Results 1 to 7 of 7

Math Help - Finding the Velocity Vector and Speed after t seconds

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    79

    Finding the Velocity Vector and Speed after t seconds

    This is to do with mechanics, which as far as I know is pre-calc, not pre-algebra.

    Question: A particle has an initial velocity of i+2j and is accelerating in the direction i+j. If the magnitude of the acceleration is 5\sqrt{2}, find the velocity vector and the speed of the particle after 2 seconds.

    Answer:
    velocity vector: 11i+12j
    speed of particle: \sqrt{265} ms^{-1}

    I am literally stumped on how to achieve this answer. If you can tell me what to do to get there, I will try working it out .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,037
    Thanks
    883
    Quote Originally Posted by Viral View Post
    This is to do with mechanics, which as far as I know is pre-calc, not pre-algebra.

    Question: A particle has an initial velocity of i+2j and is accelerating in the direction i+j. If the magnitude of the acceleration is 5\sqrt{2}, find the velocity vector and the speed of the particle after 2 seconds.

    Answer:
    velocity vector: 11i+12j
    speed of particle: \sqrt{265} ms^{-1}
    direction of the acceleration vector is \arctan(1) = 45^\circ

    since the acceleration vector has magnitude 5\sqrt{2}, each of its components has magnitude 5.

    a = 5i + 5j

    v(t) = (1 + 5t)i + (2 + 5t)j

    v(2) = 11i + 12j

    |v(2)| = \sqrt{11^2 + 12^2} = \sqrt{265}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    79
    Ehh, that seems too condensed. Can you expand on the v(t) and v(2) parts, and is there a way without using arctan? We haven't had to use trigonometry for these questions, although we have used pythagoras ( \sqrt{i^{2}+j^{2}}).

    I worked out the 5i and 5j components somehow, but that's as far as I got.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,037
    Thanks
    883
    Quote Originally Posted by Viral View Post
    Ehh, that seems too condensed. Can you expand on the v(t) and v(2) parts, and is there a way without using arctan? We haven't had to use trigonometry for these questions, although we have used pythagoras ( \sqrt{i^{2}+j^{2}}).

    I worked out the 5i and 5j components somehow, but that's as far as I got.
    you should know that a = 5i+5j forms a 45-45-90 triangle, so there is no need for the arctangent function in this case.

    the basic velocity kinematics equation for constant acceleration is v = v_o + at

    component-wise, for your problem, that would be

    v_x = v_{ox} + a_x t = 1 + 5t

    so, velocity as a function of time in the x-direction is (1+5t)i

    v_y = v_{oy} + a_y t = 2 + 5t

    and, velocity as a function of time in the y-direction is (2+5t)j

    the overall velocity vector represented in component form is

    v(t) = (1+5t)i + (2+5t)j

    that's about as easy as I can put it for you without teaching about two-dimensional kinematics in total.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2009
    Posts
    79
    Hmm, I just tried working it out again, without the arctan stuff and got the answer right:

    1). v=u+at=11i+12j
    2). \sqrt{11^{2}+12^{2}}=\sqrt{265}ms^{-1}

    But now I don't see where the direction of acceleration came into play at all.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,037
    Thanks
    883
    Quote Originally Posted by Viral View Post
    Hmm, I just tried working it out again, without the arctan stuff and got the answer right:

    1). v=u+at=11i+12j
    2). \sqrt{11^{2}+12^{2}}=\sqrt{265}ms^{-1}

    But now I don't see where the direction of acceleration came into play at all.
    what if (i + j) was not the direction of the acceleration vector?

    how, then, would one determine the value of each component?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2009
    Posts
    79
    That's what I don't understand, since I didn't use it at all (well, I obviously did, but the value of 1 doesn't change my final answer).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 11
    Last Post: April 15th 2011, 01:34 PM
  2. Replies: 5
    Last Post: January 10th 2011, 02:33 PM
  3. [SOLVED] Finding the speed from the velocity
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: February 8th 2009, 10:56 AM
  4. speed velocity
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: February 23rd 2007, 09:35 AM
  5. velocity and speed
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 24th 2006, 06:59 AM

Search Tags


/mathhelpforum @mathhelpforum