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

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

In calculus, the derivative provides a powerful tool for analyzing the instantaneous rate of change of a function. The study of straight-line motion applies this concept to describe how an object moves along a line over time. The fundamental idea is that the three key quantities of motion—position, velocity, and acceleration—are intrinsically linked through the process of differentiation.

If we know the position of a particle at any given time, we can determine its instantaneous velocity by finding the derivative of the position function. Velocity describes not only how fast the particle is moving but also in which direction. By taking the derivative again, we can find the particle's acceleration, which describes the rate at which its velocity is changing. This topic establishes the foundational relationships that allow us to analyze and interpret the motion of a particle using the tools of differential calculus.

Key Formulas

The relationships between position, velocity, and acceleration are defined by the derivative. Let $x(t)$ be the position of a particle moving along a line at time $t$.

• Velocity as the Derivative of Position

  The instantaneous velocity of the particle, denoted by $v(t)$, is the derivative of the position function $x(t)$ with respect to time.

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

• Acceleration as the Derivative of Velocity

  The instantaneous acceleration of the particle, denoted by $a(t)$, is the derivative of the velocity function $v(t)$ with respect to time. It is also the second derivative of the position function $x(t)$.

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

• Speed

  Speed is the magnitude of velocity and is always a non-negative value. It is defined as the absolute value of the velocity function.

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

Understanding Direction and Changes in Speed

A critical aspect of motion analysis is distinguishing between velocity and speed, and understanding the conditions under which a particle's speed is increasing or decreasing. These concepts rely on interpreting the signs of the velocity and acceleration functions.

Direction of Motion

The sign of the velocity function determines the direction in which the particle is moving along a line (e.g., the x-axis).

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

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

• If $v(t)=0$, the particle is momentarily at rest.

Speeding Up and Slowing Down

To determine if the particle's speed is increasing or decreasing, one must compare the signs of velocity and acceleration at a given time.

• Speed is increasing when $v(t)$ and $a(t)$ have the same sign.

  • This occurs if both are positive ($v(t)>0$ and $a(t)>0$), meaning the particle is moving right and its velocity is becoming more positive.

  • This also occurs if both are negative ($v(t)<0$ and $a(t)<0$), meaning the particle is moving left and its velocity is becoming more negative (e.g., changing from -2 to -5).

• Speed is decreasing (slowing down) when $v(t)$ and $a(t)$ have opposite signs.

  • This occurs if $v(t)>0$ and $a(t)<0$, meaning the particle is moving right but its velocity is decreasing (approaching zero).

  • This also occurs if $v(t)<0$ and $a(t)>0$, meaning the particle is moving left but its velocity is increasing (approaching zero from the negative side).

Core Concepts & Rules

• Velocity is the Rate of Change of Position: The velocity function, $v(t)$, is the first derivative of the position function, $x(t)$.

• Acceleration is the Rate of Change of Velocity: The acceleration function, $a(t)$, is the first derivative of the velocity function, $v(t)$, and the second derivative of the position function, $x(t)$.

• Speed is Magnitude: Speed is the absolute value of velocity, $speed=|v(t)|$, and describes how fast the particle is moving without regard to direction.

• Sign of Velocity Indicates Direction: A positive velocity implies movement in the positive direction (e.g., right or up). A negative velocity implies movement in the negative direction (e.g., left or down).

• Determining Change in Speed: The particle's speed is increasing when velocity and acceleration have the same sign. The speed is decreasing when they have opposite signs.

Step-by-Step Example 1: Analyzing Motion from a Position Function

A particle moves along the x-axis so that its position at any time $t\ge 0$ is given by the function $x(t)=\frac{1}{3}{t}^3-3t^2+8t+2$. Analyze the motion of the particle at $t=2$.

Step 1: Find the velocity function, $v(t)$.

The velocity is the first derivative of the position function.

$$v(t)=x^{\prime }(t)=\frac{d}{dt}\left(\frac{1}{3}{t}^3-3t^2+8t+2\right)$$

$$v(t)=t^2-6t+8$$

Step 2: Find the acceleration function, $a(t)$.

The acceleration is the first derivative of the velocity function.

$$a(t)=v^{\prime }(t)=\frac{d}{dt}\left(t^2-6t+8\right)$$

$$a(t)=2t-6$$

Step 3: Evaluate position, velocity, and acceleration at $t=2$.

• Position: $x(2)=\frac{1}{3}(2{)}^3-3(2{)}^2+8(2)+2=\frac{8}{3}-12+16+2=\frac{26}{3}$

• Velocity: $v(2)=(2{)}^2-6(2)+8=4-12+8=0$

• Acceleration: $a(2)=2(2)-6=4-6=-2$

Step 4: Determine the speed and direction of motion at $t=2$.

• Speed: The speed is $|v(2)|$. Since $v(2)=0$, the speed is 0.

• Direction: Since $v(2)=0$, the particle is momentarily at rest. It is not moving to the left or right at this instant.

Step 5: Determine if the particle's speed is increasing or decreasing at $t=2$.

To analyze this, we must compare the signs of $v(t)$ and $a(t)$. At $t=2$, $v(2)=0$, which is a point where the particle could be changing direction. Because the velocity is zero, the speed is neither increasing nor decreasing at this exact instant. However, we can analyze the motion just before and after $t=2$.

• For $t<2$ (e.g., $t=1.9$), $v(1.9)=(1.9{)}^2-6(1.9)+8=3.61-11.4+8=0.21>0$.

• For $t>2$ (e.g., $t=2.1$), $v(2.1)=(2.1{)}^2-6(2.1)+8=4.41-12.6+8=-0.19<0$.

• At $t=2$, $a(2)=-2<0$.

Since the velocity changes from positive to negative at $t=2$, the particle is changing direction from right to left.

Step-by-Step Example 2: Analyzing Motion from a Velocity Graph

The graph below shows the velocity $v(t)$ of a particle moving along the x-axis for $0\le t\le 10$. The graph consists of line segments.

(Imagine a graph of v(t) that starts at (0,0), goes up to (2,4), then down to (6,-4), then up to (10,0).)

1. When is the particle moving to the left?

The particle moves to the left when its velocity is negative ($v(t)<0$).

• Analysis: Looking at the graph, the function $v(t)$ is below the t-axis on the interval $(4,10)$.

• Conclusion: The particle is moving to the left for $4<t<10$.

2. What is the particle's acceleration at $t=3$?

Acceleration is the derivative of velocity, $a(t)=v^{\prime }(t)$. On a graph of $v(t)$, the acceleration is the slope of the line.

• Analysis: At $t=3$, the particle is on the line segment connecting the points $(2,4)$ and $(6,-4)$. We find the slope of this segment.

• Calculation:

  $$a(3)=\text{slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{-4-4}{6-2}=\frac{-8}{4}=-2$$

• Conclusion: The acceleration at $t=3$ is $a(3)=-2$.

3. When is the particle's speed increasing?

Speed increases when $v(t)$ and $a(t)$ have the same sign.

• Analysis:

  • Case 1: $v(t)>0$ and $a(t)>0$.

    • $v(t)>0$ on $(0,4)$.

    • $a(t)=v^{\prime }(t)>0$ (slope is positive) on $(0,2)$.

    • Both conditions are met on the interval $(0,2)$.

  • Case 2: $v(t)<0$ and $a(t)<0$.

    • $v(t)<0$ on $(4,10)$.

    • $a(t)=v^{\prime }(t)<0$ (slope is negative) on $(2,6)$.

    • Both conditions are met on the interval $(4,6)$.

• Conclusion: The particle's speed is increasing on the intervals $(0,2)$ and $(4,6)$. This can be visualized as the times when the graph of $v(t)$ is moving away from the t-axis.

Using Your Calculator

For functions that are difficult to differentiate by hand, a graphing calculator is an essential tool for finding numerical values of velocity and acceleration.

Suppose a particle's position is given by $x(t)=t^2\cos (t)$. Find the velocity, acceleration, and speed at $t=3$.

1. Storing the Function:

It is efficient to store the position function in your calculator.

• Enter Y_1 = X^2 cos(X).

2. Finding Velocity at $t=3$:

Velocity is $v(t)=x^{\prime }(t)$. Use the numerical derivative command to find $x^{\prime }(3)$.

• On the home screen, use the numerical derivative function (often nDeriv( on TI-84 or found in the MATH menu).

• The syntax is typically nDeriv(function, variable, value).

• Enter: nDeriv(Y_1, X, 3)

• The result will be approximately $v(3)\approx -4.738$.

3. Finding Acceleration at $t=3$:

Acceleration is $a(t)=v^{\prime }(t)=x^{\prime \prime }(t)$. You can find this by taking the numerical derivative of the velocity function.

• Method A (Nested Derivatives): Use the numerical derivative command on the velocity.

  • Enter: nDeriv(nDeriv(Y_1, X, X), X, 3)

  • This command calculates the derivative of the derivative of $Y_1$ at $X=3$.

• Method B (Graphing Velocity): This is often more reliable.

  • In the Y= editor, set Y_2 = \text{nDeriv}(Y_1, X, X)`. This graphs the velocity function. * Now, use the calculator's $dy/dx tool (often in the $CALC$ menu) on the graph of $Y_2$ at $X=3$.

• Either method will give the result $a(3)\approx -11.698$.

4. Finding Speed at $t=3$:

Speed is $|v(3)|$.

• From Step 2, we found $v(3)\approx -4.738$.

• Speed $=|-4.738|=4.738$.

AP Exam Quick Hit

Common Question Types

• Analysis from an Equation: Given a position function $x(t)$, you will be asked to find the velocity $v(t)$ and acceleration $a(t)$ at a specific time $c$. You may also be asked to determine if the particle is speeding up or slowing down at $t=c$ by comparing the signs of $v(c)$ and $a(c)$.

• Analysis from a Graph: Given the graph of a particle's velocity $v(t)$, you will be asked to identify intervals where the particle is moving left/right ($v(t)<0$ or $v(t)>0$), find the acceleration at a point (slope of the graph), and determine when the speed is increasing (graph moving away from the t-axis) or decreasing (graph moving toward the t-axis).

• Analysis from a Table: Given a table of values for velocity $v(t)$ at various times, you may be asked to approximate the acceleration at a time $t=c$. This is done by calculating the average rate of change of velocity over the smallest interval containing $c$. For example, $a(c)\approx \frac{v(c+\Delta t)-v(c-\Delta t)}{2\Delta t}$.

Common Mistakes

• Confusing Speed and Velocity: A common error is to state that a particle is slowing down simply because its velocity is decreasing. For example, if velocity changes from $v=-2$ to $v=-4$, the velocity has decreased, but the speed (from $2$ to $4$) has increased. Always compare the signs of $v(t)$ and $a(t)$ to determine if speed is increasing or decreasing.

• Confusing Position and Direction: Students sometimes incorrectly use the sign of the position function $x(t)$$ to determine the direction of motion. The direction of motion is determined only by the sign of the velocity function $v(t)$. A particle can be moving left ($v(t)<0$) even when its position is positive ($x(t)>0$).

• Assuming Positive Acceleration Means Speeding Up: A positive acceleration $a(t)>0$ does not automatically mean the particle is speeding up. If the velocity $v(t)$ is negative at that time, the particle is actually slowing down because the signs are opposite.

• Forgetting Absolute Value for Speed: When asked for the speed of a particle, students may provide the velocity. If the velocity is negative, this is an incorrect answer. Speed must be non-negative.