A cameraman on a pickup truck is traveling westward at20 km/hwhile he records a cheetah that is moving westward 30 km/hfaster than the truck. Suddenly, the cheetah stops, turns, and then runs at 45 km/h eastward, as measured by a suddenly nervous crew member who stands alongside the cheetah’s path. The change in the animal’s velocity takes 2.0 s. What are the (a) magnitude and (b)direction of the animal’s acceleration according to the cameraman and the (c)magnitude and (d) direction according to the nervous crew member?

$$\begin{gathered} v_{TG}=-20\mathrm{km}/\mathrm{h} \\ =-5.6\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} v_{CTi}=-30\mathrm{km}/\mathrm{h} \\ =-8.3\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} v_{CGf}=45\mathrm{km}/\mathrm{h} \\ =12.5\mathrm{m}/\mathrm{s} \end{gathered}$$

When two frames of reference Aand Bare moving relative to each other at a constant velocity, the velocity of a particle Pas measured by an observer in frame Ausually differs from that measured in frame B. The two measured velocities are related by

${\overrightarrow{\mathbf{v}}}_{\mathbf{P}\mathbf{A}}\mathbf{=}{\overrightarrow{\mathbf{v}}}_{\mathbf{P}\mathbf{B}}\mathbf{+}{\overrightarrow{\mathbf{v}}}_{\mathbf{B}\mathbf{A}}$

Here ${\overrightarrow{\mathbf{v}}}_{\mathbf{B}\mathbf{A}}$is the velocity of B with respect to A.

Use the relative motion concept and find the final velocity of the cheetah relative to the truck ${\overrightarrow{\mathbf{v}}}_{\mathbf{c}\mathbf{}\mathbf{T}\mathbf{}\mathbf{f}}$ . By using the expression of acceleration, find the acceleration of the cheetah relative truck (cameraman) ${\overrightarrow{\mathbf{a}}}_{\mathbf{c}\mathbf{}\mathbf{T}\mathbf{}}$. Similarly, find the initial velocity of the cheetah relative to the ground ${\overrightarrow{\mathbf{}\mathbf{v}}}_{\mathbf{cGi}\mathbf{}}$and then use the expression to find its acceleration.

Formula:

$$\begin{gathered} v_{cTf}=v_{cGf}+v_{GT} \\ {v}_{cGi}=v_{cTi}+v_{TG} \\ \overrightarrow{a}=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{\overrightarrow{t_2}-\overrightarrow{t_1}} \end{gathered}$$

According to the relative motion equation

$$\begin{gathered} v_{cTf}=v_{cGf}+v_{GT} \\ =v_{cGf}-v_{TG} \\ =\left(12.5\mathrm{m}/\mathrm{s}\right)-\left(-5.6\mathrm{m}/\mathrm{s}\right) \\ =18.1\mathrm{m}/\mathrm{s} \end{gathered}$$

The expression of the acceleration is,

role="math" localid="1660901231505" $$\begin{gathered} {\overrightarrow{a}}_{CT}=\frac{{\overrightarrow{v}}_{CTf}-{\overrightarrow{v}}_{CTi}}{t_2-t_1} \\ =\frac{\left(18.1\mathrm{m}/\mathrm{s}\right)-\left(-8.3\mathrm{m}/\mathrm{s}\right)}{2.0} \\ =13.2\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration is $13.2\mathrm{m}/{\mathrm{s}}^2$.

The direction of the animal’s acceleration according to truck is positive that is east.

According to the relative motion

$$\begin{gathered} v_{CGi}=v_{CTi}+v_{TG} \\ =\left(-8.30\mathrm{m}/\mathrm{s}\right)+\left(-5.6\mathrm{m}/\mathrm{s}\right) \\ =-13.9\mathrm{m}/\mathrm{s} \end{gathered}$$

The expression of the acceleration is,

$$\begin{gathered} a_{CG}=\frac{v_{CGf}-v_{CGi}}{t_2-t_1} \\ =\frac{\left(12.5\mathrm{m}/\mathrm{s}\right)-\left(-13.9\mathrm{m}/\mathrm{s}\right)}{2.0\mathrm{s}} \\ =13.2\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the magnitude of the animal’s acceleration is $13.2\mathrm{m}/{\mathrm{s}}^2$.

The direction of the animal’s acceleration according to the nervous crew member (ground) is positive that is east.