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okay

okay so after i take the derivative of that i subtitute 2 in it right?

yes c:

thanks :)

@mathsmind here it is

so whats the problem?

i need to solve 8/(2^2 + 8)^3/2 or should i just leave it like that ?

well either u work this out with ur head or use the calculator to get the final answer...

I did use the calculation but i'm not getting a right answer

are u there?

ok

now solve for v(2), and a(2) you'll get the answer

so i should subtitute 2 into t?

yes

okay

i got 0

\[v(2)=\frac{(2)}{\sqrt{(2)^2+8}}\]

\[v(2)=\frac{2}{2\sqrt{1+2}}=\frac{1}{\sqrt{3}}\]

\[m.s^{-1}\]

what does m.s -1 mean?

it means meter per a second, u have to put units

okay

so bow i have to solve the problem above?

\[a(2) = \frac{1}{((2)^2+8)^{\frac{1}{2}}}-\frac{(2)^2}{((2)^2+8)^{\frac{3}{2}}}\]

okay so i need to solve that

a(2)=0.192 m/s^2

okay

but u really need to know how to differentiate those simple equations ok

okay