1. I have moved this thread to calculus, since no one is interested in the per-calculus solution.

CB

2. ## Re: particle motion

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.
Can u do the problem

Sent from my SGH-I747M using Tapatalk

3. ## Re: particle motion

Originally Posted by abdurrahmanzafar
Can u do the problem

Sent from my SGH-I747M using Tapatalk
why are you reviving a thread that is over 7 years old?

4. ## Re: particle motion

Originally Posted by skeeter
why are you reviving a thread that is over 7 years old?
I'm new to this app honestly i don't really no what to do i just needed help on that problem

Sent from my SGH-I747M using Tapatalk

5. ## Re: particle motion

The derivative of the position function $d=t^2-8t+15$ is velocity ... the particle is at rest when velocity is zero.

Find $v(t)=d’(t)$, set equal to zero, and solve for $t$.

6. ## Re: particle motion

Originally Posted by skeeter
The derivative of the position function $d=t^2-8t+15$ is velocity ... the particle is at rest when velocity is zero.

Find $v(t)=d’(t)$, set equal to zero, and solve for $t$.
Got it thanks

Sent from my SGH-I747M using Tapatalk

7. ## Re: particle motion

Without using Calculus- since the position function $x= t^2- 8t+ 15$, is quadratic complete the square:
$x= t^2- 8t+ 16- 16+ 15= (t- 4)^2- 1$. That is a parabola, opening upwards, with vertex at (4, -1). If t< 4, as t increases, x goes downward. The velocity is negative. If t< 4, as t increases, x goes upward. The velocity is positive. At the vertex, t= 4, the velocity is 0.

8. ## Re: particle motion

Originally Posted by HallsofIvy
Without using Calculus- since the position function $x= t^2- 8t+ 15$, is quadratic complete the square:
$x= t^2- 8t+ 16- 16+ 15= (t- 4)^2- 1$. That is a parabola, opening upwards, with vertex at (4, -1). If t< 4, as t increases, x goes downward. The velocity is negative. If t< 4, as t increases, x goes upward. The velocity is positive. At the vertex, t= 4, the velocity is 0.
Thanks a bunch for the answer

Sent from my SGH-I747M using Tapatalk

Page 2 of 2 First 12