91Ó°ÊÓ

$$\begin{gathered} x_{g0}=270 m \\ {x}_{r0}=-35.0 m \\ {V}_g=20.0 m/s \end{gathered}$$

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using the formula for the second kinematic equation, the equation for the position of both the red and the green car can be written.By solving these equations simultaneously, the acceleration of the red car can be found.

Formula:

The displacement in kinematic equation can be expressed as,

$\mathrm{Δ³\ae }={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$

The position of the cars as function of time is given by

${\mathrm{x}}_{\mathrm{g}}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{g}0}+{\mathrm{v}}_{\mathrm{g}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{g}}{\mathrm{t}}^2$

As acceleration ${\mathrm{a}}_{\mathrm{g}}=0\mathrm{m}/{\mathrm{s}}^2$, position ${\mathrm{x}}_{\mathrm{g}0}=270 \mathrm{m}$at time$\mathrm{t}=0 \mathrm{s}$

$=(270 \mathrm{m})-\frac{20 \mathrm{m}}{\mathrm{s}}× \mathrm{t}$

As position ${\mathrm{x}}_{\mathrm{r}0}=-35.0 \mathrm{m},{\mathrm{v}}_{\mathrm{r}}=0\mathrm{m}/\mathrm{s}$ at time$\mathrm{t}=0 \mathrm{s}$

$$\begin{gathered} {\mathrm{x}}_{\mathrm{r}}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{r}0}+{\mathrm{v}}_{\mathrm{r}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{r}}{\mathrm{t}}^2 \\ {\mathrm{x}}_{\mathrm{r}}\left(\mathrm{t}\right)=(-35.0 \mathrm{m})+\frac{1}{2}×{\mathrm{a}}_{\mathrm{r}}×{\mathrm{t}}^2 \end{gathered}$$

The graph shows two cars to pass each other at. This implies

$(270 \mathrm{m})-\frac{20 \mathrm{m}}{\mathrm{s}}12.0 \mathrm{s}=(-35.0 \mathrm{m})+\frac{1}{2}{\mathrm{a}}_{\mathrm{r}}(12.0{)}^2$

This equation gives${\mathrm{a}}_{\mathrm{r}}=0.90 \mathrm{m}/{\mathrm{s}}^2$

Therefore, the acceleration of red car, role="math" localid="1656995848006" ${\mathrm{a}}_{\mathrm{r}}=0.90 \mathrm{m}/{\mathrm{s}}^2$.