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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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1so whats the problem?

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

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

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

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

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

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

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0what does m.s 1 mean?

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

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

mathsmind
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0okay so i need to solve that

mathsmind
 one year 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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