P01R2015 – Graphical Subtraction of Vectors

By definition, the position vector $\overrightarrow{r}$ points from the origin to the location of the particle it describes the position of. Thus, we draw the vectors $\overrightarrow{r_A}$, $\overrightarrow{r_B}$ and $\overrightarrow{r_C}$ extending from the origin to points $A$, $B$, and $C$ respectively.

The displacement vector $\mathrm{\Delta }{\overrightarrow{r}}_{A\to B}={\overrightarrow{r}}_B-{\overrightarrow{r}}_A$ can be constructed by flipping the vector $\overrightarrow{r_A}$ and placing its tail at the tip of $\overrightarrow{r_B}$ as shown below. The displacement vector $\mathrm{\Delta }{\overrightarrow{r}}_{A\to B}$ is the obtained by connecting the tail of $\overrightarrow{r_B}$ to the tip of $-\overrightarrow{r_A}$. You can then slide this vector so that it originates at point $A$ and terminates at point $B$. Alternatively, this can be simplified and generalized to simply drawing the vector that starts at point $A$ and ends at point $B$ thus describing the displacement $A\to B$.

We then use the simplified method to draw the displacement vector $\mathrm{\Delta }{\overrightarrow{r}}_{B\to C}$ and directly draw the vector that extends from point $B$ to point $C$ and is therefore equal to ${\overrightarrow{r}}_C-{\overrightarrow{r}}_B$.

The average velocities are defined by

\begin{aligned}
{\overrightarrow{v}}_A&=\frac{\mathrm{\Delta }{\overrightarrow{r}}_{A\to B}}{\mathrm{\Delta }t}=\frac{\mathrm{\Delta }{\overrightarrow{r}}_{A\to B}}{1}=\mathrm{\Delta }{\overrightarrow{r}}_{A\to B} \\
\\
{\overrightarrow{v}}_B&=\frac{\mathrm{\Delta }{\overrightarrow{r}}_{B\to C}}{\mathrm{\Delta }t}=\frac{\mathrm{\Delta }{\overrightarrow{r}}_{B\to C}}{1}=\mathrm{\Delta }{\overrightarrow{r}}_{B\to C}
\end{aligned}

Since $\mathrm{\Delta }t=1\ s$, we find that the average velocity vectors ${\overrightarrow{v}}A$ and ${\overrightarrow{v}}_B$ have the same magnitude as the displacement vectors $\mathrm{\Delta }{\overrightarrow{r}}_{A\to B}$ and $\mathrm{\Delta }{\overrightarrow{r}}_{B\to C}$ respectively. Thus, we draw the vectors ${\overrightarrow{v}}_A$ and ${\overrightarrow{v}}_B$ as follows

By definition the average acceleration at point $B$ is given by

{\overrightarrow{a}}_{avg}=\frac{{\overrightarrow{v}}_B-{\overrightarrow{v}}_A}{\mathrm{\Delta }t}={\overrightarrow{v}}_B-{\overrightarrow{v}}_A

We start by constructing the vector ${\overrightarrow{v}}_B-{\overrightarrow{v}}_A$ by flipping ${\overrightarrow{v}}_A$ into $-{\overrightarrow{v}}_A$ and then adding $-{\overrightarrow{v}}_A$ to ${\overrightarrow{v}}_B$ as shown below.

We then note that because $\mathrm{\Delta }t=1\ s$, the two vectors ${\overrightarrow{a}}{avg}$ and ${\overrightarrow{v}}_B-{\overrightarrow{v}}_A$ have the same magnitude and that the average acceleration vector ${\overrightarrow{a}}_{avg}$ should be drawn as follows

This result is immediately derived from the definition of the average velocity as shown below

{\overrightarrow{v}}_A=\frac{\mathrm{\Delta }{\overrightarrow{r}}_{A\to B}}{\mathrm{\Delta }t}=\frac{{\overrightarrow{r}}_B-{\overrightarrow{r}}_A}{\mathrm{\Delta }t}\ \ \ \ \ \Leftrightarrow \ \ \ \ \ \boxed{{\overrightarrow{r}}_B-{\overrightarrow{r}}_A={\overrightarrow{v}}_A\cdot \mathrm{\Delta }t}

This shows that the displacement vector ${\overrightarrow{r}}_B-{\overrightarrow{r}}_A$ changes by an amount ${\overrightarrow{v}}_A\cdot \mathrm{\Delta }t$ and can therefore be interpreted as the fact that over time, velocity changes the position of the particle by an amount equal to its displacement.

This results is immediately derived from the definition of the average acceleration as shown below

{\overrightarrow{a}}_A=\frac{\mathrm{\Delta }\overrightarrow{v}}{\mathrm{\Delta }t}=\frac{{\overrightarrow{v}}_B-{\overrightarrow{v}}_A}{\mathrm{\Delta }t}\ \ \ \ \ \Leftrightarrow \ \ \ \ \ \boxed{{\overrightarrow{v}}_B-{\overrightarrow{v}}_A={\overrightarrow{a}}_A\cdot \mathrm{\Delta }t}

This shows that the velocity changes by an amount ${\overrightarrow{v}}_B-{\overrightarrow{v}}_A$ which is equal to ${\overrightarrow{a}}_A\cdot \mathrm{\Delta }t$ and can therefore be interpreted as the fact that over time, acceleration changes the velocity of the particle.

By applying the definition of average velocity, we obtain the following two results for $\mathrm{\Delta }t=1\ s$ and $\mathrm{\Delta }t=3\ s$ respectively.

If $\mathrm{\Delta }t=1\ s$:

{\overrightarrow{v}}_A=\frac{\mathrm{\Delta }{\overrightarrow{r}}_{A\to B}}{\mathrm{\Delta }t}=\frac{{\overrightarrow{r}}_B-{\overrightarrow{r}}_A}{\mathrm{\Delta }t}={\overrightarrow{r}}_B-{\overrightarrow{r}}_A

If $\mathrm{\Delta }t=3\ s$:

{\overrightarrow{v}}A=\frac{\mathrm{\Delta }{\overrightarrow{r}}_{A\to B}}{\mathrm{\Delta }t}=\frac{{\overrightarrow{r}}_B-{\overrightarrow{r}}_A}{\mathrm{\Delta }t}=\frac{{\overrightarrow{r}}_B-{\overrightarrow{r}}_A}{3}

In the first case ${\overrightarrow{v}}_A$ and ${\overrightarrow{r}}_B-{\overrightarrow{r}}_A$ have the same magnitude and in the second case ${\overrightarrow{v}}_A$ is a third of the length of ${\overrightarrow{r}}_B-{\overrightarrow{r}}_A$. We draw the vector ${\overrightarrow{v}}_A$ for each situation below.