Use the position function to find the velocity and acceleration at time t = 2. S(t) = sqrt(t^2 + 8).

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Use the position function to find the velocity and acceleration at time t = 2. S(t) = sqrt(t^2 + 8).

Mathematics
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@zepdrix can u help?
The position function is given by \(\large s(t)=\sqrt{t^2+8}\). The derivative function of position is velocity. \(\large s'(t)=v(t)\). The second dervative of position, which is the first derivative of velocity, is the acceleration function. \(\large s''(t)=v'(t)=a(t)\).
So I'll help you along with the first part, I'll bet you can figure out the acceleration based on this. \(\large s'(t)=\dfrac{t}{\sqrt{t^2+8}}\) This is our velocity function, so let's write it like this, \(\large v(t)=\dfrac{t}{\sqrt{t^2+8}}\) They want us to find the velocity at time t=2. \(\large v(\color{royalblue}{2})=\dfrac{\color{royalblue}{2}}{\sqrt{\color{royalblue}{2}^2+8}}\)

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Understand the process? Now take the derivative of your velocity function to find your acceleration function. \(\large v'(t)=a(t)\) and then again, plug t=2 into it to find the acceleration at that time.
okay
so i'm taking the derivative of sqrt(t^2 + 8) right? then i plus in 2 ? @zepdrix
You/I took the derivative of sqrt(t^2 + 8) to get velocity. Now take the derivative of velocity to get acceleration. So you're now taking the derivative of \[\large v(t)=\dfrac{t}{\sqrt{t^2+8}}\]
okay so after i take the derivative of that i subtitute 2 in it right?
yes c:
thanks :)
@mathsmind here it is
so i found the derivative of t^2/sqrt(t^2 + 8) and i got 8/t^2 + 8)^3/2 so i substitute the 2 in the t and i got 8/(2^2 + 8)^3/2 which i'm trying to solve now
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
@zepdrix i'm stuck on this one can u help?
are u there?
\[s(t) = \sqrt{t^2 + 8}\] \[v(t)=\frac{t}{\sqrt{t^2+8}}\] \[a(t) = \frac{1}{(t^2+8)^{\frac{1}{2}}}-\frac{t^2}{(t^2+8)^{\frac{3}{2}}}\]
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

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