91Ó°ÊÓ

When two frames of reference A and B are moving relative to each other at a constant velocity, the velocity of a particle P as measured by an observer in frame A usually differs from that measured in frame B. The two measured velocities are related by

${\stackrel{\mathbf{â‡\P Ä}}{\mathbf{V}}}_{\mathbf{P}\mathbf{A}}\mathbf{=}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{V}}}_{\mathbf{P}\mathbf{B}}\mathbf{+}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{V}}}_{\mathbf{B}\mathbf{A}}$

Here${\stackrel{\mathbf{â‡\P Ä}}{\mathbf{V}}}_{\mathbf{B}\mathbf{A}}$ V⇶ÄPA=V⇶ÄPB+V⇶ÄBAis the velocity of B with respect to A.

Using the concept of relative velocity, drawing the figure according to the directions of the car, and using trigonometry, we can find the angle between the velocity and the line of sight between the two cars. Cars are moving with constant velocity hence there is no change in the velocity of motorists relative to the police car and the angle between velocity found and the line of sight between the two cars.

Formula:

$$\begin{gathered} {\stackrel{â‡\P Ä}{\mathrm{V}}}_{\mathrm{PA}}={\stackrel{â‡\P Ä}{\mathrm{V}}}_{\mathrm{PB}}+{\stackrel{â‡\P Ä}{\mathrm{V}}}_{\mathrm{BA}} \\ \tan \mathrm{θ}=\frac{oppositeside}{adjacentside} \end{gathered}$$

The motorist is moving with respect to the ground ${\stackrel{â‡\P Ä}{V}}_{Mg}$in the downward direction and police is moving with respect to the ground ${\stackrel{â‡\P Ä}{V}}_{Mg}$in the negative x axis as shown in figure. By relative motion expression we get,

$\begin{array}{rcl}{\stackrel{â‡\P Ä}{V}}_{MP}& =& {\stackrel{â‡\P Ä}{V}}_{Mg}-{\stackrel{â‡\P Ä}{V}}_{Pg}\\ & =& \left(-60km/h\right)\hat{j}-\left(-80km/h\right)\hat{i}\\ & =& \left(80km/h\right)\hat{i}-\left(60km/h\right)\hat{j}\end{array}$

From the figure, the velocity of motorist with respect to the police car ${\stackrel{â‡\P Ä}{V}}_{MP}$vector is acting in the fourth quadrant, then by using trigonometry we can find the angle made by${\stackrel{â‡\P Ä}{V}}_{MP}$relative positive x-axis is,

$\begin{array}{rcl}\tan \mathrm{θ}& =& \frac{{\stackrel{â‡\P Ä}{V}}_{Mpy}}{{\stackrel{â‡\P Ä}{V}}_{Mpx}}\\ & =& \frac{-60km/h}{80km/h}\\ \mathrm{θ}& =& {\tan }^{-1}\left(\frac{-60km/h}{80km/h}\right)\\ & =& -36.8^o\end{array}$

The vector pointing from car to another is,

$\stackrel{â‡\P Ä}{r}=\left(800m\right)\hat{i}-\left(600m\right)\hat{j}$

Therefore, the angle is,

$\begin{array}{rcl}\tan \mathrm{θ}& =& \frac{-600m}{800m}\\ \mathrm{θ}& =& {\tan }^{-1}\left(\frac{-600m}{800m}\right)\\ & =& -36.8^o\end{array}$

The velocity and must be point in the same direction, hence angle between them is

The two cars maintain their velocities; hence there is no change in velocity of motorist relative to the police ${\stackrel{â‡\P Ä}{V}}_{MP}$and angle between velocity ${\stackrel{â‡\P Ä}{V}}_{MP}$and the line of sight between the two cars.