onegirl
Use the position function to find the velocity and acceleration at time t = 2. S(t) = sqrt(t^2 + 8).
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onegirl
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@zepdrix can u help?
zepdrix
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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. \(\large s''(t)=v'(t)=a(t)\).
zepdrix
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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.
\(\large v(\color{royalblue}{2})=\dfrac{\color{royalblue}{2}}{\sqrt{\color{royalblue}{2}^2+8}}\)
zepdrix
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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.
onegirl
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okay
onegirl
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so i'm taking the derivative of sqrt(t^2 + 8) right? then i plus in 2 ? @zepdrix
zepdrix
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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 \[\large v(t)=\dfrac{t}{\sqrt{t^2+8}}\]
onegirl
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okay so after i take the derivative of that i subtitute 2 in it right?
zepdrix
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yes c:
onegirl
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thanks :)
onegirl
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@mathsmind here it is
onegirl
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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
mathsmind
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so whats the problem?
onegirl
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i need to solve 8/(2^2 + 8)^3/2 or should i just leave it like that ?
mathsmind
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well either u work this out with ur head or use the calculator to get the final answer...
onegirl
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I did use the calculation but i'm not getting a right answer
onegirl
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@zepdrix i'm stuck on this one can u help?
onegirl
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are u there?
onegirl
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@zepdrix
mathsmind
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\[s(t) = \sqrt{t^2 + 8}\]
\[v(t)=\frac{t}{\sqrt{t^2+8}}\]
\[a(t) = \frac{1}{(t^2+8)^{\frac{1}{2}}}-\frac{t^2}{(t^2+8)^{\frac{3}{2}}}\]
onegirl
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ok
mathsmind
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now solve for v(2), and a(2) you'll get the answer
onegirl
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so i should subtitute 2 into t?
mathsmind
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yes
onegirl
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okay
onegirl
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i got 0
mathsmind
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\[v(2)=\frac{(2)}{\sqrt{(2)^2+8}}\]
mathsmind
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\[v(2)=\frac{2}{2\sqrt{1+2}}=\frac{1}{\sqrt{3}}\]
mathsmind
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\[m.s^{-1}\]
onegirl
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what does m.s -1 mean?
mathsmind
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it means meter per a second, u have to put units
onegirl
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okay
onegirl
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so bow i have to solve the problem above?
mathsmind
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\[a(2) = \frac{1}{((2)^2+8)^{\frac{1}{2}}}-\frac{(2)^2}{((2)^2+8)^{\frac{3}{2}}}\]
onegirl
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okay so i need to solve that
mathsmind
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a(2)=0.192 m/s^2
onegirl
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okay
mathsmind
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but u really need to know how to differentiate those simple equations ok
onegirl
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okay