91Ó°ÊÓ

It is given that,

$$\begin{gathered} {\stackrel{→}{\mathrm{v}}}_0=4.0\hat{\mathrm{i}}-2.0\hat{\mathrm{j}}+3.0\hat{\mathrm{k}} \\ {\stackrel{→}{\mathrm{v}}}_{\mathrm{f}}=-2.0\hat{\mathrm{i}}-2.0\hat{\mathrm{j}}+5.0\hat{\mathrm{k}} \\ \mathrm{t}=4\mathrm{s} \end{gathered}$$

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

Formulae:

The average acceleration can be written as,

${\stackrel{→}{a}}_{avg}=\frac{∆\stackrel{→}{v}}{t}$ (i)

Where $∆\stackrel{→}{v}$is the average velocity which is given by,

$∆\stackrel{→}{v}={\stackrel{→}{v}}_f-{\stackrel{→}{v}}_0$ (ii)

Where, ${\stackrel{→}{v}}_0$and ${\stackrel{→}{v}}_0$are the initial and final velocities

The magnitude of the acceleration is

$a_{avg}=\sqrt{a_x^2+a_y^2}$ (iii)

The direction of the acceleration is,

$\tan \mathrm{θ}=\frac{a_y}{a_x}$

Substituting the values of in equation (ii), the average velocity can be written as,

$∆\stackrel{→}{v}=(2.0\hat{i}-2.0\hat{j}+5.0\hat{k})-(4.0\hat{i}-2.0\hat{j}+3.0\widehat{k)}$

Substituting in equation (i), the acceleration will be,

$$\begin{gathered} {\stackrel{→}{\mathrm{a}}}_{\mathrm{avg}}=\frac{∆\stackrel{→}{\mathrm{v}}=(2.0\hat{\mathrm{i}}-2.0\hat{\mathrm{j}}+5.0\hat{\mathrm{k}})-(4.0\hat{\mathrm{i}}-2.0\hat{\mathrm{j}}+3.0\widehat{\mathrm{k})}}{4} \\ \mathrm{So}, \\ {\stackrel{→}{\mathrm{a}}}_{\mathrm{avg}}=(-1.5\hat{\mathrm{i}}+0.5\hat{\mathrm{k}})\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The magnitude of the acceleration using equation (iii) can be written as,

$$\begin{gathered} {\stackrel{→}{\mathrm{a}}}_{\mathrm{avg}}=\sqrt{1.5^2+0.5^2} \\ {\stackrel{→}{\mathrm{a}}}_{\mathrm{avg}}=1.6\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Using equation (iv) the direction is can be written as,

$$\begin{gathered} \mathrm{³Ù²¹²Ôθ}=\frac{0.5}{-1.5} \\ \mathrm{Î\pm }=-{18}^{°} \end{gathered}$$

So, the angle would be measured in the clockwise direction with respect to negative x axis. The angle can be written as $180-18={162}^{°}$with respect to positive x axis in the counter clockwise direction.