Acceleration as the Second Derivative of Position

Position, Velocity, and Acceleration

A position function describes where an object is along a line at each time. Differentiating that function gives the object's instantaneous velocity; differentiating once more gives its instantaneous acceleration. This connects the second derivative to a physical quantity: it measures how quickly velocity is changing.

Let \(s(t)\) denote the signed position of an object at time \(t\). If position is measured in meters and time in seconds, then \(s(t)\) has units of meters. Its first derivative has units of meters per second, and its second derivative has units of meters per second squared.

Position, velocity, and acceleration are functions of time. The values \(s(t_0)\), \(v(t_0)\), and \(a(t_0)\) describe the position, velocity, and acceleration at a particular instant \(t_0\). In one-dimensional motion, the signs of velocity and acceleration refer to the chosen positive direction. A negative velocity means motion in the negative direction; a negative acceleration means velocity is decreasing, not necessarily that the object is slowing down.

Why Acceleration Is the Second Derivative

The derivative of position is velocity because velocity is defined as the instantaneous rate of change of position. For a small time interval \(h\), the average velocity from \(t\) to \(t+h\) is the change in position divided by elapsed time. Taking the limit as the interval shrinks gives the instantaneous velocity:

Acceleration is, in turn, the instantaneous rate of change of velocity. Applying the same definition to the velocity function gives:

Since \(v(t)=s'(t)\), differentiating velocity differentiates position a second time. Therefore \(a(t)=s''(t)\). This is the reason the second derivative of position represents acceleration: it is the derivative of the rate of change of position.

The units follow from the meaning of each derivative. A change in position is measured in meters, and a change in time in seconds, so the rate of position change has units \(\text{m}/\text{s}\). A change in velocity is measured in \(\text{m}/\text{s}\); dividing by a change in time gives \((\text{m}/\text{s})/\text{s}=\text{m}/\text{s}^2\).

Polynomial Position Models

For a polynomial position function, use the power rule term by term to find velocity, then differentiate the result to find acceleration. Each differentiation reduces the power of \(t\) by one, and constant terms disappear. Remember that if \(t\) represents time in seconds, the coefficients must give each term in \(s(t)\) units of position.

Sinusoidal Position Models

Sinusoidal functions can model repeated motion. When differentiating a sine or cosine with an input such as \(\omega t\), the chain rule contributes a factor of \(\omega\) each time. As a result, the second derivative includes a factor of \(\omega^2\). Here \(\omega\) is the angular frequency; when \(t\) is in seconds, it has units of inverse seconds.

Exponential Position Models

Exponential models also require careful use of the chain rule. The exponent must be dimensionless. If time is measured in seconds, a constant multiplying \(t\) in the exponent must have units of inverse seconds. Each derivative brings down that constant as a factor.

Keep the Derivatives and Their Meanings Distinct

The functions \(s\), \(v\), and \(a\) answer different questions. Position says where the object is. Velocity says how quickly and in which direction position is changing. Acceleration says how quickly velocity is changing. In particular, the sign of acceleration alone does not determine whether speed is increasing: that depends on how the velocity and acceleration signs compare.

For example, if velocity is positive and acceleration is negative, the object moves in the positive direction while its velocity decreases. If velocity is negative and acceleration is also negative, the object moves in the negative direction while its velocity becomes more negative. In both cases acceleration is negative, but its effect on speed differs.

Another common error in a sinusoidal model is to include the frequency factor only once when finding acceleration. For \(s(t)=4\cos(\frac{\pi}{3}t)\), writing \(a(t)=-\frac{4\pi}{3}\cos(\frac{\pi}{3}t)\) is incorrect: that expression is missing a second factor of \(\frac{\pi}{3}\). Differentiating velocity shows the correction explicitly: \(a(t)=-\frac{4\pi}{3}\cos(\frac{\pi}{3}t)\cdot\frac{\pi}{3}=-\frac{4\pi^2}{9}\cos(\frac{\pi}{3}t)\). The chain rule applies again on the second differentiation.

A Reliable Calculation Process

Given an explicit position function, keep the calculation and interpretation organized:

In AP Calculus problems, a numerical result should be accompanied by its units and a sentence that connects the value to the situation. Writing \(a(2)=0.75e\) without units or interpretation is incomplete when acceleration is asked for in context. Writing \(a(2)=0.75e\text{ m/s}^2\) identifies the physical quantity and its scale.