91Ó°ÊÓ

Let the initial point of Carol is Y

The velocity of Carol is ${\left(v_i\right)}_c=36.0m/h$

The position of Carol is role="math" localid="1648444538537" ${\left(x_i\right)}_c=2.4m$

Similarly,
The initial point of Ann is X

The velocity of Ann is ${\left(v_i\right)}_A=50m/h$

The position of Ann is ${\left(x_i\right)}_A=0m$
The time at this instant is ${\left(t_i\right)}_A=0.5h$

Let the point of crossing is Z. At this point let the time is $t$
The position of Carol at this point is ${\left(x_f\right)}_c$

The position of Ann at this point is ${\left(x_f\right)}_A$

Write the equation of motion at point Z for Carol

${\left(x_f\right)}_c={\left(x_i\right)}_c+{\left(v_i\right)}_c\left[t-{\left(t_i\right)}_c\right]$
Insert the known variables in the above expression,

localid="1648445871907" $$\begin{gathered} {\left(x_f\right)}_c=2.4+36\left[t-0\right] \\ {\left(x_f\right)}_c=2.4+36t......\left(1\right) \end{gathered}$$

Similarly, write the equation of motion for Ann at point Z

$$\begin{gathered} {\left(x_f\right)}_A={\left(x_i\right)}_A+{\left(v_i\right)}_A\left[t-{\left(t_i\right)}_A\right] \\ {\left(x_f\right)}_A=0+50\left[t-0.5\right] \\ {\left(x_f\right)}_A=50t-25.......\left(2\right) \end{gathered}$$

Compare the equations (1) and (2)
$$\begin{gathered} 2.4+36t=50t-25 \\ 27.4=14t \\ t=2h \end{gathered}$$
Thus, after 2 hours, Ann's car would cross Carol's car.

The position of crossing is determined by substituting the value of $t$in equation (1 ) or (2),

Consider equation (1)

$$\begin{gathered} {\left(x_f\right)}_c=2.4+36\left(2\right) \\ {\left(x_f\right)}_c=74.4m \end{gathered}$$
Thus, Ann's car would cross Caol's car at 74.4 m.

The position-time graph of Carol and Ann's car is shown below. The red line shows the change in position of Carol's car and the blue line shows the change in position of Ann's Car. They both meet at the time of 2 hours at a distance of 74.4 meters.