Results 1 to 3 of 3

Math Help - Integration, Acceleration, Velocity and time

  1. #1
    Newbie
    Joined
    Jan 2008
    Posts
    11

    Integration, Acceleration, Velocity and time

    The question states

    An object is moving in a straight line through a fixed point A.
    When the object first passes through A it is travelling at 5cm per sec and the timing beings.
    The acceleration of the object is given by

    a= 6t-16
    where acceleration is in cm/sec^2 at time t seconds after the object first passes through A.
    Find the distance of the object from A the second time it comes to rest.

    So i first integrated to get V=3t-16+c
    since t=0 , V=5 then c=5
    so V=3t-16+c

    Then i integrated again to get s = t^3 -8t^2 + 5t + c
    I thought at
    time = 0, distance would equal 0 therefore making c=0 aswell, but when i look at the answer its telling me s=3, so i just assumed that to be right and decided to think about it later as i knew when working out t for when V=0 it wouldn't matter, so i got x= 1/3 and 5 which is correct according to the answer.

    then it says in the answer s(5) = -50cm

    but when i input 5 into the equation s = t^3 -8t^2 + 5t + 3 it gives me -47
    thus bringing me to the 3 again,

    Where does the 3 come from? or is the answer book wrong? am i over looking something?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by immunofort View Post
    The question states

    An object is moving in a straight line through a fixed point A.
    When the object first passes through A it is travelling at 5cm per sec and the timing beings.
    The acceleration of the object is given by

    a= 6t-16
    where acceleration is in cm/sec^2 at time t seconds after the object first passes through A.
    Find the distance of the object from A the second time it comes to rest.

    So i first integrated to get V=3t-16+c
    since t=0 , V=5 then c=5
    so V=3t-16+c

    Then i integrated again to get s = t^3 -8t^2 + 5t + c
    I thought at
    time = 0, distance would equal 0 therefore making c=0 aswell, but when i look at the answer its telling me s=3, so i just assumed that to be right and decided to think about it later as i knew when working out t for when V=0 it wouldn't matter, so i got x= 1/3 and 5 which is correct according to the answer.

    then it says in the answer s(5) = -50cm

    but when i input 5 into the equation s = t^3 -8t^2 + 5t + 3 it gives me -47
    thus bringing me to the 3 again,

    Where does the 3 come from? or is the answer book wrong? am i over looking something?
    x = t^3 - 8t^2 + 5t NOT x = t^3 - 8t^2 + 5t + 3.

    I don't know where the 3 is coming from but it should not be in your expression for x since x(0) = 0 NOT 3.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2008
    Posts
    11

    thanks

    Ok thanks for that, Just needed some confirmation from somebody who was a bit more experienced than me. I guess it must be fault on the exam sheet.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: December 6th 2010, 02:52 AM
  2. Replies: 10
    Last Post: February 20th 2010, 03:34 PM
  3. Replies: 1
    Last Post: September 17th 2009, 03:51 AM
  4. Velocity and Acceleration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 1st 2008, 07:30 PM
  5. Replies: 9
    Last Post: October 4th 2007, 05:54 AM

Search Tags


/mathhelpforum @mathhelpforum