Question 1127316
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x'(t)\ =\ v(t)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(t)\ +\ C\ =\ \int\,x'(t)\ =\ \int\,\frac{3}{\sqrt{t}}\ =\ 6\sqrt{t}\ +\ C]


But *[tex \Large x(1)\ =\ 11] so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\sqrt{1}\ +\ C\ =\ 11]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C\ =\  5]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(t)\ =\ 6\sqrt{t}\ +\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x{'}{'}(t)\ =\ v'(t)\ =\ \frac{d}{dt}\,\frac{3}{\sqrt{t}}\ =\ -\frac{3}{2t^{\small{\frac{3}{2}}}] 
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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