MHB 2.1.2 AP Calculus Exam particle at rest

[karush]

View attachment 9511
ok I chose e for the zeros

 

[GJA]

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?

 

[karush]

I'll probably be studying that this spring semester!

 

[HallsofIvy]

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

 

[skeeter]

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?

 

[karush]

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.

 

[skeeter]

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?

 

[karush]

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

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

 

[skeeter]

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 ...