Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

onegirl Group Title

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

  • one year ago
  • one year ago

  • This Question is Closed
  1. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @zepdrix can u help?

    • one year ago
  2. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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. \(\large s''(t)=v'(t)=a(t)\).

    • one year ago
  3. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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. \(\large v(\color{royalblue}{2})=\dfrac{\color{royalblue}{2}}{\sqrt{\color{royalblue}{2}^2+8}}\)

    • one year ago
  4. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  5. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

    • one year ago
  6. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so i'm taking the derivative of sqrt(t^2 + 8) right? then i plus in 2 ? @zepdrix

    • one year ago
  7. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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 \[\large v(t)=\dfrac{t}{\sqrt{t^2+8}}\]

    • one year ago
  8. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  9. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes c:

    • one year ago
  10. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks :)

    • one year ago
  11. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @mathsmind here it is

    • one year ago
  12. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  13. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so whats the problem?

    • one year ago
  14. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  15. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  16. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  17. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @zepdrix i'm stuck on this one can u help?

    • one year ago
  18. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    are u there?

    • one year ago
  19. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @zepdrix

    • one year ago
  20. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  21. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

    • one year ago
  22. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  23. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so i should subtitute 2 into t?

    • one year ago
  24. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes

    • one year ago
  25. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

    • one year ago
  26. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i got 0

    • one year ago
  27. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  28. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  29. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  30. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    what does m.s -1 mean?

    • one year ago
  31. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  32. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

    • one year ago
  33. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so bow i have to solve the problem above?

    • one year ago
  34. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  35. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    okay so i need to solve that

    • one year ago
  36. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    a(2)=0.192 m/s^2

    • one year ago
  37. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

    • one year ago
  38. mathsmind Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  39. onegirl Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.