A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Abhisar

  • one year ago

position vector of a particle is given by \(\sf r = r_0(1-at)t\) , where t is the time and a as well as \(\sf r_0\) are constant. How much distance is covered by the particle in returning to the starting point?

  • This Question is Closed
  1. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ash2326

  2. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @radar @waterineyes

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

    @pinkbubbles

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry i'm not sure @SolomonZelman

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

    It's ok c:

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

    @Jhannybean !!

  7. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @midhun.madhu1987 @mathstudent55

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

    @ParthKohli

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so you want to solve for when \(r(t)=0\implies t-at^2=1\implies t(at-1)=0\) so $$t=0\text{ or }t=\frac1a$$ so presumably we're taking the \(t=1/a\) root, and the distance covered is given by: $$\begin{align*}s&=\frac12\int_0^t \|r'(\tau)\|\,d\tau\end{align*}$$ i.e. half the total distance, but notice our path is completely collinear, and we know we reach our furthest point at \(t=1/(2a)\) (midpoint of the roots) so our distance on the return is equal to the distance going: $$d=\left\|r\left(\frac1{2a}\right)-r(0)\right\|=\|r_0\|\cdot\frac1{2a}\left(1-\frac12\right)=\frac{\|r_0\|}{4a}$$

  10. 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.