- #1
karush
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View attachment 9511
ok I chose e for the zeros
ok I chose e for the zeros
2.1.2 AP Calculus Exam particle at rest
In summary, the conversation discusses the relationship between a particle's location and velocity, and the use of calculus to determine velocity given a position function. The formula for the position function is given as $x(t) = (t-a)(t-b) = t^2 - (a + b)t + ab$, and the formula for velocity is $x'(t) = 2t - (a + b)$. The conversation also mentions the use of symmetry in determining the vertex of a parabola and the value of $t$ when the particle is at rest, which is $\dfrac{a+b}{2}$. The conversation concludes with a recommendation to refer to a provided pdf for more information on particle problems in calculus.
- #2
GJA
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Hi karush,
$x(t)$ represents the particle's location at time $t$, not its velocity. Hence, $t=a,b$ give the time values when the particle is located at $x=0$.
Do you know which tool from calculus will tell us about the particle's velocity, given that we know its position function?
$x(t)$ represents the particle's location at time $t$, not its velocity. Hence, $t=a,b$ give the time values when the particle is located at $x=0$.
Do you know which tool from calculus will tell us about the particle's velocity, given that we know its position function?
- #3
karush
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I'll probably be studying that this spring semester!
- #4
HallsofIvy
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If you are not now taking a Calculus class, where did you get this problem?
- #5
skeeter
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karush said:
wait ... after posting quite a few AP calculus type motion problems, are you saying you don’t know the derivative of a position function is a velocity function?
- #6
karush
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expanding we have
$x(t) = (t - a)(t - b) = t^2 - (a + b)t + ab$
the velocity will be dx(x(t)) so
$x'(t) = 2t - (a + b)$
particle will be at rest when x'(t) = 0 so if rewriting we have
$t-\dfrac{(a + b)}{2}=0$
thus
$\dfrac{(a + b)}{2}$ is when the particle is at rest which is b.
$x(t) = (t - a)(t - b) = t^2 - (a + b)t + ab$
the velocity will be dx(x(t)) so
$x'(t) = 2t - (a + b)$
particle will be at rest when x'(t) = 0 so if rewriting we have
$t-\dfrac{(a + b)}{2}=0$
thus
$\dfrac{(a + b)}{2}$ is when the particle is at rest which is b.
Last edited:
- #7
skeeter
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visualize the graph of the position function $x=(t-a)(t-b)$, a open upward parabola with zeros at $t=a$ and $t=b$.
given the symmetry of the parabola, we know the vertex is midway between the zeros and the slope at the vertex equals zero $\implies v(t)=0$.
the value of $t$ midway between $a$ and $b$ would be $\dfrac{a+b}{2}$, correct?
given the symmetry of the parabola, we know the vertex is midway between the zeros and the slope at the vertex equals zero $\implies v(t)=0$.
the value of $t$ midway between $a$ and $b$ would be $\dfrac{a+b}{2}$, correct?
- #8
karush
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strangely, I don't remember doing particle on x-axis problems when I took calculus...
but it never was a strong spot...
43/365
but it never was a strong spot...
43/365- #9
skeeter
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https://mathhelpboards.com/calculus-10/293-ap-calculus-exam-t-v-t-t-8-a-26976.html
post #5, provided a link to a pdf I strongly recommend you have a look at ...
post #5, provided a link to a pdf I strongly recommend you have a look at ...
1. What is a particle at rest in the context of the AP Calculus Exam?
A particle at rest in the context of the AP Calculus Exam refers to a point on a graph or a function where the velocity is equal to 0. This means that the particle is not moving and is at a stationary position.
2. How is a particle at rest related to the concept of derivatives?
In calculus, the derivative of a function at a certain point represents the instantaneous rate of change of that function at that point. Therefore, when a particle is at rest, its velocity is 0 and the derivative of its position function is also 0.
3. Can a particle be at rest and still have a non-zero acceleration?
Yes, a particle can be at rest and still have a non-zero acceleration. This can happen when the particle is at a turning point on a graph, where its velocity is momentarily 0 but its acceleration is not 0.
4. How is a particle at rest represented on a position-time graph?
A particle at rest is represented on a position-time graph as a horizontal line, where the position of the particle does not change over time.
5. What is the significance of identifying a particle at rest on the AP Calculus Exam?
Identifying a particle at rest is important in solving problems involving motion and understanding the behavior of a function. It also helps in determining the maximum and minimum values of a function, which is crucial in optimization problems.
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