2.1.2 AP Calculus Exam particle at rest

• MHB
• karush
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.
karush
Gold Member
MHB
View attachment 9511
ok I chose e for the zeros

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?

I'll probably be studying that this spring semester!

If you are not now taking a Calculus class, where did you get this problem?

karush said:
I'll probably be studying that this spring semester!

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?

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.

Last edited:
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?

strangely, I don't remember doing particle on x-axis problems when I took calculus...

but it never was a strong spot...43/365

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 ...

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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