Results 1 to 3 of 3

Math Help - Kinematic #3

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    68

    Kinematic #3

    Can someone help me with this problem. Thank you very much! Here is what I did so far.

    Question: A train is traveling at 100 m/s. The engineer applies the brake b/c he sees a bike on the tracks ahead. The brakes caused a acceleration of -2.2 m/s^2. The enginner applies the brake when the train is 960 m from the bike. How long does it take the train to reach the bike?

    Here is what I got so far...
    it is asking for time, so I got the equation:
    x = x not + v not t + 1/2 a t^2
    = 960 + 100t + 1/2 (-2.2)t^2
    I use -b +/- squareroot b^2 -4ac / 2a and got a huge number which is incorrect.
    Please help me, Im really confused.
    Last edited by mr fantastic; October 5th 2009 at 03:24 AM. Reason: Re-titled post
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,698
    Thanks
    454
    Quote Originally Posted by jsu03 View Post
    Can someone help me with this problem. Thank you very much! Here is what I did so far.

    Question: A train is traveling at 100 m/s. The engineer applies the brake b/c he sees a bike on the tracks ahead. The brakes caused a acceleration of -2.2 m/s^2. The enginner applies the brake when the train is 960 m from the bike. How long does it take the train to reach the bike?

    Here is what I got so far...
    it is asking for time, so I got the equation:
    x = x not + v not t + 1/2 a t^2
    = 960 + 100t + 1/2 (-2.2)t^2
    I use -b +/- squareroot b^2 -4ac / 2a and got a huge number which is incorrect.
    Please help me, Im really confused.
    \Delta x = v_0t - \frac{1}{2}at^2

    960 = 100t - 1.1t^2

    1.1t^2 - 100t + 960 = 0

    t = \frac{100 \pm \sqrt{(-100)^2 - 4(1.1)(960)}}{2(1.1)}

    two solutions for t ... choose the one that makes sense in the context of the problem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    From
    New Delhi
    Posts
    153
    Quote Originally Posted by jsu03 View Post
    Can someone help me with this problem. Thank you very much! Here is what I did so far.

    Question: A train is traveling at 100 m/s. The engineer applies the brake b/c he sees a bike on the tracks ahead. The brakes caused a acceleration of -2.2 m/s^2. The enginner applies the brake when the train is 960 m from the bike. How long does it take the train to reach the bike?

    Here is what I got so far...
    it is asking for time, so I got the equation:
    x = x not + v not t + 1/2 a t^2
    = 960 + 100t + 1/2 (-2.2)t^2
    I use -b +/- squareroot b^2 -4ac / 2a and got a huge number which is incorrect.
    Please help me, Im really confused.
    Dear, the eq of motion used here is
    S= ut+(1/2) at^2
    on substituting values
    960= 100t + (1/2) (-2.2)t^2
    or 1.1 t^2 - 100t + 960 = 0

    D=b^2-4ac
    = 10000-4224
    = 5776
    sqrt(D) =sqrt(5776) =76
    now you can time.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Two kinematic problems.
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: September 25th 2009, 03:12 PM
  2. Kinematic #2
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: September 7th 2009, 05:20 PM
  3. Kinematic #1
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 7th 2009, 11:44 AM
  4. Physics Help-Kinematic Equations
    Posted in the Math Topics Forum
    Replies: 7
    Last Post: February 6th 2008, 11:04 AM

Search Tags


/mathhelpforum @mathhelpforum