21. Antiderivatives, Areas and the FTC

c. From Acceleration to Velocity and Position

1. In \(1\) Dimension

We know that the derivative of the position is the velocity and the derivative of the velocity is the acceleration. Conversely, the antiderivative of the velocity is the position, up to an additive constant, and the antiderivative of the acceleration is the velocity, up to an additive constant. We can determine the constants by specifying initial conditions for the position and/or velocity.

If a car starts at time \(t=2\,\text{hr}\) at the position \(x(2)=3\,\text{km}\) and has velocity \(v(t)=3t^2+2t\,\dfrac{\text{km}}{\text{hr}}\), where is the car at \(t=3\,\text{hr}\)?

The position is an antiderivative of the velocity. So \[ x(t)=t^3+t^2+C \] To find \(C\), we use the initial condition: \[ x(2)=2^3+2^2+C=3 \qquad \Longrightarrow \qquad C=3-8-4=-9 \] So the position at time \(t\) is: \[ x(t)=t^3+t^2-9 \] and the position at \(t=3\,\text{hr}\) is: \[ x(3)=3^3+3^2-9=27+9-9=27\,\text{km} \]

You can also practice computing the velocity and position from the acceleration by using the following Maplet (requires Maple on the computer where this is executed):

From Acceleration to Velocity and Position   Rate It