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Math Help - Velocity and Acceleration

  1. #1
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    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 f(t)=-16t^2+v0t+s0?
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  2. #2
    Super Member Aryth's Avatar
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    We need to build backwards.

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

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

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

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

    = v = at + C

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

    v(0) = a(0) + C = C

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

    v(t) = at

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

    v(t) = s'(t) = at

    We integrate:

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

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

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

    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 -32 \ \frac{ft}{s^2}):

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

    The general equation is:

    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).
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