Question

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?

Short answer

(a) Magnitude of displacement is$56.6\mathrm{m}$

(b) Direction of displacement is $-45°$(North of West).

(c) Magnitude of average velocity is$1.89\mathrm{m}/\mathrm{s}$

(d) Direction of average velocity is $-45°$(North of West).

(e) Magnitude of average acceleration is$0.471\mathrm{m}/{\mathrm{s}}^2$

(f) Direction of average acceleration is$45°$(North of East).

Step-by-step solution

Given information

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

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

Understanding the concept

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

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

(a) To find the magnitude of displacement

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,

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

(b) To find the direction of displacement

Direction of displacement is given by equation (ii)

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

(c) To find the magnitude of average velocity

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

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

(d) To find the direction of average velocity

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

(e) Determining the magnitude of average acceleration

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:

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

(f) To find thedirection of average acceleration

Direction of an average acceleration is,

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