91Ó°ÊÓ

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

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

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

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

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

Here ${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{B}\mathbf{A}}$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 ${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{c}\mathbf{}\mathbf{T}\mathbf{}\mathbf{f}}$ . By using the expression of acceleration, find the acceleration of the cheetah relative truck (cameraman) ${\stackrel{\mathbf{→}}{\mathbf{a}}}_{\mathbf{c}\mathbf{}\mathbf{T}\mathbf{}}$. Similarly, find the initial velocity of the cheetah relative to the ground ${\stackrel{\mathbf{→}}{\mathbf{}\mathbf{v}}}_{\mathbf{cGi}\mathbf{}}$and then use the expression to find its acceleration.

Formula:

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

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

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.