# velocity Question

• Feb 22nd 2010, 04:29 PM
mneox
velocity Question
if s = 2t^3 - 5t^2 - 2 and t=2, is s increasing or decreasing?

i don't really get what im supposed to do here. how do i find which one it is?
i know how to find avg. V (or V) and avg. a (or a), but what do i do here?

thanks for any help, this question is stumping me
• Feb 22nd 2010, 04:42 PM
skeeter
Quote:

Originally Posted by mneox
if s = 2t^3 - 5t^2 - 2 and t=2, is s increasing or decreasing?

i don't really get what im supposed to do here. how do i find which one it is?
i know how to find avg. V (or V) and avg. a (or a), but what do i do here?

thanks for any help, this question is stumping me

if s(t) is increasing, then s'(t) > 0

if s(t) is decreasing, then s'(t) < 0

so ... find s'(2) and answer the question.
• Feb 22nd 2010, 04:52 PM
Quote:

Originally Posted by mneox
if s = 2t^3 - 5t^2 - 2 and t=2, is s increasing or decreasing?

i don't really get what im supposed to do here. how do i find which one it is?
i know how to find avg. V (or V) and avg. a (or a), but what do i do here?

thanks for any help, this question is stumping me

Hi mneox,

You have 2 choices,
depending whether you have covered differentiation or not.

Differentiation:

calculate the derivative

$\displaystyle \frac{d}{dt}=f'(t)=\left(2t^3-5t^2-2\right)=6t^2-10t$

Now evaluate f'(2)

f(t) is increasing if f'(t) >0
f(t) is decreasing if f'(t)<0

Alternative

Work with the graph and examine f(2).
See attached sketch.
• Feb 22nd 2010, 05:04 PM
mneox
Quote:

Originally Posted by skeeter
if s(t) is increasing, then s'(t) > 0

if s(t) is decreasing, then s'(t) < 0

so ... find s'(2) and answer the question.

Ohh okay, thank you very much.

I also want to ask, does a particle reverse its direction every time there is a root?
• Feb 23rd 2010, 07:17 AM
skeeter
Quote:

Originally Posted by mneox
Ohh okay, thank you very much.

I also want to ask, does a particle reverse its direction every time there is a root?

a particle reverses its direction every time s'(t) = v(t) changes sign.