Straight-Line Motion: Connecting Position, Velocity, and Acceleration - AP Calculus AB Study Guide

The Core Idea: Straight-Line Motion: Connecting Position, Velocity, and Acceleration

Calculus provides a powerful framework for describing motion. For an object moving along a straight line, its position can be described by a function of time, typically denoted as $s(t)$ or $x(t)$. The fundamental concept of this topic is that the derivative allows us to analyze this motion precisely. The instantaneous rate of change of the particle's position with respect to time is its instantaneous velocity. In turn, the instantaneous rate of change of the particle's velocity is its acceleration.

By applying the principles of differentiation, we can move from a function describing where an object is to functions describing how fast it is moving and how its speed is changing. This topic establishes the foundational relationships between position, velocity, and acceleration, which are essential for solving problems involving motion.

Key Formulas

The relationships between position, velocity, speed, and acceleration are defined by derivatives. Let $s(t)$ be the position of a particle at time $t$.

• Velocity: The instantaneous velocity, $v(t)$, is the derivative of the position function.

  $$v(t)=s^{\prime }(t)$$

• Acceleration: The instantaneous acceleration, $a(t)$, is the derivative of the velocity function (which is also the second derivative of the position function).

  $$a(t)=v^{\prime }(t)=s^{\prime \prime }(t)$$

• Speed: The speed of the particle is the absolute value (or magnitude) of its velocity. Speed is always a non-negative quantity.

  $$\text{speed}=|v(t)|$$

Understanding Velocity vs. Speed

A critical distinction in the study of motion is the difference between velocity and speed. These two concepts are related but are not interchangeable.

• Velocity includes direction. The sign of the velocity function indicates the direction of motion along an axis (e.g., the x-axis).

  • If $v(t)>0$, the particle is moving in the positive direction (e.g., to the right or up).

  • If $v(t)<0$, the particle is moving in the negative direction (e.g., to the left or down).

  • If $v(t)=0$, the particle is momentarily at rest. This is often where the particle changes direction.

• Speed is a scalar quantity and does not include direction. It tells you how fast the particle is moving, regardless of its direction. Since speed is defined as $\text{speed}=|v(t)|$, it can never be negative. For example, a velocity of $-10\text{ m/s}$ and a velocity of $+10\text{ m/s}$ both correspond to a speed of $10\text{ m/s}$.

Core Concepts & Rules

• The position of a particle moving on a line is given by a function of time, $s(t)$ or $x(t)$.

• The first derivative of the position function, $s^{\prime }(t)$, gives the instantaneous velocity function, $v(t)$.

• The second derivative of the position function, $s^{\prime \prime }(t)$, gives the instantaneous acceleration function, $a(t)$. This is also the first derivative of the velocity function, $v^{\prime }(t)$.

• The sign of $v(t)$ indicates the direction of the particle's motion.

• The particle is at rest when its velocity is zero, i.e., $v(t)=0$.

• The speed of the particle is the absolute value of its velocity, $\text{speed}=|v(t)|$.

• The speed of a particle is increasing when its velocity and acceleration have the same sign ($v(t)>0$ and $a(t)>0$, or $v(t)<0$ and $a(t)<0$).

• The speed of a particle is decreasing when its velocity and acceleration have opposite signs ($v(t)>0$ and $a(t)<0$, or $v(t)<0$ and $a(t)>0$).

Step-by-Step Example 1: Finding Velocity, Acceleration, and Speed

A particle moves along the x-axis so that its position at any time $t\ge 0$ is given by the function $x(t)=t^3-9t^2+24t+2$.

(a) Find the velocity and acceleration functions.

(b) What is the velocity of the particle at $t=3$?

(c) What is the speed of the particle at $t=3$?

(d) What is the acceleration of the particle at $t=3$?

Solution

(a) Find the velocity and acceleration functions.

• The velocity function $v(t)$ is the first derivative of the position function $x(t)$.

  $$v(t)=x^{\prime }(t)=\frac{d}{dt}(t^3-9t^2+24t+2)$$

  $$v(t)=3t^2-18t+24$$

• The acceleration function $a(t)$ is the first derivative of the velocity function $v(t)$.

  $$a(t)=v^{\prime }(t)=\frac{d}{dt}(3t^2-18t+24)$$

  $$a(t)=6t-18$$

(b) Find the velocity at $t=3$.

• Substitute $t=3$ into the velocity function $v(t)$.

  $$v(3)=3(3{)}^2-18(3)+24$$

  $$v(3)=3(9)-54+24$$

  $$v(3)=27-54+24=-3$$

• The velocity at $t=3$ is $-3$. This means the particle is moving to the left.

(c) Find the speed at $t=3$.

• Speed is the absolute value of velocity.

  $$\text{speed}=|v(3)|=|-3|=3$$

• The speed at $t=3$ is $3$.

(d) Find the acceleration at $t=3$.

• Substitute $t=3$ into the acceleration function $a(t)$.

  $$a(3)=6(3)-18$$

  $$a(3)=18-18=0$$

• The acceleration at $t=3$ is $0$.

Step-by-Step Example 2: Exam-Style Application

Using the same particle motion from Example 1, with position $x(t)=t^3-9t^2+24t+2$:

(a) At what times $t$ is the particle momentarily at rest?

(b) On what open time intervals is the particle moving to the right?

(c) At $t=1$, is the speed of the particle increasing or decreasing? Justify your answer.

Solution

(a) Find when the particle is at rest.

• The particle is at rest when its velocity is zero. From Example 1, we know $v(t)=3t^2-18t+24$.

• Set $v(t)=0$ and solve for $t$.

  $$3t^2-18t+24=0$$

  $$3(t^2-6t+8)=0$$

  $$3(t-2)(t-4)=0$$

• The solutions are $t=2$ and $t=4$. The particle is at rest at these two times.

(b) Find when the particle is moving to the right.

• The particle moves to the right when its velocity is positive, $v(t)>0$.

• We use the points where $v(t)=0$ (from part a) to test intervals on a sign chart for $v(t)$. The critical points are $t=2$ and $t=4$. We only consider $t\ge 0$.

  Interval	Test Value ($t$)	$v(t)=3(t-2)(t-4)$	Sign of $v(t)$	Direction
  $(0,2)$	$t=1$	$3(1-2)(1-4)=3(-1)(-3)=9$	$+$	Right
  $(2,4)$	$t=3$	$3(3-2)(3-4)=3(1)(-1)=-3$	$-$	Left
  $(4,\infty )$	$t=5$	$3(5-2)(5-4)=3(3)(1)=9$	$+$	Right

• The particle is moving to the right on the intervals $(0,2)$ and $(4,\infty )$.

(c) At $t=1$, is the speed increasing or decreasing?

• To determine if speed is increasing or decreasing, we must compare the signs of velocity and acceleration at $t=1$.

• Find the velocity at $t=1$:

  $$v(1)=3(1{)}^2-18(1)+24=3-18+24=9$$

• Find the acceleration at $t=1$:

  $$a(t)=6t-18$$

  $$a(1)=6(1)-18=-12$$

• At $t=1$, $v(1)=9$ (positive) and $a(1)=-12$ (negative).

• Justification: Because the velocity and acceleration have opposite signs at $t=1$, the speed of the particle is decreasing at that time.

Using Your Calculator

For functions where finding the derivative by hand is complex or for questions on the calculator-active section of the exam, your calculator is an essential tool. The primary function to use is the numerical derivative command.

Problem: The position of a particle is given by $s(t)=t^2\cos (t)$ for $t\ge 0$. Find the acceleration of the particle at t=2.5`. **Method (TI-84 Style Commands):** 1. **Enter the position function:** Press `Y=` and enter `Y1 = X^2 * cos(X)`. (Use $X as the variable).

2. Find acceleration: Acceleration is the second derivative of position. You can find this from the home screen using the numerical derivative command, nDeriv (found under MATH -> $8$).

   • To find the first derivative (velocity) at $t=2.5$, the syntax is nDeriv(Y1, X, 2.5).

   • To find the second derivative (acceleration), you must take the derivative of the derivative. The syntax is:

     nDeriv(nDeriv(Y1, X, X), X, 2.5)

   • This command tells the calculator to take the numerical derivative of the function `nDeriv(Y1, X, X)(which represents $v(t)) with respect to $X$, evaluated at $X=2.5‘.3.\ast \ast ExecuteandInterpret:\ast \ast Pressing‘ENTER‘willyieldaresultofapproximately$-13.566$. So, $a(2.5)\approx -13.566$.

This method is highly efficient for finding the value of a derivative at a point without having to compute the derivative function by hand, which in this case would require two applications of the product rule.

AP Exam Quick Hit

Common Question Types

• Finding Motion Properties from a Position Function: Given $x(t)$, you will be asked to find $v(t)$, $a(t)$, or speed at a specific time $t=c$. This is a direct test of the derivative relationships.

  • Example: "A particle's position is given by $x(t)=5t^3-t^2+3$. Find the acceleration of the particle at $t=2$."

• Analyzing the Direction of Motion: Given $x(t)$, you will be asked to find the time intervals when the particle is moving left, moving right, or is at rest. This requires finding $v(t)$, setting it to zero, and analyzing its sign.

  • Example: "For the particle with position $x(t)=t^3-6t^2+5$, find all time intervals $t\ge 0$ during which the particle is moving to the left."

• Determining if Speed is Increasing or Decreasing: You will be given a position function and a time $t=c$ and asked if the speed is increasing or decreasing. This requires finding the signs of both $v(c)$ and $a(c)$ and comparing them.

  • Example: "At $t=1$, is the speed of the particle with position $x(t)=e^{2t}\sin (t)$ increasing or decreasing? Give a reason for your answer."

Common Mistakes

• Confusing Speed and Velocity: Reporting a negative value for speed. Remember, speed is $\text{speed}=|v(t)|$ and must be greater than or equal to zero.

• Confusing "At Rest" with "Zero Acceleration": A particle is at rest when $v(t)=0$. A particle has zero acceleration when $a(t)=0$, which means its velocity is not changing at that instant (it could be at a local maximum or minimum velocity). The particle can still be moving when acceleration is zero.

• Incorrectly Justifying Increasing/Decreasing Speed: A common mistake is to state that speed is increasing simply because acceleration is positive. This is incorrect. Speed increases only when velocity and acceleration have the same sign. Your justification must mention the signs of both$v(t)$ and $a(t)$.

• Forgetting to Justify: On free-response questions, an answer without a justification (e.g., "The speed is decreasing because $v(t)>0$ and $a(t)<0$") will not receive full credit. You must show your work for finding the values or signs of velocity and acceleration.