# Math Help - particle motion

1. ## particle motion

a particle's motion is described by the equation d=t^2-8t+15 where d and t are measured in meters and seconds. show that the particle is at rest when t=4.

2. Again, if you know derivatives, then you can take the derivative of d(t) to get
$d(t) = 2t - 8$
... and then plug in t = 4 to see if the particle is indeed at rest.

Or, mimicking your previous post, find d(4) and d(4 + h), and plug into
$\lim_{h \to 0} \dfrac{d(4 + h) - d(4)}{h}$
and take the limit.

3. Originally Posted by euclid2
a particle's motion is described by the equation d=t^2-8t+15 where d and t are measured in meters and seconds. show that the particle is at rest when t=4.
The graph of position against time is a parabola, and the particle is stationary when the tangent to this parabola is horizontal, which is the vertex of the parabola. So you need to show that the point on the parabola corresponding to t=4 is the vertex.

CB

4. Originally Posted by eumyang
Again, if you know derivatives, then you can take the derivative of d(t) to get
$d(t) = 2t - 8$
... and then plug in t = 4 to see if the particle is indeed at rest.

Or, mimicking your previous post, find d(4) and d(4 + h), and plug into
$\lim_{h \to 0} \dfrac{d(4 + h) - d(4)}{h}$
and take the limit.
i got f(4)=7 and f(4+h)=h^2+7 so limh-->0 h^2+7-7/h limh-->0 h^2/h = 0 so the answer to this one is also 0?

5. What function are you using? You said that $f(t)= t^2-8t+15$ so that f(4)= 16- 32+ 15= 1, not 7.
And f(4+h) is NOT $h^2+ 7$. Where did you get that?

Since $f(t)= t^2-8t+15$, $f(4+h)= (4+ h)^2- 8(4+ h)+ 15$
$f(4+h)= 16+ 8h+ h^2- 32- 8h+ 15= h^2+ 8h+ 1$

$\frac{f(4+h)- f(4)}{h}= \frac{h^2+ 8h+ 1- 1}{h}= \frac{h^2+ 8h}{h}$
and, for $h\ne 0$, that is h+ 8. Take the limit of that as h goes to 0.

6. Originally Posted by euclid2
i got f(4)=7 and f(4+h)=h^2+7 so limh-->0 h^2+7-7/h limh-->0 h^2/h = 0 so the answer to this one is also 0?
So is this a calculus question after all?

Please tell, so I can move it to the appropriate forum.

CB

7. for a particle to be in rest its velocity should be zero for finding velocity we need to integrate velocity which comes out to be 2t-8
v=2t-8
v(4)=2(4)-8
=8-8
=0
hence proved

8. Originally Posted by grgrsanjay
for a particle to be in rest its velocity should be zero for finding velocity we need to integrate velocity which comes out to be 2t-8
v=2t-8
v(4)=2(4)-8
=8-8
=0
hence proved
"Need" is wrong here, "can" is appropriate, in fact we don't need calculus at all.

Also velocity is the derivative of position

CB

9. Originally Posted by grgrsanjay
for a particle to be in rest its velocity should be zero for finding velocity
No, you need to differentiate the position function to get velocity.

we need to integrate velocity which comes out to be 2t-8
v=2t-8
v(4)=2(4)-8
=8-8
=0
hence proved
And it was not clear whether the OP could use "rules" of differentiation like that. Since this was posted in "Precalculus" it is likely that he knows only the "limit of the difference quotient" definition of derivative- or that he could find where the slope of the tangent line is 0 by completing the square as Captain Black suggested.

10. Technically, I am not sure if it should be moved to the calculus section. I haven't yet learned derivatives so i thought pre-calculus was appropriate. Although it is a "calculus course". Thank you everyone for your help!

11. Originally Posted by HallsofIvy
What function are you using? You said that $f(t)= t^2-8t+15$ so that f(4)= 16- 32+ 15= 1, not 7.
And f(4+h) is NOT $h^2+ 7$. Where did you get that?

Since $f(t)= t^2-8t+15$, $f(4+h)= (4+ h)^2- 8(4+ h)+ 15$
$f(4+h)= 16+ 8h+ h^2- 32- 8h+ 15= h^2+ 8h+ 1$

$\frac{f(4+h)- f(4)}{h}= \frac{h^2+ 8h+ 1- 1}{h}= \frac{h^2+ 8h}{h}$
and, for $h\ne 0$, that is h+ 8. Take the limit of that as h goes to 0.
Maybe I am missing something but how does $f(4+h)= 16+ 8h+ h^2- 32- 8h+ 15= h^2+ 8h+ 1$? wouldn't the 8h's cancel and leave us with h^2-1

12. \begin{aligned}
f(4 + h) &= (4 + h)^2 - 8(4 + h) + 15 \\
&= 16 + 8h + h^2 - 32 - 8h + 15 \\
&= h^2 + 8h - 8h + 16 - 32 + 15 \\
&= h^2 - 1
\end{aligned}

You're right, looks like a typo.

Furthermore, f(4) = -1, not 1.

13. Originally Posted by euclid2
a particle's motion is described by the equation d=t^2-8t+15 where d and t are measured in meters and seconds. show that the particle is at rest when t=4.
As mentioned elsewhere the the curve represented by the equation of displacement against time is a parabola and the particle is at rest when the tangent to the parabola is horizontal and this occurs at the vetex of the parabola.

The value of $t$ corresponding to the vertex is the mid point between the roots of the quadratic obtained by setting displacement to zero:

$t^2-8t+15=0$

Applying the quadratic formula to this quadratic give the roots as $t=4 \pm 1$

QED

14. Originally Posted by eumyang
\begin{aligned}
f(4 + h) &= (4 + h)^2 - 8(4 + h) + 15 \\
&= 16 + 8h + h^2 - 32 - 8h + 15 \\
&= h^2 + 8h - 8h + 16 - 32 + 15 \\
&= h^2 - 1
\end{aligned}

You're right, looks like a typo.

Furthermore, f(4) = -1, not 1.
but then we get

limh-->0 h(4+h)-h(4)/h

limh-->0 h^2-1-1/h

limh--> h-2

limh--> (0)-2

=-2

which doesn't make sense?? because it's not at rest??

15. Nope. Careful with your signs...
$\lim_{h \to 0} \dfrac{d(4 + h) - d(4)}{h}$

$= \lim_{h \to 0} \dfrac{h^2 - 1 - (-1)}{h}$

$= \lim_{h \to 0} \dfrac{h^2 - 1 + 1}{h}$

$= \lim_{h \to 0} \dfrac{h^2}{h}$

$= \lim_{h \to 0} h$

$= 0$

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