Question

Help!

Answer 1

is your function like this:
\[\Large s\left( t \right) = \sqrt {2t + 1} \]

Answer 2

Yes

Answer 3

In order to find the acceleration, we have to compute the second derivative of that function s(t)

Answer 4

I can rewrite your function as below:
\[\Large s\left( t \right) = \sqrt {2t + 1} = {\left( {2t + 1} \right)^{1/2}}\]
so the first derivative is:
\[\Large \frac{{ds}}{{dt}} = \frac{1}{2} \cdot {\left( {2t + 1} \right)^{\frac{1}{2} - 1}} \cdot 2 = ...?\]
please complete

Answer 5

do you guys know what website I can take screen shots with?

Answer 6

Michele can you help me with a question???

Answer 7

When I derive it I got 1/sqr root(2t+1)

Answer 8

correct!

Answer 9

Please don't ask questions on my question :)

Answer 10

And I would plug in 4 into this?

Answer 11

\[\Large \frac{{ds}}{{dt}} = \frac{1}{2} \cdot {\left( {2t + 1} \right)^{\frac{1}{2} - 1}} \cdot 2 = \frac{1}{{\sqrt {2t + 1} }}\]

Answer 12

Sorry but Michele can you help me with a question I will fan and medal

Answer 13

now, we have to compute the second derivative of your starting function, namely we have to compute this:
\[\Large \frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left( {\frac{1}{{\sqrt {2t + 1} }}} \right)\]

Answer 14

Okay I got -1/(2t+1)^(3/2)

Answer 15

we can rewrite that last expression as below:
\[\Large \begin{gathered}
\frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left( {\frac{1}{{\sqrt {2t + 1} }}} \right) = \hfill \\
\hfill \\
= \frac{d}{{dt}}{\left( {2t + 1} \right)^{ - 1/2}} = - \frac{1}{2} \cdot {\left( {2t + 1} \right)^{ - \frac{1}{2} - 1}} \cdot 2 = ...? \hfill \\
\end{gathered} \]

Answer 16

correct!
Now, you can set t=4, into that last second derivative

Answer 17

Okay I got -1/27

Answer 18

that's right!

Answer 19

Thank you so much for your help!!

Answer 20

:)