The Velocity function is:

v(t) = 3e^{-t}- 2 and the initial position is s(0) = 0.

Here are my calculations:

s(t) = 3t(e^{-t}) - 2t + C

Since s(0) = 0 we have c=0, Thus,

s(t) = 3t(e^{-t}) - 2t + 0

Did I do this one correct?

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- Dec 1st 2012, 10:11 AMJDSDetermine the position function given the velocity function
The Velocity function is:

v(t) = 3e^{-t}- 2 and the initial position is s(0) = 0.

Here are my calculations:

s(t) = 3t(e^{-t}) - 2t + C

Since s(0) = 0 we have c=0, Thus,

s(t) = 3t(e^{-t}) - 2t + 0

Did I do this one correct? - Dec 1st 2012, 10:39 AMrousseauRe: Determine the position function given the velocity function
There looks to be a mistake involving the integral of e^-t

- Dec 1st 2012, 10:40 AMJDSRe: Determine the position function given the velocity function
I thought as much, I was not sure how to go about doing that

- Dec 1st 2012, 10:52 AMJDSRe: Determine the position function given the velocity function
so then would this be correct? s(t) = 3t(e

^{t}) - 2t + 0 - Dec 1st 2012, 11:11 AMPlatoRe: Determine the position function given the velocity function
- Dec 1st 2012, 11:16 AMrousseauRe: Determine the position function given the velocity function
Can you do the derivative of $\displaystyle e^{-x}$? if so think backwards and try and figure out what the integral would be.

- Dec 1st 2012, 11:18 AMJDSRe: Determine the position function given the velocity function
I do not understand.... in the question it says s(0) = 0 so would c not equal 0? and I also do not understand how you got -3e^-t.....It looks to me like the derivative of that would not be 3e^-t.......perhaps I need more coffee :(

- Dec 1st 2012, 11:21 AMrousseauRe: Determine the position function given the velocity function
I suggest going back to basics and just looking up what the derivative is of a function of the type [e^{ax}] and then reasoning backwards to figure out what the integral, knowing that integration is the opposite of differentiation.

- Dec 1st 2012, 11:27 AMPlatoRe: Determine the position function given the velocity function
- Dec 1st 2012, 11:28 AMJDSRe: Determine the position function given the velocity function
thanks, to both of you, and I did just that rousseau, sometimes all this calculus just jiggles around for me lol, but I do see, now, how the derivative of e^-x would in fact equal -e^(-x)..thanks! so that now makes since.....

-3e^-t - 2t + C However I still am unsure why you got a 3 in place of the C(Constant) when the constant was defined as 0 (at least for the starting point)

Sorry if I sound like an idiot, teaching calculus to oneself is quite a task! - Dec 1st 2012, 11:33 AMJDSRe: Determine the position function given the velocity function
correction, I guess I am no longer teaching myself, you are all teaching me, which is greatly appreciated!!!

- Dec 1st 2012, 11:36 AMJDSRe: Determine the position function given the velocity function
- Dec 1st 2012, 11:49 AMPlatoRe: Determine the position function given the velocity function
- Dec 1st 2012, 12:20 PMJDSRe: Determine the position function given the velocity function
okay, so then you are saying that s(3) = 3 = t which makes that the constant?

- Dec 1st 2012, 12:21 PMJDSRe: Determine the position function given the velocity function
I understand that s(0) = 0 is the starting point, but how did we arrive at the constant being equal to 3, that is the part Im just not quite getting