Question

Two cars approach a street corner at right angles to each other (Fig. 3-47). Car 1 travels at a speed${\mathbf{v}}_{\mathbf{1}\mathbf{E}}=\mathbf{35}\mathbf{km}\mathbf{/}\mathbf{h}$relative to the earth, and car 2 at${\mathbf{v}}_{\mathbf{2}\mathbf{E}}=\mathbf{55}\mathbf{km}\mathbf{/}\mathbf{h}$. What is the relative velocity of car 1 as seen by car 2? What is the velocity of car 2 relative to car 1?

Short answer

The velocity of car 1 relative to car 2 is $65.2\frac{\mathrm{km}}{\mathrm{h}}$at $57.5°$west of north, and that of car 2 relative to car 1 is $65.2\frac{\mathrm{km}}{\mathrm{h}}$at $32.5°$south of east.

Step-by-step solution

Step 1. Given data

The velocity on one object relative to the second object is the difference between the velocity vectors of the two objects.

Given data:

The speed of the first car relative to the ground is $v_{1\mathrm{E}}=35\frac{\mathrm{km}}{\mathrm{h}}$along the north.

The speed of the second car relative to the ground is $v_{2\mathrm{E}}=55\frac{\mathrm{km}}{\mathrm{h}}$along the east.

Assumptions:

You can assume that the positive x-direction is east, and the positive y direction is the north.

Let $v_{12}$be the relative velocity of car 1 as seen by car 2, and $v_{21}$be the velocity of car 2 relative to car 1.

Step 2. Representations of the velocity vectors

Now, the representations of the velocity vectors are shown below as follows:

Step 3. Finding the magnitude of the relative velocities

Here, the magnitude of the relative velocities $v_{12}$and $v_{21}$are the same.

$\begin{array}{c}{v}_{12}=v_{21}\\ =\sqrt{{\left(v_{1\mathrm{E}}\right)}^2+{\left(v_{2\mathrm{E}}\right)}^2}\\ =\sqrt{{\left(35\frac{\mathrm{km}}{\mathrm{h}}\right)}^2+{\left(55\frac{\mathrm{km}}{\mathrm{h}}\right)}^2}\\ =65.2\frac{\mathrm{km}}{\mathrm{h}}\end{array}$

Now, from the first figure,

$\begin{array}{c}\tan \theta =\frac{v_{2\mathrm{E}}}{v_{1\mathrm{E}}}\\ \theta ={\tan }^{-1}\left(\frac{55\frac{\mathrm{km}}{\mathrm{h}}}{35\frac{\mathrm{km}}{\mathrm{h}}}\right)\\ =57.5°\end{array}$

From the second figure,

$\begin{array}{c}\tan \varphi =\frac{v_{1\mathrm{E}}}{v_{2\mathrm{E}}}\\ \varphi ={\tan }^{-1}\left(\frac{35\frac{\mathrm{km}}{\mathrm{h}}}{55\frac{\mathrm{km}}{\mathrm{h}}}\right)\\ =32.5°\end{array}$

Hence, the velocity of car 1 relative to car 2 is $65.2\frac{\mathrm{km}}{\mathrm{h}}$at $57.5°$west of north, and that of car 2 relative to car 1 is $65.2\frac{\mathrm{km}}{\mathrm{h}}$at $32.5°$south of east.