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

ksaimouli Group Title

The position of a particle moving along a line is given by

  • one year ago
  • one year ago

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

    \[s(t)=2t^3-24t^2+90t+7\]

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

    for what values t is the speed of particle increasing

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

    \[s'(t) \implies velocity \space \space s''(t) \implies acceleration\]

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

    ya i got a(t)=0 i got 4

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

    as critical point

    • one year ago
  6. ceb105 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    ds/dt=6t^2-48t+90 thus d^2s/dt^2=12t-48 => aceleration is greater than zero when t>4

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

    well first get the turning point, you mean s'(t)=0

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

    no i mean s''(t) gives acceleration =0 will give us information about velocity

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

    yes so you got 4 right

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

    yes

    • one year ago
  11. ceb105 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    So yea it is when t>4 that the acceleration is postivei .e. speed increases

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

    yes

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

    no that not the answer i am looking for

    • one year ago
  14. ceb105 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Well its the right one...

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

    but the answer is 3<t<4 and t>5 ( i was shocked !!!!

    • one year ago
  16. ceb105 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    is it possible that this is wrong?

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

    @hartnn

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

    |dw:1360515901447:dw| graph increasing the speed inceases when the grapg rises so x<a and x>b

    • one year ago
  19. ceb105 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Okay well 3 and 5 are the critical points of the velocity

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

    |dw:1360516136831:dw|

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

    |dw:1360516160914:dw|

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

    But why i need the critical points of velocity when they ask for velocity

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

    This is no calculator question so i cannot graph

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

    they did not say the rate of change in velocity nor speed just velocity increasing

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

    yeah i know what i am saying is\[s'(t)=0\]

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

    i e 0,3,5

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

    then how did they get 4

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

    so you get 3 and 5 right

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

    yes

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

    speed =|velocity| \[|6t^2-8t+15|>0\] i would like to explain grapghically 4 a sec

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

    i mean\[|t^2-8t+15|\]

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

    its actually velocity=I speedI

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

    velocity is vector where as speed is scalar

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

    similar quest http://math.stackexchange.com/questions/116405/the-position-of-a-particle-moving-along-a-line-is-given-by-2t3-24t290t-7

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

    i agree with person 3

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

    i found out that they used velocity and acceleration derivatives and they set them =0 and found the answer but i have learned that to find graph of y' use y'' to find max or min

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

    yes so we found \[|t^2-8t+15|>0 \implies t<3 ,t>5\] but also for - velocity we put \[-(t^2-8t+15)\] its positive for \[-(2t-8)>0 \implies t<4\] (3,4) and x>5

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

    did the link help

    • 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.