1. ## 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 $f(t)=-16t^2+v0t+s0?$

2. 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).