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#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 2,678

$x(t)$ 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?

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- #3

- Jan 31, 2012

- 2,678

I'll probably be studying that this spring semester!

- Jan 29, 2012

- 1,151

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

- Mar 1, 2012

- 655

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?I'll probably be studying that this spring semester!

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- #6

- 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**.

$x(t) = (t - a)(t - b) = t^2 - (a + b)t + ab$

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

Last edited:

- Mar 1, 2012

- 655

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?

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- #8

- Jan 31, 2012

- 2,678

but it never was a strong spot...

43/365

- Mar 1, 2012

- 655

post #5, provided a link to a pdf I strongly recommend you have a look at ...