91Ó°ÊÓ

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

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

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

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

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

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

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)

$$\begin{gathered} \mathrm{³Ù²¹²ÔÎ\pm }=\frac{40}{-40} \\ \mathrm{θ}=-45°\left(\mathrm{North}\mathrm{of}\mathrm{West}\right) \end{gathered}$$

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,

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

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,

$$\begin{gathered} \mathrm{³Ù²¹²Ôθ}=\frac{\mathrm{y}}{\mathrm{x}} \\ \mathrm{³Ù²¹²Ôθ}=\frac{0.333}{0.333} \\ \mathrm{θ}=45°\left(\mathrm{North}\mathrm{ofEast}\right) \end{gathered}$$