21. Antiderivatives, Areas and the FTC

c. From Acceleration to Velocity and Position

2. Motion with Constant Acceleration

We frequently need to consider the situation where the acceleration of an object is constant. For example, an object falling near the surface of the Earth feels a gravitational field which produces a constant acceleration \[ g=32\,\dfrac{\text{ft}}{\text{sec}^2} =9.8\,\dfrac{\text{m}}{\text{sec}^2} =980\,\dfrac{\text{cm}}{\text{sec}^2} \] We start by looking at motion with constant acceleration along a line and then look at \(2\) dimensional motion on the next page.

Consider an object which moves with constant acceleration \(a(t)=a\). If the object starts at \(x(0)=x_o\) with velocity \(v(0)=v_o\), find its position and velocity at time \(t\).

The velocity is an antiderivative of the acceleration: \[ v(t)=a t+C \] To find \(C\), we use the initial condition: \[ v(0)=C=v_o \] So the velocity is \[ v(t)=a t+v_o \] The position is an antiderivative of the velocity: \[ x(t)=\dfrac{1}{2}a t^2+v_o t+K \] To find \(K\), we use the initial condition: \[ x(0)=K=x_o \] So the position is \[ x(t)=\dfrac{1}{2}a t^2+v_o t+x_o \] We summarize:

If an object starts at \(x(0)=x_o\) with velocity \(v(0)=v_o\), and moves with constant acceleration \(a(t)=a\), then its position and velocity at time \(t\) are: \[\begin{aligned} x(t)&=x_o+v_o t+\dfrac{1}{2}a t^2 \\ v(t)&=v_o+a t \end{aligned}\]

Although these are the formulas most people remember, we may not have initial conditions at \(t=0\) as in the next problem. So we need to go back to the basic antiderivatives.