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

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zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1The 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
 2 years ago
Best ResponseYou've already chosen the best response.1So 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
 2 years ago
Best ResponseYou've already chosen the best response.1Understand 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
 2 years ago
Best ResponseYou've already chosen the best response.0so i'm taking the derivative of sqrt(t^2 + 8) right? then i plus in 2 ? @zepdrix

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1You/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
 2 years ago
Best ResponseYou've already chosen the best response.0okay so after i take the derivative of that i subtitute 2 in it right?

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0so 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
 2 years ago
Best ResponseYou've already chosen the best response.1so whats the problem?

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0i need to solve 8/(2^2 + 8)^3/2 or should i just leave it like that ?

mathsmind
 2 years ago
Best ResponseYou've already chosen the best response.1well either u work this out with ur head or use the calculator to get the final answer...

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0I did use the calculation but i'm not getting a right answer

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0@zepdrix i'm stuck on this one can u help?

mathsmind
 2 years ago
Best ResponseYou've already chosen the best response.1\[s(t) = \sqrt{t^2 + 8}\] \[v(t)=\frac{t}{\sqrt{t^2+8}}\] \[a(t) = \frac{1}{(t^2+8)^{\frac{1}{2}}}\frac{t^2}{(t^2+8)^{\frac{3}{2}}}\]

mathsmind
 2 years ago
Best ResponseYou've already chosen the best response.1now solve for v(2), and a(2) you'll get the answer

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0so i should subtitute 2 into t?

mathsmind
 2 years ago
Best ResponseYou've already chosen the best response.1\[v(2)=\frac{(2)}{\sqrt{(2)^2+8}}\]

mathsmind
 2 years ago
Best ResponseYou've already chosen the best response.1\[v(2)=\frac{2}{2\sqrt{1+2}}=\frac{1}{\sqrt{3}}\]

mathsmind
 2 years ago
Best ResponseYou've already chosen the best response.1it means meter per a second, u have to put units

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0so bow i have to solve the problem above?

mathsmind
 2 years ago
Best ResponseYou've already chosen the best response.1\[a(2) = \frac{1}{((2)^2+8)^{\frac{1}{2}}}\frac{(2)^2}{((2)^2+8)^{\frac{3}{2}}}\]

onegirl
 2 years ago
Best ResponseYou've already chosen the best response.0okay so i need to solve that

mathsmind
 2 years ago
Best ResponseYou've already chosen the best response.1but u really need to know how to differentiate those simple equations ok
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