1. ## Particle motion

Particle moves along vertical axis, position at any time t greater than or equal to 0 is given by the function $\displaystyle s(t) = \frac{1}{3}t^3-3t^2+8t-4$,$\displaystyle s$ is centimenters and $\displaystyle t$ is seconds.

a. Find the displacement during the first 6 seconds

b. Find the average velocity during the first 6 seconds

c. Find expressions for velocity and acceleration at time $\displaystyle t$

d. For what values of $\displaystyle t$ is the particle moving downward?

Here are my answers...are they correct? thanks.

a. 12

b. -10

c. $\displaystyle v(t) = t^2-6t+8$
$\displaystyle a(t)=2t-6$

d. $\displaystyle t$ is less than 0? I'm not sure how to do this one..

2. Originally Posted by live_laugh_luv27
Particle moves along vertical axis, position at any time t greater than or equal to 0 is given by the function $\displaystyle s(t) = \frac{1}{3}t^3-3t^2+8t-4$,$\displaystyle s$ is centimenters and $\displaystyle t$ is seconds.

a. Find the displacement during the first 6 seconds

b. Find the average velocity during the first 6 seconds

c. Find expressions for velocity and acceleration at time $\displaystyle t$

d. For what values of $\displaystyle t$ is the particle moving downward?

Here are my answers...are they correct? thanks.

a. 12 ok

b. -10 no ... avg v = displacement/time

c. $\displaystyle v(t) = t^2-6t+8$ ok
$\displaystyle a(t)=2t-6$ ok

d. $\displaystyle t$ is less than 0? I'm not sure how to do this one..

find the interval where v(t) < 0
...

3. Originally Posted by skeeter
...

so,
b would be 2 sec/centimeter?

and would d be all real numbers less than 0.64?

4. Originally Posted by live_laugh_luv27
so,
b would be 2 sec/centimeter? cm/sec

and would d be all real numbers less than 0.64?
$\displaystyle t^2 - 6t + 8 = 0$

$\displaystyle (t - 2)(t - 4) = 0$

$\displaystyle v(t) < 0$ on the interval $\displaystyle 2 < t < 4$