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

virtus Group Title

if displacement is given by x=ln(t^2 +3) where x is in cm and t is in seconds, find the exact time when the particle is 5cm to the right of the origin

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. Callum29 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Solve the following: \[\ln(t^2 + 3) = 5\] \[t^2 + 3 - e^5 = 0 \Longrightarrow t = \pm \sqrt{e^5 - 3} \]

    • 2 years ago
  2. Callum29 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    If it did't start at the origin, say it started at the point a where\[P>a>0\] and where \[P\] is a distance of interest, then the exact time it would take for the particle to reach P having started at a is calculate by solving: \[P-a = \ln (t^2 + 3)\]

    • 2 years ago
  3. Callum29 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    didn't*

    • 2 years ago
  4. virtus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    callum 29 how to i go from where you left me at?

    • 2 years ago
  5. virtus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    *do i go

    • 2 years ago
  6. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    what is the answer?

    • 2 years ago
  7. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    12.0587 seconds?

    • 2 years ago
  8. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I have to go now, later guys :)

    • 2 years ago
  9. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I have to go now, later guys :)

    • 2 years ago
  10. Callum29 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Well \[e^5\] is a constant, approximately equal to 148.413 so we have two solutions: \[t = \pm \sqrt{148.413 - 3} = \pm \sqrt{145.413} = \pm 12.18249 = \pm 12.18\] to 2 d.p. Obviously, there are two solutions, however only one makes sense. You cannot have a negative value for time - otherwise you are going back in time! So you jut take the positive one.

    • 2 years ago
  11. Callum29 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    I think the negative time corresponds to the time taken by the particle to move from the point 5cm away from the origin back to the origin, in re-wind if you like. It tells you it takes 12.18 seconds to get there and 12.18 seconds to come back...

    • 2 years 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.