In the overhead view of Fig.4-47, Jeeps P and B race along straight lines, across flat terrain, and past stationary border guard A. Relative to the guard, B travels at a constant speed of $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$, at the angle${\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{30}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$. Relative to the guard, P has accelerated from rest at a constant rate of $\mathbf{0}\mathbf{.}\mathbf{400}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$at the angle ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{60}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$At a certain time during the acceleration, P has a speed of$\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. At that time, what are the (a) magnitude and (b) direction of the velocity of P relative to B and the (c) magnitude and (d) direction of the acceleration of P relative to B?

1. The velocity of B with respect to A $\left(V_{BA}\right)\mathbf{=}\mathbf{20}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$

2. The acceleration of P with respect to A $\left(a_{PA}\right)=0.4m/s^2$

3. The velocity of P with respect to $\left(V_{PA}\right)=40m/s$

The relative velocity of an object is defined as its velocity in relation to some other observer Using the relative motion concept, the magnitude and direction of vectors can be calculated. The velocities of jeeps are added vectorially to find the magnitude and directions of the motion of both jeeps relative to each other.

Formulae:

Magnitude of the vector $\stackrel{\xcancel{}}{V}$is,

$\sqrt{V_x^2-V_y^2}$

The direction of vector is given by,

$\tan \theta =\frac{V_y}{V_x}$

The velocity of P with respect to A can be written in the vector notation as,

$$\begin{gathered} \overrightarrow{V}\theta i_{PA}=V_{PA}\cos {\theta }_1\hat{j}+{}_{PA}\sin {}_1 \\ =\left(40m/s\right)\cos {60}^{°}\hat{i}+\left(40m/s\right)\sin {60}^{°}\hat{j} \end{gathered}$$

The velocity of B with respect to A can be written in the vector notation as,

$$\begin{gathered} \overrightarrow{V}\theta i_{BA}=V_{BA}\cos {\theta }_2\hat{j}+{}_{BA}\sin {}_2 \\ =\left(20m/s\right)\cos \left({30}^{°}\right)\hat{i}+\left(20m/s\right)\sin \left({30}^{°}\right)\hat{j} \end{gathered}$$

Now, velocity of P with respect to B is

$$\begin{gathered} {\overrightarrow{V}}_{BA}={\overrightarrow{V}}_{PA}-{\overrightarrow{V}}_{BA} \\ =\left[\left(40m/s\right)\cos {60}^{°}\hat{i}+\left(40m/s\right){\sin }^{°}\hat{j}\right]-\left[\left(20m/s\right)\cos \left({30}^{°}\right)\hat{i}+\left(20m/s\right)\sin \left({30}^{°}\right)\hat{j}\right] \\ =\left(20m/s\right)\hat{i}+\left(34.6m/s\right)\hat{j}-\left(17.32m/s\right)\hat{i}+\left(10m/s\right)\hat{j} \\ =\left(2.68m/s\right)\hat{i}+\left(24.6m/s\right)\hat{j} \end{gathered}$$

The magnitude of velocity of P with respect to B is

$$\begin{gathered} {\mathit{V}}_{\mathbf{P}\mathbf{B}}\mathbf{=}\sqrt{{{\mathbf{V}}_{\mathbf{P}\mathbf{B}\mathbf{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{V}}_{\mathbf{P}\mathbf{B}\mathbf{y}}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\sqrt{{\left(2.68m/s\right)}^{\mathbf{2}}\mathbf{+}{\left(24.m/s\right)}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{75}\mathbf{}\mathit{m}\mathbf{/}\mathit{s} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\approx }\mathbf{24}\mathbf{.}\mathbf{8}\mathbf{}\mathit{m}\mathbf{/}\mathit{s} \end{gathered}$$

Therefore, the magnitude of velocity of P with respect to B is $\mathbf{24}\mathbf{.}\mathbf{8}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$

The direction of velocity of P with respect to B is given by,

$$\begin{gathered} \tan \theta =\frac{V_{PBy}}{V_{PBx}} \\ \tan \theta =\left(\frac{24.60m/s}{2.68m/s}\right) \\ \theta ={\tan }^{-1}\left(\frac{24.60m/s}{2.68m/s}\right) \\ =83.{78}^{°} \\ \approx 83.8^{°} \end{gathered}$$

Therefore, the direction of velocity of P with respect to B is $\mathbf{83}\mathbf{.}{\mathbf{8}}^{\mathbf{°}}$north of east.

The acceleration of P with respect to B is,

$$\begin{gathered} {\overrightarrow{a}}_{PB}={\overrightarrow{a}}_{PA}-{\overrightarrow{a}}_{BA} \\ =\left(0.4m/s^2\right)\cos {60}^{°}\hat{i}+\left(0.4m/s^2\right)\sin {60}^{°}\hat{j}-0 \\ =\sqrt{{\left(0.2m/s^2\right)}^2+{\left(0.346m/s^2\right)}^2} \\ =0.4m/s^2 \end{gathered}$$

The magnitude of the acceleration of P with respect to B is

$$\begin{gathered} a_{PB}=\sqrt{{\left(0.2m/s^2\right)}^2+{\left(0.346m/s^2\right)}^2} \\ =0.4m/s^2 \end{gathered}$$

Therefore, the acceleration of P with respect to B is $0.4m/s^2$.

The direction of acceleration of P with respect to B is given by

$$\begin{gathered} \tan \theta =\frac{a_{PBy}}{a_{PBx}} \\ =\left(\frac{0.346m/s^2}{0.2m/s^2}\right) \\ ={60}^{°} \end{gathered}$$

The direction of acceleration of P with respect to B is ${60}^{°}northofeast$.