At one instant a bicyclist is $\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$due east of a park’s flagpole, going due south with a speed of$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. Then$\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$later, the cyclist is$\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$due north of the flagpole, going due east with a speed of$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$.For the cyclist in this$\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$interval, what are the(a)magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity and the (e) magnitude and (f) direction of the average acceleration?

Considering $\hat{\mathrm{i}}$pointed at east and $\hat{\mathrm{j}}$pointed north then,

$$\begin{gathered} \overrightarrow{{\mathrm{r}}_0}=\left(-40\mathrm{m}\right)\hat{\mathrm{i}} \\ \overrightarrow{\mathrm{r}}=\left(-40\mathrm{m}\right)\hat{\mathrm{j}} \\ \overrightarrow{{\mathrm{v}}_0}=\left(-10\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}} \\ \mathrm{v}=\left(-10\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}} \end{gathered}$$

This problem deals with a simple algebraic operation that involves calculation of magnitude and direction of position vector, average velocity and the acceleration at given time. Here, to find all these quantities, Pythagorean theorem can be used.

Formulae

$$\begin{gathered} ∆\overrightarrow{r}=\overrightarrow{r}-\overrightarrow{r_0}\left(i\right) \\ \tan \theta =\frac{y}{x}\left(ii\right) \\ {\overrightarrow{v}}_{avg}=\frac{∆\overrightarrow{r}}{t}\left(iii\right) \\ {\overrightarrow{a}}_{avg}=\frac{\overrightarrow{v}-{\overrightarrow{v}}_0}{t}\left(iv\right) \end{gathered}$$

Substituting the given values of ${\overrightarrow{r}}_0$and $\overrightarrow{r}$in equation (i), the displacement is given by,

$∆\overrightarrow{r}=-40\hat{i}+40\hat{j}$

So magnitude of displacement will be,

$$\begin{gathered} ∆\mathrm{r}=\sqrt{{\left(-40\right)}^2+{40}^2} \\ ∆\mathrm{r}=56.6\mathrm{m} \end{gathered}$$

Direction of displacement is given by equation (ii)

$$\begin{gathered} \mathrm{tan\alpha }=\frac{40}{-40} \\ \mathrm{\theta }=-45°\left(\mathrm{North}\mathrm{of}\mathrm{West}\right) \end{gathered}$$

Substituting the magnitude of the displacement in equation (iii), the average velocity can be written as,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{avg}}=\frac{∆\mathrm{r}}{\mathrm{t}} \\ {\mathrm{v}}_{\mathrm{avg}}=\frac{56.6}{30} \\ \mathrm{Thus},{\mathrm{v}}_{\mathrm{avg}}=1.89\mathrm{m}/\mathrm{s} \end{gathered}$$

The direction of average velocity is the same as direction of displacement, that is,$-45°$ (North of West).

Magnitude of average acceleration is given by,

$$\begin{gathered} {\mathrm{a}}_{\mathrm{avg}}=\frac{\overrightarrow{\mathrm{v}}-{\overrightarrow{\mathrm{v}}}_0}{\mathrm{t}} \\ {\mathrm{a}}_{\mathrm{avg}}=\frac{10\hat{\mathrm{i}}+10\hat{\mathrm{j}}}{30} \\ {\mathrm{a}}_{\mathrm{avg}}=0.333\hat{\mathrm{i}}+0.333\hat{\mathrm{j}} \end{gathered}$$

So, magnitude is as follow:

$$\begin{gathered} {\mathrm{a}}_{\mathrm{avg}}=\sqrt{0.{333}^2+0.{333}^2} \\ {\mathrm{a}}_{\mathrm{avg}}=0.471\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Direction of an average acceleration is,

$$\begin{gathered} \mathrm{tan\theta }=\frac{\mathrm{y}}{\mathrm{x}} \\ \mathrm{tan\theta }=\frac{0.333}{0.333} \\ \mathrm{\theta }=45°\left(\mathrm{North}\mathrm{ofEast}\right) \end{gathered}$$