91Ó°ÊÓ

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,

$\tan \mathrm{θ}=\frac{V_y}{V_x}$

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

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

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

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

Now, velocity of P with respect to B is

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

The magnitude of velocity of P with respect to B is

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

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,

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

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,

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

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

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

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