4.5 Relative motion in one and two dimensions  (Page 2/8)

Adding the vectors, we find ${\overrightarrow{v}}_{\text{PE}}=8\text{m}\text{/}\text{s}\widehat{i}\text{,}$ so the person is moving 8 m/s east with respect to Earth. Graphically, this is shown in [link] .

Relative velocity in two dimensions

We can now apply these concepts to describing motion in two dimensions. Consider a particle P and reference frames S and $S^{\prime },$ as shown in [link] . The position of the origin of $S^{\prime }$ as measured in S is ${\overrightarrow{r}}_{S^{\prime }S},$ the position of P as measured in $S^{\prime }$ is ${\overrightarrow{r}}_{P{S}^{\prime }},$ and the position of P as measured in S is ${\overrightarrow{r}}_{PS}.$

From [link] we see that

The relative velocities are the time derivatives of the position vectors. Therefore,

The velocity of a particle relative to S is equal to its velocity relative to $S^{\prime }$ plus the velocity of $S^{\prime }$ relative to S .

We can extend [link] to any number of reference frames. For particle P with velocities ${\overrightarrow{v}}_{PA},{\overrightarrow{v}}_{PB}\text{,}\text{and}{\overrightarrow{v}}_{PC}$ in frames A , B , and C ,

We can also see how the accelerations are related as observed in two reference frames by differentiating [link] :

We see that if the velocity of $S^{\prime }$ relative to S is a constant, then ${\overrightarrow{a}}_{S^{\prime }S}=0$ and

This says the acceleration of a particle is the same as measured by two observers moving at a constant velocity relative to each other.

Motion of a car relative to a truck

A truck is traveling south at a speed of 70 km/h toward an intersection. A car is traveling east toward the intersection at a speed of 80 km/h ( [link] ). What is the velocity of the car relative to the truck?

Strategy

First, we must establish the reference frame common to both vehicles, which is Earth. Then, we write the velocities of each with respect to the reference frame of Earth, which enables us to form a vector equation that links the car, the truck, and Earth to solve for the velocity of the car with respect to the truck.

Solution

The velocity of the car with respect to Earth is ${\overrightarrow{v}}_{\text{CE}}=80\text{km}\text{/}\text{h}\widehat{i}.$ The velocity of the truck with respect to Earth is ${\overrightarrow{v}}_{\text{TE}}=\mathrm{-70}\text{km}\text{/}\text{h}\widehat{j}\text{.}$ Using the velocity addition rule, the relative motion equation we are seeking is

${\overrightarrow{v}}_{\text{CT}}={\overrightarrow{v}}_{\text{CE}}+{\overrightarrow{v}}_{\text{ET}}.$

Here, ${\overrightarrow{v}}_{\text{CT}}$ is the velocity of the car with respect to the truck, and Earth is the connecting reference frame. Since we have the velocity of the truck with respect to Earth, the negative of this vector is the velocity of Earth with respect to the truck: ${\overrightarrow{v}}_{\text{ET}}=\text{−}{\overrightarrow{v}}_{\text{TE}}.$ The vector diagram of this equation is shown in [link] .

We can now solve for the velocity of the car with respect to the truck:

$\left|{\overrightarrow{v}}_{\text{CT}}\right|=\sqrt{(80.0\text{km}\text{/}{\text{h})}^2+(70.0\text{km}\text{/}{\text{h})}^2}=106.\text{km}\text{/}\text{h}$

and

$\theta ={\text{tan}}^{\mathrm{-1}}\left(\frac{70.0}{80.0}\right)=41.2\text{°}\text{north of east.}$

Significance

Drawing a vector diagram showing the velocity vectors can help in understanding the relative velocity of the two objects.

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