# Thread: Need help with this velocity question

1. ## Need help with this velocity question

Hi can someone help me figure this out:

position of a particle (s) at time t:
s = 2t^3 + 18t^2 - 672t

I found the total distance traveled during the first 19 secs is 7448 (right?)

I need help figuring this out:

the particle is at rest after ___ seconds. <- is it 8? the answer doesn't work

also, The particle is moving forward in the time interval t>8 and backward in the time interval t<8, is this right?

thanks!

2. Hello, coldfire!

I will assume that: $t \geq 0$

Position $(s)$ of a particle at time $t$: . $s(t) \:= \:2t^3 + 18t^2 - 672t$

I found the total distance traveled during the first 19 secs is 7448, right? . . . . no

I need help figuring this out:
The particle is at rest after ___ seconds.
. . Is it 8? . . .the answer doesn't work . . . . yes, it does

Also, the particle is moving forward in the time interval t > 8
and backward in the time interval t < 8, is this right? . . . . Yes!

The position function is: . $s(t) \;=\;2t^3 + 18t^2 - 672t$

The velocity function is: . $v(t) \:=\:s'(t) \:=\:6t^2 + 36t - 672$

The velocity is zero when: . $6t^2 + 36t - 672 \:=\:0 \quad\Rightarrow\quad t^2 + 6t - 112 \:=\:0$

. . Factor:. $(t-8)(t+14) \:=\:0 \quad\Rightarrow\quad t \;=\;8,\:{\color{red}\rlap{/////}}-14$

Let's examine the behavior of the particle.

At $t = 0:\;\;s(0) = 0,\;v(0) = -672$
. . It is at the origin and is moving to the left.

At $t = 8:\;\;s(8) = -3200,\;v(8) = 0$
. . It has moved 3200 units to the left ... and has stopped.
An instant later, it is moving to the right.

At $t=19:\;s(19) = +7448,\;v(19) = 2178$
. . It has moved: . $7448 - (-3200) \:=\:10,648$ units to the right.

It moved a total of: . $32,000 + 10,648 \;=\;\boxed{13,048\text{ units}}$

3. ohh I get it now!

that was very helpful, thank you!

(answer was 13,848 by the way, I saw where you made a small mistake in the addition)

So it's basically the same thing as saying:

$|s(8)-s(0)| + |s(19)-s(8)|$?