 Gregory Jones

2022-01-01

David is driving a steady 25.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.00 m/s2 at the instant when David passes. How far does Tina travel before David is passed, and what speed does she travel at? sirpsta3u

Write the expression for distance travelled by an object in a time t, and solve for t.
$d=ut+\frac{1}{2}a{t}^{2}$

$t=25s$
The distance in which Tina drive before passing David can be calculated using the equation (1).
$d=0+\frac{1}{2}\left(2\frac{m}{{s}^{2}}\right){\left(25s\right)}^{2}$

The speed of Tina when she passes David can be calculated as follows.
$v=at$ Eliza Beth13

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, ${v}_{0}=25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}$, $a=2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$, 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, $v=25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}$. 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:
$2{v}^{2}=ax$
Substituting the given values:
$2\left(25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}{\right)}^{2}=\left(2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}\right)x$
Solving for $x$:
$x=\frac{2\left(25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}{\right)}^{2}}{2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}}$
Simplifying:
$x=\frac{2\left(625\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}\right)}{2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}}$
$x=\frac{1250\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}}{2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}}$
$x=625\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}×\frac{1}{2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}}$
$x=312.5\phantom{\rule{0.167em}{0ex}}{\text{m}}^{2}/{\text{s}}^{2}×\frac{1\phantom{\rule{0.167em}{0ex}}\text{m}}{{\text{s}}^{2}}$
$x=312.5\phantom{\rule{0.167em}{0ex}}\text{m}$
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, ${v}_{0}=0\phantom{\rule{0.167em}{0ex}}\text{m/s}$, $a=2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$, $t$ is the time taken by Tina to pass David, and $v$ is the final velocity.
Substituting the value of $t$:
$v=0\phantom{\rule{0.167em}{0ex}}\text{m/s}+2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}×\left(\frac{312.5\phantom{\rule{0.167em}{0ex}}\text{m}}{25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}}\right)$
Simplifying:
$v=0\phantom{\rule{0.167em}{0ex}}\text{m/s}+2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}×12.5$
$v=0\phantom{\rule{0.167em}{0ex}}\text{m/s}+25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}$
$v=25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}$
Therefore, Tina travels at a speed of 25.0 m/s. Tina travels 156.25 m at a speed of 25.0 m/s.
Explanation:
The initial velocity of Tina, , is zero as she is sitting at rest. The acceleration, ${a}_{\text{Tina}}$, is given as $2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$, and the final velocity of Tina, , is the same as David's velocity, which is $25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}$.
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:

$25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}=0+2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}·t$
Simplifying the equation, we have:
$25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}=2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}·t$
To find $t$, we divide both sides of the equation by $2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$:
$\frac{25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}}{2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}}=t$
$t=12.5\phantom{\rule{0.167em}{0ex}}\text{s}$
Now that we have the time, we can find the distance Tina traveled during this time using the equation of motion $d={v}_{\text{initial}}·t+\frac{1}{2}a·{t}^{2}$:

Substituting the known values:
${d}_{\text{Tina}}=0·12.5\phantom{\rule{0.167em}{0ex}}\text{s}+\frac{1}{2}·2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}·\left(12.5\phantom{\rule{0.167em}{0ex}}\text{s}{\right)}^{2}$
Simplifying the equation, we have:
${d}_{\text{Tina}}=0+\frac{1}{2}·2.00\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}·156.25\phantom{\rule{0.167em}{0ex}}{\text{s}}^{2}$
${d}_{\text{Tina}}=156.25\phantom{\rule{0.167em}{0ex}}\text{m}$
Therefore, Tina travels a distance of $156.25\phantom{\rule{0.167em}{0ex}}\text{m}$ before David passes her. Tina's speed at that point is the same as David's speed, which is $25.0\phantom{\rule{0.167em}{0ex}}\text{m/s}$.

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