1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help on Physics

  1. Jun 25, 2006 #1
    The acceleration of a motorcycle is given by ax(t) = At - Bt2, where A = 1.70 m/s3 and B = 0.170 m/s4. The motorcycle is at rest at the origin at time t = 0.

    (a) Find its position as a function of time t.

    Find its velocity as a function of time t.

    b. Calculate the max velocity it attains. I already calculated the max velocity it attains which is 28.33 m/s.
     
  2. jcsd
  3. Jun 25, 2006 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Hi there musicman and welcome to PF,

    What are your thoughts on this question? Would you mind posting what you have attempted thus far?
     
  4. Jun 25, 2006 #3
    For the first question it looks like you are just supposed to integrate and sketch some graphs, but, as Hootenanny has already asked, what have you done so far?
     
  5. Jun 25, 2006 #4
    ok i got:

    v - v' = integral(At - Bt^2)dt
    v = (1/2)At^2 - (1/3)Bt^3 + v'

    x - x' = integral( v dt )
    x = (1/6)At^3 - (1/12)Bt^4 + v't + x'
     
  6. Jun 25, 2006 #5

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's correct, as the motocycle starts from rest your v' or constant of integration drops out.
    This is also correct, again as your motocycle begins from rest at the origin your terms v' and x' drop out.
     
  7. Jun 25, 2006 #6
    so would i write those exact functions in the space provided? or solve for an answer? and do i leave the apostrophes after the variables or no
     
  8. Jun 25, 2006 #7

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, they can be simplified a bit, but will your tutor accept unsimplified answers? As a said before, the v', v't and x' terms drop out because of the intial conditions imposed by the question;
     
  9. Jun 25, 2006 #8
    yea, they gotta be simplified, thanks for the help.
     
  10. Jun 25, 2006 #9

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Have you attempted to simplify them? How are you getting on with your second question?
     
  11. Jun 25, 2006 #10
    v = (1/2)At^2 - (1/3)Bt^3 + v' was this the one as position as a function of time or velocity? the velocity right? because x is displacement
     
  12. Jun 25, 2006 #11

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is a function of velocity, this is a result of integrating acceleration.
     
  13. Jun 25, 2006 #12
    ok right after it says find its velocity as a funtion of time it says (m/s) after the blank, so are they askign for a number or just the equation v=(1/2)At^2-(1/3)Bt^3...and isnt it ok for me to take off the +v' sicne it is zero?
     
  14. Jun 25, 2006 #13

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Is a function a number of an equation?:wink:
    Yes, in fact you should do as I have said on numerous occasions previously.
     
  15. Jun 25, 2006 #14
    i wrote an actual number for position as a function of time and velocity as a function of time and got it wrong
     
  16. Jun 25, 2006 #15

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Damn webassign homework :grumpy: . Sorry, I'm not getting at you. What exactly did you write?
     
  17. Jun 25, 2006 #16
    13.12 m for the position as a fucntion of time and 17.34 m/s as the velocity as a function of time.
     
  18. Jun 25, 2006 #17

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Where did you get those numbers from? There is no way you can calculate them; besides the question asks for a function, not a numerical answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help on Physics
  1. Physics help. (Replies: 4)

  2. Physics Help (Replies: 1)

  3. Physics Help (Replies: 2)

Loading...