Results 1 to 7 of 7

Math Help - magnitude of acceleration

  1. #1
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1

    magnitude of acceleration

    Hello All:

    Here's a problem I am ashamed to admit has me stymied. I am sure someone here has an idea.

    "Professor lives S miles from the university and it takes him T minutes to drive from home to work. Prove that at some instant the magnitude of the acceleration of his car is at least \frac{4S}{T^{2}}.

    I think we can use the MVT.

    Wouldn't he be accelerating to the midpoint and then decelerating?.

    I was thinking perhaps the max acceleration would be at S/4.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,117
    Thanks
    382
    Awards
    1
    Quote Originally Posted by galactus View Post
    Hello All:

    Here's a problem I am ashamed to admit has me stymied. I am sure someone here has an idea.

    "Professor lives S miles from the university and it takes him T minutes to drive from home to work. Prove that at some instant the magnitude of the acceleration of his car is at least \frac{4S}{T^{2}}.

    I think we can use the MVT.

    Wouldn't he be accelerating to the midpoint and then decelerating?.

    I was thinking perhaps the max acceleration would be at S/4.
    Assuming that the professor ends his/her trip at 0 m/s. (I suppose he could crash his car at the end of the problem, but then we can calculate the deceleration at the end and find out it is enormously larger than the given answer. ) So the least possible acceleration will be given by a constant acceleration to the half-way point and the same magnitude constant deceleration to the destination. (That this gives the "minimum possible maximum" acceleration is guaranteed by the MVT.)

    So the acceleration may be calcuated by
    d = \frac{1}{2}at^2

    The professor travels S/2 m to get to the half-way point, in a time T/2.

    Thus
    \frac{S}{2} = \frac{1}{2}a \left ( \frac{T}{2} \right )^2

    Solve for a. This is the least possible maximum acceleration to get to the destination on time.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by galactus View Post
    Wouldn't he be accelerating to the midpoint and then decelerating?.
    You're right about that. As we're looking for the least acceleration, we will take the movement uniformly increasing and decreasing. (You can easily prove this)

    Here's the V-t graph:



    Area under the V-t graph gives the distance travelled.
    Then, the area from 0 to T under the graph is S.

    Area = \frac{1}{2} T \cdot V_{max} = S

    S = \frac{T\cdot V_{max}}{2}

    V_max = \frac{2S}{T}

    Slope of the lines (which also means the derivative) gives the acceleration.

    Slope = a = \frac{\Delta y}{\Delta x} = \frac{V_{max}-0}{\frac{T}{2}-0}

    a = \frac{2\cdot V_{max}}{T}

    Plug the V_{max} we found before,
    a = \frac{2\cdot \frac{2S}{T}}{T}

    a = \frac{4S}{T^2}

    Edit: topsquark was faster :P
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I knew I'd feel stupid. I forgot all about the \frac{1}{2}at^{2}

    I was also trying to use the MVT which wasn't necessary.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,117
    Thanks
    382
    Awards
    1
    Quote Originally Posted by galactus View Post
    I knew I'd feel stupid. I forgot all about the \frac{1}{2}at^{2}

    I was also trying to use the MVT which wasn't necessary.

    Thanks.
    The MVT is needed to prove that this is the minimum necessary acceleration, so I wouldn't say that it wasn't necessary. I'd say it's rather critical. (Though it admittedly doesn't help the actual calculation.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    That's pretty close to what wingless showed, isn't it?.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,117
    Thanks
    382
    Awards
    1
    Quote Originally Posted by galactus View Post
    That's pretty close to what wingless showed, isn't it?.
    Yeah, I guess you are right about that.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Magnitude of acceleration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 23rd 2010, 12:55 PM
  2. Need help with Magnitude!!!
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 18th 2010, 09:29 PM
  3. Magnitude of acceleration help
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: February 11th 2010, 04:46 PM
  4. magnitude
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 9th 2009, 10:55 AM
  5. Replies: 0
    Last Post: March 21st 2009, 04:40 AM

Search Tags


/mathhelpforum @mathhelpforum