91Ó°ÊÓ

Mass of automobile, $M_1=1000kg$

Acceleration of automobile, $a=4.0m/s^2$.

Mass of truck, $M_2=2000kg$

Speed of truck,localid="1657266200854" $v_2=8.0m/s.$

Using the kinematics equation, we can find the location of the automobileas well as the truck at the given time. From that, we can find the center of mass of the system and also the speed of the center of mass.

Formula:

$$\begin{gathered} x_{com}=\frac{m_1{x}_1+m_2{x}_2+....}{m_1+m_2+m_3+....} \\ v=\frac{dis\tan ce}{time} \\ s=v_{o}t+\frac{1}{2}a{t}^2 \end{gathered}$$

The automobile and truck are initially at rest.

So, at the given time, $t=3.0sec$ the location of the automobile can be found by using the kinematic equation,

$$\begin{gathered} x_1=v_{o}t+\frac{1}{2}a{t}^2 \\ {x}_1=0+\frac{1}{2}(4.0){(3.0)}^2{x}_1=18m \end{gathered}$$

The speed oftheautomobile is

$v_1=at=(4.0)(3.0)=12.0m/s$

At the same time the position ofthetruck is

$x_2=vt=(8.0)(3.0)=24m$

The speed of truck is ,

$v_2=8.0m/s.$$v_2=8.0m/s.$

The location of center of mass is

$$\begin{gathered} x_{com}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2} \\ =\frac{\left(1000kg\right)\left(18m\right)+\left(2000kg\right)\left(24m\right)}{(1000kg+2000kg)} \\ =22m \end{gathered}$$

Thus, the distance of the center of mass from traffic signal is 22m.

The speed of center of mass,

$$\begin{gathered} v_{com}=\frac{m_1{v}_1+m_2+v_2}{m_1+m_2} \\ =\frac{\left(1000kg\right)(12m/s)+\left(2000kg\right)(8m/s)}{(1000kg+2000kg)} \\ =9.3m/s \end{gathered}$$

Thus, the velocity of the center of mass is 9.3m/s.