A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

How would you find the acceleration, velocity, and speed of particle with position function r(t)=2i+6tj+12t^2k at the point (2,0,0)?

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Velocity= <0,6,0> speed is the magnitude of V which is 6 and acceleration is 0 at (2,0,0)

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you just need to take the derivative once to get the velocity, and derive the position function twice to get the acceleration, and then just substitute for the point (2,0,0)

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    v=6j+24tk v(2,0,0)=0 a=24k a(2,0,0)=0

  4. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.