Question

$$\begin{gathered} \mathbf{A}\mathbf{}\mathbf{proton}\mathbf{}\mathbf{initially}\mathbf{}\mathbf{has}\mathbf{}\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{then}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}\mathbf{}\mathbf{later}\mathbf{}\mathbf{has}\mathbf{}\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{} \\ \mathbf{(}\mathbf{in}\mathbf{}\mathbf{meters}\mathbf{}\mathbf{per}\mathbf{}\mathbf{second}\mathbf{)}\mathbf{.}\mathbf{}\mathbf{For}\mathbf{}\mathbf{that}\mathbf{}\mathbf{,}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}\mathbf{}\mathbf{what}\mathbf{}\mathbf{are}\mathbf{}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{the}\mathbf{}\mathbf{proton}\mathbf{’}\mathbf{s}\mathbf{}\mathbf{average}\mathbf{}\mathbf{acceleration}\mathbf{}{\overrightarrow{\mathbf{a}}}_{\mathbf{avg}\mathbf{}}\mathbf{in} \\ \mathbf{unit}\mathbf{}\mathbf{vector}\mathbf{}\mathbf{notation}\mathbf{.}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{the}\mathbf{}\mathbf{magnitude}\mathbf{}\mathbf{}{\overrightarrow{\mathbf{a}}}_{\mathbf{avg}\mathbf{}}\mathbf{of}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{the}\mathbf{}\mathbf{angle}\mathbf{}\mathbf{between}\mathbf{}\mathbf{}{\overrightarrow{\mathbf{a}}}_{\mathbf{avg}\mathbf{}}\mathbf{and}\mathbf{}\mathbf{the}\mathbf{}\mathbf{positive}\mathbf{} \\ \mathbf{direction}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{}\mathbf{axis}\mathbf{.} \end{gathered}$$

Short answer

1. Average acceleration in unit vector notation is$(-1.5\hat{\mathrm{i}}+0.5\hat{\mathrm{k}})\mathrm{m}/{\mathrm{s}}^2$
2. Magnitude of average acceleration is$1.6\mathrm{m}/{\mathrm{s}}^2$
3. The angle between average acceleration and positive direction of x axis is$-18°$

Step-by-step solution

Given information

It is given that,

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

Determining the concept

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,

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

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

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

Where, ${\overrightarrow{v}}_0$and ${\overrightarrow{v}}_0$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,

$\tan \theta =\frac{a_y}{a_x}$

(a) To find the average acceleration

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

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

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

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

(b) To find the magnitude of average acceleration 

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

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

(c) To find the angle between average acceleration and positive direction of x axis

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

$$\begin{gathered} \mathrm{tan\theta }=\frac{0.5}{-1.5} \\ \mathrm{\alpha }=-{18}^{°} \end{gathered}$$

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.