# Analysis of position, velocity, and acceleration

• May 1st 2009, 12:07 PM
Kaitosan
Analysis of position, velocity, and acceleration
I've been having a hard time with this graph problem. Go to this link and scroll down to #3 -

http://www.collegeboard.com/prod_dow..._bc_frq_01.pdf

(a.) Hm. I'm going to say that the velocity is increasing at t=2 because the acceleration graph basically traces the value of the slope of v(t). The graph is positive (and thus velocity increases) at t=2. Am I right?

(b.) Maybe t=12? It looks like, at t=12, there's net acceleration of zero...?

(c.) I'm pretty sure it's t=6 because it satisfies the first derivative test.

(d.) I don't think there's any time. Just my hunch. Probably because there's not enough negative element in a definite integral of a(t) to cancel the initial velocity or something, I don't know.

I'm somewhat weak in analyzing f, f', and f'' graphs. So I have some questions.

(1.) Definite integrating f gives the area. Definite integrating f' gives the total change of distance (aka distance traveled and displacement, depending on the situation) or the change of quantity in a rate. But what about definite integrating f''? What does it give, exactly?

(2.) f'' represents the value of the slope of f', right? So if the slope of f' is zero, f''=0 right? Ok, so. But it's common knowledge that f'' represents how the rate (f') changes. So, to be honest, it doesn't make sense to me that f''=0 if the slope of f' is zero because the rate of change of f' from, say, the x-interval of (a,b) with a local extremum in between would be either negative (concave down) or positive (concave up). But I'm sure I'm missing something simple here.

• May 1st 2009, 12:27 PM
skeeter
Quote:

Originally Posted by Kaitosan
I've been having a hard time with this graph problem. Go to this link and scroll down to #3 -

http://www.collegeboard.com/prod_dow..._bc_frq_01.pdf

(a.) Hm. I'm going to say that the velocity is increasing at t=2 because the acceleration graph basically traces the value of the slope of v(t). The graph is positive (and thus velocity increases) at t=2. Am I right?

correct ... v(t) and a(t) are both > 0

(b.) Maybe t=12? It looks like, at t=12, there's net acceleration of zero...?

correct, because $\textcolor{red}{55 + \int_0^{12} a(t) \, dt = 0}$

(c.) I'm pretty sure it's t=6 because it satisfies the first derivative test.

yes, absolute max velocity = $\textcolor{red}{55 + \int_0^6 a(t) \, dt}$

(d.) I don't think there's any time. Just my hunch. Probably because there's not enough negative element in a definite integral of a(t) to cancel the initial velocity or something, I don't know.

absolute min velocity = $\textcolor{red}{55 + \int_0^{16} a(t) \, dt > 0}$

.
• May 1st 2009, 01:25 PM
Kaitosan
Thank you Skeeter!

Definite integrating an acceleration function gives you the net change (unless you include the absolute value sign) of velocity... right?
• May 1st 2009, 01:47 PM
redsoxfan325
Yes.