91Ó°ÊÓ

The speed of David's car is $v_D=30m/s$

The acceleration of Tina's car is $a_T=2m/s^2$

The initial velocity of Tina's car is ${\left(v_i\right)}_T=0m/s$

Write the equation of motion for David,
$$\begin{gathered} s=v_{D}t \\ s=30t..........\left(1\right) \end{gathered}$$

Write the equation of motion for Tina

localid="1648466776525" $$\begin{gathered} s={\left(v_i\right)}_{T}t+\frac{1}{2}{a}_T{t}^2 \\ s=0+\frac{1}{2}\left(2\right)t^2 \\ s=t^2.......\left(2\right) \end{gathered}$$
Since, both the cars pass each other at a point, therefore, equate the equation (1) and (2)

$$\begin{gathered} 30t=t^2 \\ t=30sec \end{gathered}$$

At this time, both the cars would pass each other.

To determine the distance covered by Tina's Car when it passes David's car, insert the value of t= 30 sec in equation (2)
localid="1648467343719" $$\begin{gathered} s={\left(30\right)}^2 \\ s=900m \end{gathered}$$

Thus, Tina drives for 900 m before crossing David

Write the equation of motion,

$$\begin{gathered} {{\left(v_f\right)}_T}^2-{{\left(v_i\right)}_T}^2=2as \\ {{\left(v_f\right)}_T}^2-0=2\left(2\right)\left(900\right) \\ {\left(v_f\right)}_T=\sqrt{3600} \\ {\left(v_f\right)}_T=60m/sec \end{gathered}$$
Thus, the speed of Tina as she passes Davis is 60 m/sec