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[SOLVED] 2.1.216 AP Calculus Exam particle at rest

karush

Well-known member
Jan 31, 2012
2,678

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
271
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

Well-known member
Jan 31, 2012
2,678
I'll probably be studying that this spring semester!
 

HallsofIvy

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MHB Math Helper
Jan 29, 2012
1,151
If you are not now taking a Calculus class, where did you get this problem?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655
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

Well-known member
Jan 31, 2012
2,678
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.
 
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skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655
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

Well-known member
Jan 31, 2012
2,678
strangely, I don't remember doing particle on x axis problems when I took calculus....

but it never was a strong spot...:rolleyes:


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