Position, Velocity, and Acceleration as Derivatives
This lesson shows how derivatives turn a position function into velocity and acceleration functions, explains the difference between speed and velocity, and covers how to tell when an object speeds up or slows down using sign charts and worked examples.
How Derivatives Describe Motion
In calculus, the motion of an object moving along a straight line is described by three related functions of time: position, velocity, and acceleration. What makes this topic a natural fit for calculus is that each of these quantities is the derivative of the one before it. Once you know the position function \(s(t)\), differentiating it once and then twice gives you everything you need to know about how the object is moving.
The Three Functions and Their Derivatives
Let \(s(t)\) represent the position of an object at time \(t\). Then:
Velocity is the rate of change of position: \(v(t) = s'(t) = \dfrac{ds}{dt}\).
Acceleration is the rate of change of velocity: \(a(t) = v'(t) = s''(t) = \dfrac{d^2s}{dt^2}\).
In words, velocity tells you how fast position is changing, and acceleration tells you how fast velocity is changing. Since acceleration is the derivative of the derivative of \(s(t)\), it is called the second derivative of position.
Speed vs. Velocity
Velocity can be positive, negative, or zero. A positive value means the object is moving in the direction chosen as positive (often to the right or upward); a negative value means it is moving in the opposite direction. Speed, on the other hand, is never negative, it is just the magnitude of velocity: \(\)speed\( = |v(t)|\). So an object with \(v(t) = -5\) meters per second is moving in the negative direction, but its speed is \(5\) meters per second.
Worked Example: Position to Velocity to Acceleration
Suppose the position of a particle moving along a line is given by \(s(t) = t^3 - 6t^2 + 9t\), where \(t\) is measured in seconds and \(s\) in meters.
Step 1: Find the velocity.
\(v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)\)
Step 2: Find the acceleration.
\(a(t) = v'(t) = 6t - 12\)
Step 3: Find when the particle is at rest.
The particle is momentarily at rest when \(v(t) = 0\), which happens at \(t = 1\) and \(t = 3\). These are the critical numbers of the function \(s(t)\), and they split the motion into intervals where the particle moves forward or backward.
Position function \(s(t) = t^3 - 6t^2 + 9t\), with velocity equal to zero at \(t = 1\) and \(t = 3\).
Speeding Up or Slowing Down: Comparing Signs
Whether an object is speeding up or slowing down does not depend on the sign of acceleration alone, it depends on how the signs of velocity and acceleration compare.
Sign of \(v(t)\)
Sign of \(a(t)\)
What happens
Positive
Positive
Speeding up (moving forward, accelerating forward)
Positive
Negative
Slowing down (moving forward, decelerating)
Negative
Negative
Speeding up (moving backward, accelerating backward)
Negative
Positive
Slowing down (moving backward, decelerating)
The rule is simple: when velocity and acceleration have the same sign, speed is increasing. When they have opposite signs, speed is decreasing. For the example above, on the interval \(1 < t < 3\), \(v(t)\) is negative (the particle moves backward), while \(a(t) = 6t - 12\) changes sign at \(t = 2\), so the particle slows down then speeds up within that interval.
Why This Matters
This position, velocity, and acceleration framework is one of the clearest real-world applications of derivatives. It shows up throughout calculus wherever a quantity changes over time, from projectile motion to problems involving rates and optimization. The same idea of examining where a derivative equals zero or changes sign also underlies other calculus tools, so it is worth being comfortable with reading sign charts before moving on to more advanced applications.