Thread: Determine 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?

There looks to be a mistake involving the integral of e^-t

I thought as much, I was not sure how to go about doing that

so then would this be correct? s(t) = 3t(e^t) - 2t + 0

Originally Posted by JDS

so then would this be correct? s(t) = 3t(e^t) - 2t + 0

No it should be WHY?

Can you do the derivative of ? if so think backwards and try and figure out what the integral would be.

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

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.

Originally Posted by JDS

I do not understand.... in the question it says s(0) = 0 so would c not equal 0?

Look at reply #5. What is

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!

correction, I guess I am no longer teaching myself, you are all teaching me, which is greatly appreciated!!!

Originally Posted by Plato

Look at reply #5. What is

From what I can tell, s(0) = 0

does it not?

hmmmm....I just had a thought...but I could be wrong..... does it matter what C is as long as it is any constant?

Originally Posted by JDS

From what I can tell, s(0) = 0 does it not? I just had a thought...but I could be wrong..... does it matter what C is as long as it is any constant?

It matters very much if we are given a initial position to start with.
We were given, . So the position is 0 if t=0.

okay, so then you are saying that s(3) = 3 = t which makes that the constant?

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