1. ## Acceleration Problem

A particle moves along the curve given by s = sqrt(t+1). Find the acceleration at 2 seconds. A particle moves a long a curve.. that means it's giving me position function? And I need to take the derivative twice to get to acceleration?

Also I am working on a review hand out that my professor gave us, so instead of creating a bunch of threads could I just make one thread and post there any time I come to a problem I need explained?

Thanks!

2. Originally Posted by Hibijibi
A particle moves along the curve given by s = sqrt(t+1). Find the acceleration at 2 seconds. A particle moves a long a curve.. that means it's giving me position function? And I need to take the derivative twice to get to acceleration?
. Then plug in t = 2 to your expression.

Originally Posted by Hibijibi
Also I am working on a review hand out that my professor gave us, so instead of creating a bunch of threads could I just make one thread and post there any time I come to a problem I need explained?

Thanks!
Well the problem is that the thread gets very long and rather tedious to go through since there will be posts that have solved some of your previous problems. Just post a question per thread and it'll be easier on us to know what we're dealing with. Also, post any work that you have done too

3. Well I took the derivative twice.

s = $\sqrt{t+1}$
v = 1/(2 $\sqrt(t+1)$)
a = -1/(4 $\sqrt{t+1}^3$)

I plug in 2 for the acceleration function

-1/(4 $\sqrt{3}^3$)

My problem is the answer key says it's supposed to come out to -1/32, and I have -1/4sqrt3. Where'd I go wrong?

4. I think they made a mistake.

$-\frac{1}{4\sqrt{2}^{3}} = -\frac{1}{32}$

This is right if your position function was $s = \sqrt{t}$. It seems that they have forgotten to add + 1 to t = 2.

5. I'll make sure to point that out to my professor, thanks!