AP Calculus AB: Position, Velocity, and Acceleration

Understanding the relationship between position, velocity, and acceleration is where the abstract power of calculus becomes tangibly useful. This framework doesn't just solve textbook problems; it models the real world, from the path of a rocket to the oscillation of a spring, connecting the geometry of motion with the analysis of change through differentiation and integration.

The Foundational Relationship: Derivatives Define Motion

At the heart of this topic is a simple, powerful chain of operations: derivatives measure instantaneous rate of change. For a particle moving along a straight line, its position $s(t)$ is a function of time. The instantaneous rate at which position changes is velocity $v(t)$. Therefore, velocity is the first derivative of the position function with respect to time: $v(t)=s^{\prime }(t)$.

Following this logic, the instantaneous rate at which velocity itself changes is acceleration $a(t)$. Consequently, acceleration is the derivative of velocity and the second derivative of position: $a(t)=v^{\prime }(t)=s^{\prime \prime }(t)$. This derivative relationship is your primary tool for analysis. For example, if a particle's position is given by $s(t)=t^3-6t^2$ meters, its velocity is $v(t)=s^{\prime }(t)=3t^2-12t$ m/s, and its acceleration is $a(t)=v^{\prime }(t)=6t-12$ m/s².

The Reverse Process: Integration Recovers Information

Calculus is a two-way street. If differentiation takes you from position to velocity to acceleration, integration allows you to go backwards. Specifically, the velocity function is the integral of the acceleration function: $v(t)=\int a(t)dt+C_1$. The position function is the integral of the velocity function: $s(t)=\int v(t)dt+C_2$.

Those constants of integration, $C_1$ and $C_2$, are crucial—they represent initial conditions. You need additional information, like initial velocity $v(0)$ or initial position $s(0)$, to find their values and pin down the exact functions. This is how you move from knowing only how an object's motion is changing (acceleration) to knowing where it is at any time.

Analyzing Particle Motion: Direction, Rest, and Speed

With your functions $s(t)$, $v(t)$, and $a(t)$ in hand, you can analyze the particle's behavior in detail. The sign of the velocity indicates direction.

• When $v(t)>0$, the particle is moving forward or to the right.
• When $v(t)<0$, it is moving backward or to the left.
• A particle is at rest (i.e., changing direction) when its instantaneous velocity is zero: $v(t)=0$.

It's vital to distinguish between velocity and speed. Speed is the absolute value of velocity: $|v(t)|$. An object slows down if its speed is decreasing; it speeds up if its speed is increasing. This leads to a key test:

• Speeding Up: Velocity and acceleration have the same sign ($v(t)$ and $a(t)$ are both positive or both negative).
• Slowing Down: Velocity and acceleration have opposite signs.

For instance, if a car has a velocity of $-20$ m/s (moving left) and an acceleration of $-5$ m/s², both are negative. The acceleration is making the negative velocity more negative, so the car is moving left faster—it is speeding up.

Calculating Distance Traveled vs. Displacement

A common application is finding how far a particle travels over a time interval $[a,b]$. The displacement (net change in position) is simply ${\int }_a^{b}v(t)dt=s(b)-s(a)$. However, if a particle moves forward and then backward, the displacement will not reflect the total ground covered.

To find the total distance traveled, you must integrate the speed, not the velocity. This means you integrate the absolute value of the velocity function: $$\text{Total Distance}={\int }_a^b|v(t)|dt.$$

In practice, you find where $v(t)=0$ on the interval $[a,b]$, split the integral at those points, and integrate $v(t)$ over intervals where it is positive and $-v(t)$ over intervals where it is negative, then sum the results.

Worked Example: A particle moves with velocity $v(t)=t^2-2t-3$ on $[0,5]$. Find the total distance traveled from $t=0$ to $t=5$.

1. Find when the particle is at rest: $v(t)=0$.

$$t^2-2t-3=0⟹(t-3)(t+1)=0⟹t=3\text{ (on our interval)}.$$

1. Determine the sign of $v(t)$ on subintervals.

• On $[0,3)$: Test $t=1$. $v(1)=1-2-3=-4$. Velocity is negative.
• On $(3,5]$: Test $t=4$. $v(4)=16-8-3=5$. Velocity is positive.

1. Set up and evaluate the integral of $|v(t)|$.

$$\begin{array}{rl}\text{Total Distance}& ={\int }_0^3|v(t)|dt+{\int }_3^5|v(t)|dt\\ & ={\int }_0^3-(t^2-2t-3)dt+{\int }_3^5(t^2-2t-3)dt\\ & ={\int }_0^3(-t^2+2t+3)dt+{\int }_3^5(t^2-2t-3)dt\\ & ={\left[-\frac{t^3}{3}+t^2+3t\right]}_0^3+{\left[\frac{t^3}{3}-t^2-3t\right]}_3^5\\ & =(-9+9+9)-(0)+\left(\frac{125}{3}-25-15\right)-\left(9-9-9\right)\\ & =9+\left(\frac{125}{3}-40\right)-(-9)\\ & =9+\frac{125}{3}-40+9\\ & =\frac{125}{3}-22=\frac{125}{3}-\frac{66}{3}=\frac{59}{3}.\end{array}$$ The total distance traveled is $\frac{59}{3}$ units.

Common Pitfalls

1. Confusing Speed and Velocity in "Speeding Up" Analysis: Remember, speeding up/slowing down depends on speed ($|v|$), not the sign of velocity. The easiest check is to see if $v(t)$ and $a(t)$ have the same sign. If you only check that acceleration is positive, you will be wrong half the time.

1. Misinterpreting Signs in Integration Problems: When using integration to find position from velocity, the constant of integration is determined by the initial position, $s(0)$. Do not confuse this with the displacement over an interval. Furthermore, when calculating total distance, forgetting to take the absolute value (by splitting the integral at $v(t)=0$) will give you displacement instead.

1. Incorrectly Finding When a Particle is "At Rest": A particle is at rest specifically when its velocity is zero, $v(t)=0$. Setting acceleration to zero, $a(t)=0$, finds when velocity is at a maximum or minimum (i.e., when it is not changing), which is a different and important concept but not when the particle is stopped.

Summary

• Velocity is the derivative of position: $v(t)=s^{\prime }(t)$. Acceleration is the derivative of velocity and the second derivative of position: $a(t)=v^{\prime }(t)=s^{\prime \prime }(t)$.
• Integration, coupled with initial conditions, moves in the opposite direction: velocity is the integral of acceleration, and position is the integral of velocity.
• The sign of $v(t)$ indicates direction of motion. The particle is at rest and changes direction when $v(t)=0$.
• An object is speeding up if its velocity and acceleration have the same sign. It is slowing down if they have opposite signs.
• Total distance traveled is found by integrating the absolute value of velocity, $\int |v(t)|dt$, which typically requires finding where $v(t)=0$ and splitting the integral. This is distinct from displacement, which is $\int v(t)dt$.