particle motion

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.

eumyang

Again, if you know derivatives, then you can take the derivative of d(t) to get
$$\displaystyle d(t) = 2t - 8$$
... 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
$$\displaystyle \lim_{h \to 0} \dfrac{d(4 + h) - d(4)}{h}$$
and take the limit.

euclid2

CaptainBlack

MHF Hall of Fame
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

mr fantastic

euclid2

Again, if you know derivatives, then you can take the derivative of d(t) to get
$$\displaystyle d(t) = 2t - 8$$
... 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
$$\displaystyle \lim_{h \to 0} \dfrac{d(4 + h) - d(4)}{h}$$
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?

HallsofIvy

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

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

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

euclid2

CaptainBlack

MHF Hall of Fame
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

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

HallsofIvy

CaptainBlack

MHF Hall of Fame
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

HallsofIvy

MHF Helper
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.

euclid2

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!