David is driving a steady 25.0 m/s when he passes Tina, who is sitting

To solve this problem, we can use the equations of motion. Let's denote the distance traveled by Tina as $x$ and the time taken by Tina to pass David as $t$.
Since David is driving at a constant speed of 25.0 m/s and Tina starts from rest, the relative velocity between them is 25.0 m/s.
Using the equation of motion for distance covered during constant acceleration:
$x=v_{0}t+\frac{1}{2}a{t}^2$
where $v_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. In this case, $v_0=25.0\text{m/s}$, $a=2.00{\text{m/s}}^2$, and $x$ is the distance traveled by Tina before passing David.
Since Tina starts from rest, her initial velocity is 0. We can rewrite the equation as:
$x=\frac{1}{2}a{t}^2$
Now, we need to find the time taken by Tina to pass David. Since the relative velocity between them is constant at 25.0 m/s, we can use the equation:
$x=vt$
where $v$ is the relative velocity. In this case, $v=25.0\text{m/s}$. Rearranging the equation, we get:
$t=\frac{x}{v}$
Now, substitute this value of $t$ in the equation for $x$:
$x=\frac{1}{2}a{\left(\frac{x}{v}\right)}^2$
Simplifying this equation:
$2x=\frac{a{x}^2}{v^2}$
Rearranging the equation:
$2v^2=ax$
Substituting the given values:
$2(25.0\text{m/s}{)}^2=(2.00{\text{m/s}}^2)x$
Solving for $x$:
$x=\frac{2(25.0\text{m/s}{)}^2}{2.00{\text{m/s}}^2}$
Simplifying:
$x=\frac{2(625{\text{m}}^2/{\text{s}}^2)}{2.00{\text{m/s}}^2}$
$x=\frac{1250{\text{m}}^2/{\text{s}}^2}{2.00{\text{m/s}}^2}$
$x=625{\text{m}}^2/{\text{s}}^2\times \frac{1}{2.00{\text{m/s}}^2}$
$x=312.5{\text{m}}^2/{\text{s}}^2\times \frac{1\text{m}}{{\text{s}}^2}$
$x=312.5\text{m}$
Therefore, Tina travels 312.5 meters before passing David.
To find the speed at which Tina travels, we can use the equation of motion for velocity:
$v=v_0+at$
where $v_0$ is the initial velocity, $a$ is the acceleration, $t$ is the time, and $v$ is the final velocity. In this case, $v_0=0\text{m/s}$, $a=2.00{\text{m/s}}^2$, $t$ is the time taken by Tina to pass David, and $v$ is the final velocity.
Substituting the value of $t$:
$v=0\text{m/s}+2.00{\text{m/s}}^2\times \left(\frac{312.5\text{m}}{25.0\text{m/s}}\right)$
Simplifying:
$v=0\text{m/s}+2.00{\text{m/s}}^2\times 12.5$
$v=0\text{m/s}+25.0\text{m/s}$
$v=25.0\text{m/s}$
Therefore, Tina travels at a speed of 25.0 m/s.

Answer:
Tina travels 156.25 m at a speed of 25.0 m/s.
Explanation:
The initial velocity of Tina, $v_{\text{Tina, initial}}$, is zero as she is sitting at rest. The acceleration, $a_{\text{Tina}}$, is given as $2.00{\text{m/s}}^2$, and the final velocity of Tina, $v_{\text{Tina, final}}$, is the same as David's velocity, which is $25.0\text{m/s}$.
Using the equation of motion $v_{\text{final}}=v_{\text{initial}}+at$, we can find the time it takes for Tina to reach the velocity of David:
$v_{\text{Tina, final}}=v_{\text{Tina, initial}}+a_{\text{Tina}}·t$
$25.0\text{m/s}=0+2.00{\text{m/s}}^2·t$
Simplifying the equation, we have:
$25.0\text{m/s}=2.00{\text{m/s}}^2·t$
To find $t$, we divide both sides of the equation by $2.00{\text{m/s}}^2$:
$\frac{25.0\text{m/s}}{2.00{\text{m/s}}^2}=t$
$t=12.5\text{s}$
Now that we have the time, we can find the distance Tina traveled during this time using the equation of motion $d=v_{\text{initial}}·t+\frac{1}{2}a·t^2$:
$d_{\text{Tina}}=v_{\text{Tina, initial}}·t+\frac{1}{2}{a}_{\text{Tina}}·t^2$
Substituting the known values:
$d_{\text{Tina}}=0·12.5\text{s}+\frac{1}{2}·2.00{\text{m/s}}^2·(12.5\text{s}{)}^2$
Simplifying the equation, we have:
$d_{\text{Tina}}=0+\frac{1}{2}·2.00{\text{m/s}}^2·156.25{\text{s}}^2$
$d_{\text{Tina}}=156.25\text{m}$
Therefore, Tina travels a distance of $156.25\text{m}$ before David passes her. Tina's speed at that point is the same as David's speed, which is $25.0\text{m/s}$.