that $x_0=0$, this equation becomes $v^2-v_0^2=2ax$.
 \end{solution}
 
 \newpage
@@ -631,7 +631,7 @@ \subsection{Solutions}
 \begin{itemize}
 \item Between $t=\SI{0}{s}$ and $t=\SI{3}{s}$, position decreases quadratically, as the velocity is negative and decreasing.
 \item Between $t=\SI{3}{s}$ and $t=\SI{6}{s}$, position decreases linearly, since the velocity is negative and constant.
-\item Between $t=\SI{6}{s}$ and $t=\SI{9}{s}$, the position continues to decrease, but at a lesser rate ans the velocity approaches zero. When the velocity is zero, the position stop changing, and starts to increase quadratically as the velocity becomes positive and increasing.
+\item Between $t=\SI{6}{s}$ and $t=\SI{9}{s}$, the position continues to decrease, but at a lesser rate as the velocity approaches zero. When the velocity is zero, the position stop changing, and starts to increase quadratically as the velocity becomes positive and increasing.
 \item Between $t=\SI{9}{s}$ and $t=\SI{12}{s}$, the position continues to increase, but at a lesser rate as the velocity decreases back to zero.
 \end{itemize}
 \end{solution}