Results 1 to 2 of 2

Thread: Velocity and Acceleration

  1. #1
    Junior Member
    Joined
    Aug 2008
    Posts
    52

    Velocity and Acceleration

    An airplane taking off from a runway travels 3600 feet before lifting off. If it starts from rest, moves with constant acceleration, and makes the run in 30 seconds, with what speed does it lift off?

    I know that velocity is the first derivative of the position function, and acceleration is the second, but what is the position function?

    Is it $\displaystyle f(t)=-16t^2+v0t+s0?$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    666
    Thanks
    2
    Awards
    1
    We need to build backwards.

    We know that the acceleration, which is the derivative of the velocity, is:

    $\displaystyle a = \frac{dv}{dt} = -32 ~\frac{ft}{s^2}$

    So, when we integrate with respect to time, we get:

    $\displaystyle \int a ~dt = \int \frac{dv}{dt} ~dt$

    $\displaystyle = v = at + C$

    We know that the initial velocity is when t = 0, and in this case is actually 0, so:

    $\displaystyle v(0) = a(0) + C = C$

    So, the initial velocity $\displaystyle v_0$ is a constant, which in this case is 0, making our velocity equation:

    $\displaystyle v(t) = at$

    Now, we know that the velocity equation is the derivative of the position equation, $\displaystyle s(t)$, so we get:

    $\displaystyle v(t) = s'(t) = at$

    We integrate:

    $\displaystyle \int v(t) ~dt = \int s'(t) ~dt$

    $\displaystyle = s(t) = \frac{1}{2}at^2 + C$

    We know that the initial position, we'll call this $\displaystyle s_0$ is the position at time t = 0:

    $\displaystyle s(0) = \frac{1}{2}a(0)^2 + C = C$

    So the initial position is constant, and in this case is also 0, so we now have our final position function (Plugging in $\displaystyle -32 \ \frac{ft}{s^2}$):

    $\displaystyle s(t) = -16t^2$ (Note: This is the equation you use for your runway problem)

    The general equation is:

    $\displaystyle s(t) = s_0 + v_0t + \frac{1}{2}at^2$ (Note: This is what you use when you have either a non-zero initial height/position or a non-zero initial velocity).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. velocity and acceleration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 2nd 2011, 11:44 AM
  2. Velocity and Acceleration
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 28th 2009, 11:37 AM
  3. Velocity and acceleration
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 27th 2009, 01:54 PM
  4. Velocity and Acceleration
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: Aug 16th 2007, 05:57 AM
  5. [SOLVED] velocity acceleration
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Sep 15th 2005, 08:54 PM

Search Tags


/mathhelpforum @mathhelpforum