Physics

Chapter 3 Motion Along a Straight Line

Imron Rosyadi

Chapter 3: Motion Along a Straight Line

1. Position, Displacement, and Average Velocity
2. Instantaneous Velocity and Speed
3. Average and Instantaneous Acceleration
4. Motion with Constant Acceleration
5. Free Fall
6. Finding Velocity and Displacement from Acceleration

3.1 Position, Displacement, and Average Velocity

Learning Objectives

By the end of this section, you will be able to:

• Define position, displacement, and distance traveled.
• Calculate total displacement given position as a function of time.
• Determine total distance traveled.
• Calculate average velocity given displacement and elapsed time.

We’ll cover the foundational concepts for describing motion in a straight line.

Position: Where Are You?

• Position (\(x\)) tells you where an object is at any given time.
• It’s always relative to a frame of reference.
  • Example: A rocket’s position relative to Earth.
  • Example: A cyclist’s position relative to buildings.
• For one-dimensional motion, we often use the variable \(x\).

Cyclists in Vietnam are described by their position relative to their surroundings.

Position is the starting point for describing motion. It’s crucial to establish a consistent frame of reference, often chosen for convenience.

Displacement: Change in Position

• Displacement (\(\Delta x\)) is the change in an object’s position.
• It’s a vector quantity, meaning it has both magnitude and direction.
• Can be positive or negative depending on the chosen positive direction.

\[ \Delta x = x_f - x_0 \]

• \(x_f\): final position
• \(x_0\): initial position
• SI unit: meters (m).

A professor’s displacement is the change from initial to final position.

Displacement focuses only on the start and end points, not the path taken. This is a key distinction from distance.

Total Displacement vs. Distance Traveled

• Total Displacement (\(\Delta x_{Total}\)):
  • The sum of individual displacements. \[ \Delta x_{Total} = \sum \Delta x_i \]
  • Can be zero even if the object moved.
  • Magnitude is \(| \Delta x_{Total} |\).
• Distance Traveled (\(x_{Total}\)):
  • The total length of the path an object actually traveled.
  • Always positive. \[ x_{Total} = \sum | \Delta x_i | \]

Important

Displacement is a vector, distance is a scalar. Magnitude of total displacement \(\ne\) total distance traveled.

A classic example: walking around a track. Your displacement after one lap is zero, but your distance traveled is the length of the track.

Average Velocity: How Fast, Which Way?

• Average Velocity (\(\bar{v}\)):
  • The rate at which position changes over time.
  • A vector quantity (has magnitude and direction).
  • Calculated as total displacement divided by elapsed time.

\[ \bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_0}{t_f - t_0} \]

• \(\Delta x\): displacement
• \(\Delta t\): elapsed time
• SI unit: meters per second (m/s).

Note

Average velocity can be negative, zero, or positive depending on the displacement.

Average velocity gives an overall picture of motion, but doesn’t tell us about variations within that time interval.

Example: Delivering Flyers

Jill delivers flyers.

• Starts home (\(x_0 = 0\)).
• Leg 1: 0.5 km East (9 min).
• Leg 2: Retraces steps 0.5 km West (9 min).
• Leg 3: 1.0 km East (15 min).
• Leg 4: 1.75 km West (25 min), then stops.

Questions:

1. Total displacement?
2. Magnitude of final displacement?
3. Average velocity?
4. Total distance traveled?
5. Position vs. time graph?

Let’s break down this example step-by-step. It demonstrates the differences between displacement and distance, and how to calculate average velocity.

Example: Delivering Flyers - Solution

Strategy:

• Assign East as positive direction.
• Create a table of positions, times, and displacements.

Time \(t_i\) (min)	Position \(x_i\) (km)	Displacement \(\Delta x_i\) (km)
\(t_0 = 0\)	\(x_0 = 0\)	\(\Delta x_0 = 0\)
\(t_1 = 9\)	\(x_1 = 0.5\)	\(\Delta x_1 = 0.5\)
\(t_2 = 18\)	\(x_2 = 0\)	\(\Delta x_2 = -0.5\)
\(t_3 = 33\)	\(x_3 = 1.0\)	\(\Delta x_3 = 1.0\)
\(t_4 = 58\)	\(x_4 = -0.75\)	\(\Delta x_4 = -1.75\)

Timeline of Jill’s movements.

The table helps organize the data and calculate the individual displacements, which is crucial for answering the questions.

Example: Delivering Flyers - Calculations

1. Total Displacement: \(\Delta x_{Total} = \sum \Delta x_i = 0.5 - 0.5 + 1.0 - 1.75 = -0.75 \text{ km}\) (0.75 km West of home).

2. Magnitude of Final Displacement: \(| -0.75 \text{ km} | = 0.75 \text{ km}\).

3. Average Velocity: \(\bar{v} = \frac{\Delta x_{Total}}{\Delta t_{Total}} = \frac{-0.75 \text{ km}}{58 \text{ min}} = -0.013 \text{ km/min}\) (0.013 km/min West).

4. Total Distance Traveled: \(x_{Total} = \sum | \Delta x_i | = |0.5| + |-0.5| + |1.0| + |-1.75| = 0.5 + 0.5 + 1.0 + 1.75 = 3.75 \text{ km}\).

Notice how total displacement can be negative, indicating direction, while total distance traveled is always positive. The average velocity also reflects the overall direction of the net movement.

Example: Delivering Flyers - Graph

Jill’s position versus time graph.

This graph visually represents Jill’s journey. The slope of the line segments indicates her velocity during those intervals.

Check Your Understanding 3.1

A cyclist rides 3 km west and then turns around and rides 2 km east.

1. What is their displacement?
2. What is the distance traveled?
3. What is the magnitude of their displacement?

Think about setting up a coordinate system for direction (e.g., East = positive, West = negative) to correctly calculate displacement.

3.2 Instantaneous Velocity and Speed

Learning Objectives

By the end of this section, you will be able to:

• Explain the difference between average velocity and instantaneous velocity.
• Describe the difference between velocity and speed.
• Calculate the instantaneous velocity given the mathematical equation for the position.
• Calculate the speed given the instantaneous velocity.

We move from “overall” motion to motion at a specific moment. This is where calculus becomes incredibly useful.

Instantaneous Velocity

• Instantaneous Velocity (\(v(t)\)):
  • How fast an object is moving and in what direction at a specific instant in time.
  • It’s the limit of the average velocity as the elapsed time approaches zero.
  • Mathematically, it’s the derivative of the position function with respect to time.

\[ v(t) = \lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t} = \frac{dx(t)}{dt} \]

• On a position-versus-time graph, instantaneous velocity is the slope of the tangent line at that specific point.

This is a fundamental concept in kinematics. The tangent line at any point on the position-time graph gives the velocity at that exact moment.

Instantaneous Velocity Graphically

Instantaneous velocity as the slope of the tangent line on a position-time graph.

As the time interval \(\Delta t\) gets smaller, the average velocity (slope of the secant line) approaches the instantaneous velocity (slope of the tangent line). Where the slope is zero, the object is momentarily at rest.

Speed: How Fast?

• Speed is the magnitude of the velocity.

• It is a scalar quantity (has magnitude only, no direction).

• Average Speed (\(\bar{s}\)):

  • Total distance traveled divided by the elapsed time. \[ \bar{s} = \frac{\text{Total distance}}{\text{Elapsed time}} \]
  • Not necessarily the magnitude of the average velocity.

• Instantaneous Speed = \(|v(t)|\)

  • Always positive.

Note

In physics, speed and velocity are not interchangeable. Velocity is a vector, speed is a scalar.

If you drive 100 km and return to your starting point in 2 hours, your average velocity is 0 km/h, but your average speed is 50 km/h.

Typical Speeds

Speed	m/s	mi/h
Continental drift	\(10^{-7}\)	\(2 \times 10^{-7}\)
Brisk walk	1.7	3.9
Cyclist	4.4	10
Sprint runner	12.2	27
Rural speed limit	24.6	56
Official land speed record	341.1	763
Speed of sound at sea level	343	768
Space shuttle on reentry	7800	17,500
Escape velocity of Earth*	11,200	25,000
Orbital speed of Earth around the Sun	29,783	66,623
Speed of light in a vacuum	\(299,792,458\)	\(670,616,629\)

Table 3.1: Speeds of Various Objects

* Escape velocity is the velocity at which an object must be launched so that it overcomes Earth’s gravity and is not pulled back toward Earth.

This table gives you a sense of the vast range of speeds encountered in the universe, from geological processes to fundamental constants.

Calculating Instantaneous Velocity (Calculus)

• If the position function \(x(t)\) is given, you can find the instantaneous velocity by taking its derivative.
• For terms of the form \(At^n\) (where A is a constant and n is an integer): \[ \frac{d(At^n)}{dt} = Ant^{n-1} \]
• If \(x(t)\) is a sum of such terms, differentiate each term.

Example:

Position function: \(x(t) = (3.0 \text{ m/s})t + (0.5 \text{ m/s}^3)t^3\)

Find instantaneous velocity at \(t = 2.0 \text{ s}\).

1. Take derivative: \(v(t) = \frac{dx(t)}{dt} = (3.0 \text{ m/s}) + (0.5 \times 3 \text{ m/s}^3)t^{3-1} = (3.0 \text{ m/s}) + (1.5 \text{ m/s}^3)t^2\)

2. Substitute \(t = 2.0 \text{ s}\): \(v(2.0 \text{ s}) = (3.0 \text{ m/s}) + (1.5 \text{ m/s}^3)(2.0 \text{ s})^2 = 3.0 + 1.5(4.0) = 3.0 + 6.0 = 9.0 \text{ m/s}\)

This example shows a direct application of differentiation to find instantaneous velocity. The units also demonstrate how differentiation affects them.

Example: Position vs. Time Graph to Velocity vs. Time Graph

Position vs. Time Graph

Strategy: Calculate the slope (\(\Delta x / \Delta t\)) for each straight-line segment.

• 0 s to 0.5 s: \(v = \frac{0.5 \text{ m} - 0.0 \text{ m}}{0.5 \text{ s} - 0.0 \text{ s}} = 1.0 \text{ m/s}\)
• 0.5 s to 1.0 s: \(v = \frac{0.5 \text{ m} - 0.5 \text{ m}}{1.0 \text{ s} - 0.5 \text{ s}} = 0.0 \text{ m/s}\)
• 1.0 s to 2.0 s: \(v = \frac{0.0 \text{ m} - 0.5 \text{ m}}{2.0 \text{ s} - 1.0 \text{ s}} = -0.5 \text{ m/s}\)

This visually reinforces the idea that velocity is the slope of the position-time graph. Where the position is constant, velocity is zero. Where the position decreases, velocity is negative.

Example: Velocity vs. Time Graph

Velocity vs. Time Graph (from previous example)

This graph represents the instantaneous velocity at all times for the previous example. It directly shows when the object is moving forward, stopped, or moving backward.

Velocity vs. Speed Example

Position of a particle: \(x(t) = (3.0 \text{ m/s})t - (3.0 \text{ m/s}^2)t^2\)

1. Instantaneous velocity at \(t = 0.25 \text{ s}\), \(t = 0.50 \text{ s}\), and \(t = 1.0 \text{ s}\)?

   \(v(t) = \frac{dx(t)}{dt} = (3.0 \text{ m/s}) - (6.0 \text{ m/s}^2)t\)

   • \(v(0.25 \text{ s}) = 3.0 - 6.0(0.25) = 3.0 - 1.5 = 1.5 \text{ m/s}\)
   • \(v(0.50 \text{ s}) = 3.0 - 6.0(0.50) = 3.0 - 3.0 = 0 \text{ m/s}\)
   • \(v(1.0 \text{ s}) = 3.0 - 6.0(1.0) = 3.0 - 6.0 = -3.0 \text{ m/s}\)

2. Speed at these times?

   • Speed at \(t = 0.25 \text{ s}\): \(|1.5 \text{ m/s}| = 1.5 \text{ m/s}\)
   • Speed at \(t = 0.50 \text{ s}\): \(|0 \text{ m/s}| = 0 \text{ m/s}\)
   • Speed at \(t = 1.0 \text{ s}\): \(|-3.0 \text{ m/s}| = 3.0 \text{ m/s}\)

This example clearly distinguishes between velocity (which can be positive, negative, or zero) and speed (which is always non-negative).

Visualizing Position, Velocity, and Speed

Graphs show (a) position, (b) velocity, and (c) speed versus time.

These graphs offer a comprehensive view:

1. The peak in position means the object momentarily stopped and reversed direction.
2. Velocity is positive, then zero, then negative, consistent with position.
3. Speed is always positive, reflecting the magnitude of motion.

Check Your Understanding 3.2

The position of an object as a function of time is \(x(t) = (-3 \text{ m/s}^2)t^2\).

1. What is the velocity of the object as a function of time?
2. Is the velocity ever positive?
3. What are the velocity and speed at \(t = 1.0 \text{ s}\)?

Remember the power rule for differentiation. Pay attention to the sign of the velocity for direction.

3.3 Average and Instantaneous Acceleration

Learning Objectives

By the end of this section, you will be able to:

• Calculate the average acceleration between two points in time.
• Calculate the instantaneous acceleration given the functional form of velocity.
• Explain the vector nature of instantaneous acceleration and velocity.
• Explain the difference between average acceleration and instantaneous acceleration.
• Find instantaneous acceleration at a specified time on a graph of velocity versus time.

Acceleration describes how velocity changes, and it’s a critical concept for understanding why objects speed up, slow down, or change direction.

Average Acceleration

• Average Acceleration (\(\bar{a}\)):
  • The rate at which velocity changes over time.
  • A vector quantity (has magnitude and direction). \[ \bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_0}{t_f - t_0} \]
  • \(v_f\): final velocity
  • \(v_0\): initial velocity
  • \(\Delta t\): elapsed time
  • SI unit: meters per second squared (m/s\(^2\)).

Note

Acceleration occurs when velocity changes in magnitude (speeding up or slowing down) OR in direction, or both.

An object accelerating doesn’t necessarily mean it’s speeding up. It could be slowing down or changing direction.

Acceleration as a Vector

• Acceleration is a vector in the same direction as the change in velocity (\(\Delta v\)).
• Important: Acceleration is not always in the direction of motion.
  • If an object is slowing down, its acceleration is opposite to its direction of motion.
  • Often called “deceleration,” but in physics, we simply say “negative acceleration” if we’ve defined the positive direction as the direction of motion.

A subway train braking is accelerating in a direction opposite to its motion.

Object with positive velocity and negative acceleration slows down, stops, and reverses direction.

Understanding the vector nature of acceleration is key. A negative acceleration doesn’t always mean slowing down; it depends on the direction of velocity. If velocity is negative and acceleration is negative, the object is speeding up in the negative direction.

Example: Calculating Average Acceleration

A racehorse accelerates from rest to 15.0 m/s due west in 1.80 s. What is its average acceleration?

Strategy:

1. Define a coordinate system: East = positive, West = negative.
2. Identify knowns:
   • \(v_0 = 0 \text{ m/s}\) (starts from rest)
   • \(v_f = -15.0 \text{ m/s}\) (west is negative)
   • \(\Delta t = 1.80 \text{ s}\)
3. Identify unknown: \(\bar{a}\)
4. Use formula: \(\bar{a} = \frac{v_f - v_0}{\Delta t}\)

Solution:

\(\bar{a} = \frac{(-15.0 \text{ m/s}) - (0 \text{ m/s})}{1.80 \text{ s}} = \frac{-15.0 \text{ m/s}}{1.80 \text{ s}} = -8.33 \text{ m/s}^2\)

Significance:

The negative sign indicates the acceleration is due west. The horse increases its velocity by 8.33 m/s due west each second.

This example uses a simple one-dimensional motion to illustrate average acceleration. Pay attention to the choice of positive and negative directions.

Check Your Understanding 3.3

Protons in a linear accelerator are accelerated from rest to \(2.0 \times 10^7 \text{ m/s}\) in \(10^{-4} \text{ s}\). What is the average acceleration of the protons?

This is a direct application of the average acceleration formula.

Instantaneous Acceleration

• Instantaneous Acceleration (\(a(t)\)):
  • The acceleration at a specific instant in time.
  • It’s the limit of the average acceleration as \(\Delta t \to 0\).
  • Mathematically, it’s the derivative of the velocity function with respect to time. \[ a(t) = \frac{dv(t)}{dt} \]
• On a velocity-versus-time graph, instantaneous acceleration is the slope of the tangent line at that specific point.

Note

When instantaneous acceleration is zero, the velocity is either at a maximum or a minimum.

Similar to how velocity is the derivative of position, acceleration is the derivative of velocity. It provides detailed information about how velocity is changing moment by moment.

Instantaneous Acceleration Graphically

Instantaneous acceleration as the slope of the tangent line on a velocity-time graph.

1. Shows maximum velocity when the acceleration (slope) is zero.

2. Shows minimum velocity when the acceleration (slope) is zero.

The average acceleration over a large interval might not represent the acceleration at any specific moment.

Example: Calculating Instantaneous Acceleration

Velocity of a particle: \(v(t) = (20 \text{ m/s}^2)t - (5 \text{ m/s}^3)t^2\)

1. Functional form of acceleration:

   \(a(t) = \frac{dv(t)}{dt} = (20 \text{ m/s}^2) - (10 \text{ m/s}^3)t\)

2. Instantaneous velocity at \(t = 1, 2, 3, 5 \text{ s}\):

   • \(v(1 \text{ s}) = 20(1) - 5(1)^2 = 15 \text{ m/s}\)
   • \(v(2 \text{ s}) = 20(2) - 5(2)^2 = 40 - 20 = 20 \text{ m/s}\)
   • \(v(3 \text{ s}) = 20(3) - 5(3)^2 = 60 - 45 = 15 \text{ m/s}\)
   • \(v(5 \text{ s}) = 20(5) - 5(5)^2 = 100 - 125 = -25 \text{ m/s}\)

3. Instantaneous acceleration at \(t = 1, 2, 3, 5 \text{ s}\):

   • \(a(1 \text{ s}) = 20 - 10(1) = 10 \text{ m/s}^2\)
   • \(a(2 \text{ s}) = 20 - 10(2) = 0 \text{ m/s}^2\)
   • \(a(3 \text{ s}) = 20 - 10(3) = -10 \text{ m/s}^2\)
   • \(a(5 \text{ s}) = 20 - 10(5) = -30 \text{ m/s}^2\)

This example shows how both velocity and acceleration can change over time. It’s important to analyze the signs of both vectors to understand the motion.

Interpretation of Velocity and Acceleration Vectors

• \(t = 1 \text{ s}\): \(v(1 \text{ s}) = 15 \text{ m/s}\) (positive), \(a(1 \text{ s}) = 10 \text{ m/s}^2\) (positive).
  • Velocity and acceleration are in the same direction. The particle is speeding up.
• \(t = 2 \text{ s}\): \(v(2 \text{ s}) = 20 \text{ m/s}\) (positive), \(a(2 \text{ s}) = 0 \text{ m/s}^2\).
  • Velocity is at its maximum. Acceleration is zero, meaning velocity is momentarily constant.
• \(t = 3 \text{ s}\): \(v(3 \text{ s}) = 15 \text{ m/s}\) (positive), \(a(3 \text{ s}) = -10 \text{ m/s}^2\) (negative).
  • Velocity and acceleration are in opposite directions. The particle is slowing down.
• \(t = 5 \text{ s}\): \(v(5 \text{ s}) = -25 \text{ m/s}\) (negative), \(a(5 \text{ s}) = -30 \text{ m/s}^2\) (negative).
  • Velocity and acceleration are in the same direction. The particle has reversed direction and is speeding up in the negative direction.

This interpretation is crucial. The signs of velocity and acceleration vectors dictate whether an object is speeding up or slowing down, and in which direction.

Velocity and Acceleration Graphs

Graphs show (a) velocity and (b) acceleration versus time.

1. The peak of the velocity graph corresponds to zero acceleration. The slope of this graph is the acceleration.

2. The acceleration graph shows a linear decrease. Its values match the slopes from the velocity graph.

Check Your Understanding 3.4

An airplane lands on a runway traveling east. Describe its acceleration.

Consider the initial velocity and how it changes as the plane lands. What direction would the acceleration vector point?

Typical Accelerations

Acceleration	Value (m/s\(^2\))
High-speed train	0.25
Elevator	2
Cheetah	5
Object in a free fall without air resistance near the surface of Earth	9.8
Space shuttle maximum during launch	29
Parachutist peak during normal opening of parachute	59
F16 aircraft pulling out of a dive	79
Explosive seat ejection from aircraft	147
Sprint missile	982
Fastest rocket sled peak acceleration	1540
Jumping flea	3200
Baseball struck by a bat	30,000
Closing jaws of a trap-jaw ant	1,000,000
Proton in the large Hadron collider	\(1.9 \times 10^9\)

Table 3.2: Typical Values of Acceleration

Acceleration magnitudes vary enormously. From subtle continental drift to subatomic particle acceleration, the concept is universally applied.

3.4 Motion with Constant Acceleration

Learning Objectives

By the end of this section, you will be able to:

• Identify which equations of motion are to be used to solve for unknowns.
• Use appropriate equations of motion to solve a two-body pursuit problem.

Constant acceleration simplifies kinematics greatly, allowing us to derive powerful equations without calculus. This is a common and important scenario in physics.

Simplified Notation for Constant Acceleration

To simplify our equations for constant acceleration, we make some common assumptions:

• Initial time (\(t_0\)) = 0.
  • Elapsed time \(\Delta t = t_f - t_0 = t\).
• Initial position (\(x_0\)).
• Initial velocity (\(v_0\)).
• Final time (\(t\)).
• Final position (\(x\)).
• Final velocity (\(v\)).

With constant acceleration, the average acceleration equals the instantaneous acceleration: \[ \bar{a} = a = \text{constant} \]

Tip

When acceleration is constant, the average velocity is simply the arithmetic average of the initial and final velocities:

\[ \bar{v} = \frac{v_0 + v}{2} \]

These simplifications streamline the kinematic equations. It’s crucial to remember that these equations only apply when acceleration is constant.

Kinematic Equations for Constant Acceleration

These five equations are essential for solving constant acceleration problems:

1. Displacement from Average Velocity: \[ x = x_0 + \bar{v}t \]

2. Average Velocity (Constant Acceleration): \[ \bar{v} = \frac{v_0 + v}{2} \]

3. Final Velocity from Acceleration and Time: \[ v = v_0 + at \]

4. Final Position from Initial Position, Velocity, Acceleration, and Time: \[ x = x_0 + v_0t + \frac{1}{2}at^2 \]

5. Final Velocity from Displacement and Acceleration: \[ v^2 = v_0^2 + 2a(x - x_0) \]

It’s very useful to recognize which variables are known and unknown, and then select the equation that best fits, minimizing the need for multiple steps. Notice the absence of a specific variable in each equation, which can help in selection.

Example: Calculating Final Velocity

An airplane lands with an initial velocity of 70.0 m/s and then accelerates opposite to the motion at 1.50 m/s\(^2\) for 40.0 s. What is its final velocity?

Knowns:

• \(v_0 = 70.0 \text{ m/s}\)
• \(a = -1.50 \text{ m/s}^2\) (acceleration is opposite to motion)
• \(t = 40.0 \text{ s}\)

Unknown: \(v\)

Equation: \(v = v_0 + at\)

Solution:

\(v = 70.0 \text{ m/s} + (-1.50 \text{ m/s}^2)(40.0 \text{ s})\)

\(v = 70.0 \text{ m/s} - 60.0 \text{ m/s}\)

\(v = 10.0 \text{ m/s}\)

The final velocity is positive and less than the initial velocity, confirming the plane is slowing down but still moving forward.

Example: Calculating Displacement

A dragster accelerates from rest at 26.0 m/s\(^2\) for 5.56 s. How far does it travel in this time?

Knowns:

• \(x_0 = 0 \text{ m}\) (starts at origin)
• \(v_0 = 0 \text{ m/s}\) (starts from rest)
• \(a = 26.0 \text{ m/s}^2\)
• \(t = 5.56 \text{ s}\)

Unknown: \(x\)

Equation: \(x = x_0 + v_0t + \frac{1}{2}at^2\)

Solution:

\(x = 0 + (0 \text{ m/s})(5.56 \text{ s}) + \frac{1}{2}(26.0 \text{ m/s}^2)(5.56 \text{ s})^2\)

\(x = \frac{1}{2}(26.0)(30.9136) \text{ m}\)

\(x = 13(30.9136) \text{ m}\)

\(x \approx 402 \text{ m}\)

This displacement (402 m) is very close to one-quarter mile, which is the standard distance for drag racing, making the answer reasonable.

Example: Calculating Final Velocity (without time)

Calculate the final velocity of the dragster from the previous example without using time information.

Knowns:

• \(v_0 = 0 \text{ m/s}\)
• \(a = 26.0 \text{ m/s}^2\)
• \(x - x_0 = 402 \text{ m}\) (displacement from previous example)

Unknown: \(v\)

Equation: \(v^2 = v_0^2 + 2a(x - x_0)\)

Solution:

\(v^2 = (0 \text{ m/s})^2 + 2(26.0 \text{ m/s}^2)(402 \text{ m})\)

\(v^2 = 20904 \text{ m}^2/\text{s}^2\)

\(v = \sqrt{20904 \text{ m}^2/\text{s}^2} = 144.6 \text{ m/s}\)

\(v \approx 145 \text{ m/s}\)

This demonstrates the utility of having multiple kinematic equations. Sometimes, one equation is more direct for specific knowns and unknowns.

Check Your Understanding 3.5

A rocket accelerates at a rate of 20 m/s\(^2\) during launch. How long does it take the rocket to reach a velocity of 400 m/s?

Identify your knowns and unknown, then pick the appropriate kinematic equation.

Two-Body Pursuit Problems

• In these problems, the motion of two objects is coupled.
• The unknown you seek depends on the motion of both objects.
• Strategy:
  1. Write separate kinematic equations for each object.
  2. Identify a common parameter that links their motion (e.g., same time \(t\), same position \(x\)).
  3. Solve the equations simultaneously.

Car 1 (accelerating) pursues Car 2 (constant velocity).

These problems are more complex but follow a systematic approach. The key is to find that linking variable between the two objects.

Example: Cheetah Catching a Gazelle

A gazelle runs at a constant 10 m/s. A cheetah, starting from rest at the same instant the gazelle passes, accelerates at 4 m/s\(^2\).

1. How long does it take the cheetah to catch the gazelle?
2. What is the displacement of both animals when caught?

Strategy:

• Both start at \(x_0 = 0\).
• They have the same position \(x\) when the cheetah catches the gazelle at time \(t\).

Equations:

• Gazelle (constant velocity): \(x_g = x_0 + v_gt = v_gt\)
• Cheetah (constant acceleration): \(x_c = x_0 + v_{0c}t + \frac{1}{2}a_ct^2 = \frac{1}{2}a_ct^2\) (since \(v_{0c}=0\))

Solution:

1. Set \(x_g = x_c\):

\(v_gt = \frac{1}{2}a_ct^2\)

\(10t = \frac{1}{2}(4)t^2\)

\(10t = 2t^2\)

Since \(t \ne 0\) (they are moving), divide by \(t\):

\(10 = 2t \implies t = 5 \text{ s}\)

2. Displacement:

\(x_g = (10 \text{ m/s})(5 \text{ s}) = 50 \text{ m}\)

\(x_c = \frac{1}{2}(4 \text{ m/s}^2)(5 \text{ s})^2 = 2(25) = 50 \text{ m}\)

Both animals travel 50 meters, and it takes 5 seconds for the cheetah to catch the gazelle. The problem’s common parameter is the final position when they meet.

Check Your Understanding 3.6

A bicycle has a constant velocity of 10 m/s. A person starts from rest and begins to run to catch up to the bicycle in 30 s when the bicycle is at the same position as the person. What is the acceleration of the person?

This is another two-body problem. Think about what common quantity links the bicycle’s motion to the person’s motion at the moment they meet.

3.5 Free Fall

Learning Objectives

By the end of this section, you will be able to:

• Use the kinematic equations with the variables \(y\) and \(g\) to analyze free-fall motion.
• Describe how the values of the position, velocity, and acceleration change during a free fall.
• Solve for the position, velocity, and acceleration as functions of time when an object is in a free fall.

Free fall is a special case of motion with constant acceleration, where the acceleration is due to gravity. This is a very common and important application of kinematics.

Gravity and Free Fall

• Free Fall describes the motion of an object falling solely under the influence of gravity, where air resistance and friction are negligible.
• A remarkable fact: In the absence of air resistance, all objects fall with the same constant acceleration, regardless of their mass.
  • Demonstrated by Galileo and later on the Moon by astronaut David R. Scott.

Hammer and feather falling in air vs. vacuum.

Galileo’s experiments were foundational. Air resistance is often the reason we perceive heavier objects falling faster. In a vacuum, they fall identically.

Acceleration Due to Gravity (\(g\))

• The acceleration of free-falling objects is called acceleration due to gravity.
• Its magnitude is denoted by \(g\).
• Average value on Earth: \(g = 9.8 \text{ m/s}^2\) (or \(32.2 \text{ ft/s}^2\)).
  • We’ll use \(9.8 \text{ m/s}^2\) unless otherwise specified.
• The direction of acceleration due to gravity is always downward (towards the center of Earth).

Important

The sign of \(g\) in kinematic equations depends on your chosen coordinate system: - If upward is positive (\(+y\)), then \(a = -g = -9.8 \text{ m/s}^2\). - If downward is positive (\(+y\)), then \(a = +g = +9.8 \text{ m/s}^2\). Consistency is key!

It’s vital to be consistent with the coordinate system. Changing the positive direction reverses the sign of g.

Kinematic Equations for Free Fall (Upward Positive)

When we define the upward direction as positive (\(+y\)), the acceleration due to gravity is \(a = -g\).

The constant acceleration equations become:

1. Velocity:

   \[ v = v_0 - gt \]

2. Position:

   \[ y = y_0 + v_0t - \frac{1}{2}gt^2 \]

3. Velocity squared:

   \[ v^2 = v_0^2 - 2g(y - y_0) \]

These are the same kinematic equations as before, just with \(x\) replaced by \(y\) and \(a\) replaced by \(-g\).

Problem-Solving Strategy for Free Fall

1. Coordinate System: Decide on the positive direction (up or down). This determines the sign of \(g\).
2. Sketch: Draw a diagram to visualize the motion and define relevant points (\(y_0\), \(y\), \(v_0\), \(v\)).
3. Knowns/Unknowns: List all given values and what you need to find.
4. Equation Selection: Choose the appropriate kinematic equation (from the three shown) that contains the knowns and the single unknown.
5. Solve: Substitute values and solve for the unknown.

Following these steps systematically helps to avoid errors and build confidence in solving free fall problems.

Example: Free Fall of a Ball

A ball is thrown downward from a 98-m-high building with an initial velocity of 4.9 m/s downward.

1. How much time until the ball hits the ground?
2. What is its velocity when it hits the ground?

Strategy:

• Choose origin at top of building (\(y_0 = 0\)).
• Choose upward as positive direction. So, \(a = -g = -9.8 \text{ m/s}^2\).

Knowns:

• \(y_0 = 0 \text{ m}\)
• \(v_0 = -4.9 \text{ m/s}\) (downward is negative)
• \(y = -98 \text{ m}\) (ground is 98 m below origin)
• \(g = 9.8 \text{ m/s}^2\)

Unknowns: (a) \(t\), (b) \(v\)

Setting up the coordinate system correctly is critical here. Note the negative values for initial velocity and final position.

Example: Free Fall of a Ball - Solution

(a) Time to hit the ground:

Equation: \(y = y_0 + v_0t - \frac{1}{2}gt^2\)

\(-98 = 0 + (-4.9)t - \frac{1}{2}(9.8)t^2\)

\(-98 = -4.9t - 4.9t^2\)

Rearrange to quadratic form: \(4.9t^2 + 4.9t - 98 = 0\)

Divide by 4.9: \(t^2 + t - 20 = 0\)

Factor: \((t+5)(t-4) = 0\)

Roots: \(t = -5 \text{ s}\) or \(t = 4 \text{ s}\).

Physical meaning: \(t = 4 \text{ s}\) (time cannot be negative here).

(b) Velocity when it hits the ground:

Equation: \(v = v_0 - gt\)

\(v = -4.9 \text{ m/s} - (9.8 \text{ m/s}^2)(4 \text{ s})\)

\(v = -4.9 - 39.2 = -44.1 \text{ m/s}\)

The quadratic equation yields two roots, but only the positive time is physically relevant for this scenario. The negative velocity indicates the ball is moving downward.

Check Your Understanding 3.7

A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantity increases faster, the speed of the ice chunk or its distance traveled?

Think about which kinematic equation will allow you to solve for time directly given distance, initial velocity (starting from rest), and acceleration due to gravity. For the second question, consider how speed and distance accumulate over time.

Example: Rocket Booster

A rocket at a height of 5.0 km and velocity of 200.0 m/s releases its booster.

1. What is the maximum height the booster attains?
2. What is the velocity of the booster at a height of 6.0 km?

Strategy:

• Choose origin at the point of release (\(y_0 = 0\)).
• Choose upward as positive direction (\(a = -g = -9.8 \text{ m/s}^2\)).

Knowns (at release):

• \(y_0 = 0 \text{ m}\)
• \(v_0 = 200.0 \text{ m/s}\)
• \(g = 9.8 \text{ m/s}^2\)

The initial conditions for the booster’s free fall are established at the moment of release. The height of 5.0 km is important for the total height above ground, but for the booster’s motion relative to its release point, \(y_0 = 0\).

Example: Rocket Booster - Solution

(a) Maximum height booster attains (relative to release point):

At max height, \(v = 0\).

Equation: \(v^2 = v_0^2 - 2g(y - y_0)\)

\(0^2 = (200.0 \text{ m/s})^2 - 2(9.8 \text{ m/s}^2)(y - 0)\)

\(0 = 40000 - 19.6y\)

\(19.6y = 40000 \implies y = \frac{40000}{19.6} \approx 2040.8 \text{ m}\)

Total max height = \(5.0 \text{ km} + 2.0408 \text{ km} = 7.0408 \text{ km} \approx 7.0 \text{ km}\)

(b) Velocity at a height of 6.0 km:

This means the booster is \(1.0 \text{ km}\) above its release point (\(y = 1000 \text{ m}\)).

Equation: \(v^2 = v_0^2 - 2g(y - y_0)\)

\(v^2 = (200.0 \text{ m/s})^2 - 2(9.8 \text{ m/s}^2)(1000 \text{ m} - 0)\)

\(v^2 = 40000 - 19600 = 20400 \text{ m}^2/\text{s}^2\)

\(v = \pm \sqrt{20400} \text{ m/s} \approx \pm 142.8 \text{ m/s}\)

The maximum height calculation uses \(v=0\) as a critical point. For part (b), both positive and negative velocity solutions are physically meaningful, representing the booster’s velocity on its way up (+v) and on its way down (-v) at that altitude.

3.6 Finding Velocity and Displacement from Acceleration

Learning Objectives

By the end of this section, you will be able to:

• Derive the kinematic equations for constant acceleration using integral calculus.
• Use the integral formulation of the kinematic equations in analyzing motion.
• Find the functional form of velocity versus time given the acceleration function.
• Find the functional form of position versus time given the velocity function.

This section introduces the power of calculus to derive the kinematic equations and to handle situations where acceleration is not constant.

Kinematic Equations from Integral Calculus

• Relationship between acceleration and velocity:

  \(a(t) = \frac{dv(t)}{dt}\)

  Integrating both sides with respect to time:

  \[ v(t) = \int a(t)dt + C_1 \]

  Where \(C_1\) is the constant of integration, often found using initial velocity \(v_0\) (at \(t=0\)).

• Relationship between velocity and position:

  \(v(t) = \frac{dx(t)}{dt}\)

  Integrating both sides with respect to time:

  \[ x(t) = \int v(t)dt + C_2 \]

  Where \(C_2\) is the constant of integration, often found using initial position \(x_0\) (at \(t=0\)).

These integral definitions are the fundamental link between acceleration, velocity, and position. They are especially useful when acceleration is a function of time, not just a constant.

Deriving Constant Acceleration Equations via Integration

Assume constant acceleration: \(a(t) = a\).

1. Velocity function:

   \(v(t) = \int a \: dt + C_1 = at + C_1\)

   At \(t=0\), \(v(0) = v_0\), so \(v_0 = a(0) + C_1 \implies C_1 = v_0\). Thus, \(v(t) = v_0 + at\). (Equation 3.12)

2. Position function:

   \(x(t) = \int v(t) \: dt + C_2 = \int (v_0 + at) \: dt + C_2\)

   \(x(t) = v_0t + \frac{1}{2}at^2 + C_2\)

   At \(t=0\), \(x(0) = x_0\), so \(x_0 = v_0(0) + \frac{1}{2}a(0)^2 + C_2 \implies C_2 = x_0\).

   Thus, \(x(t) = x_0 + v_0t + \frac{1}{2}at^2\). (Equation 3.13)

This shows how the familiar kinematic equations for constant acceleration are direct consequences of these integral definitions.

Example: Motion of a Motorboat

A motorboat travels at constant velocity 5.0 m/s. It then starts to accelerate opposite to motion with \(a(t) = -\frac{1}{4}t \text{ m/s}^3\).

1. What is the velocity function?
2. At what time does velocity reach zero?
3. What is the position function?
4. Displacement from when it starts to accelerate until velocity is zero?
5. Graph velocity and position.

Strategy:

• Take \(t=0\) when acceleration begins.
• Initial velocity \(v_0 = 5.0 \text{ m/s}\).
• Initial position \(x_0 = 0\).

This is a more advanced example because acceleration is not constant. This requires direct integration.

Example: Motion of a Motorboat - Solution

(a) Velocity function:

\(v(t) = \int a(t) \: dt + C_1 = \int -\frac{1}{4}t \: dt + C_1 = -\frac{1}{8}t^2 + C_1\)

At \(t=0\), \(v(0) = 5.0 \text{ m/s}\), so \(C_1 = 5.0 \text{ m/s}\). \(v(t) = 5.0 - \frac{1}{8}t^2 \text{ (m/s)}\)

(b) Time when velocity reaches zero:

Set \(v(t) = 0\):

\(0 = 5.0 - \frac{1}{8}t^2 \implies \frac{1}{8}t^2 = 5.0 \implies t^2 = 40 \implies t = \sqrt{40} \approx 6.3 \text{ s}\)

(c) Position function:

\(x(t) = \int v(t) \: dt + C_2 = \int (5.0 - \frac{1}{8}t^2) \: dt + C_2\)

\(x(t) = 5.0t - \frac{1}{24}t^3 + C_2\)

At \(t=0\), \(x(0) = 0\), so \(C_2 = 0\).

\(x(t) = 5.0t - \frac{1}{24}t^3 \text{ (m)}\)

(d) Displacement until velocity is zero (\(t=6.3 \text{ s}\)):

\(x(6.3) = 5.0(6.3) - \frac{1}{24}(6.3)^3 \approx 31.5 - \frac{1}{24}(250.047) \approx 31.5 - 10.4 = 21.1 \text{ m}\)

Notice how each step builds upon the previous one. The initial conditions are used to find the constants of integration.

Example: Motorboat Graphs

Graphs show (a) velocity and (b) position of the motorboat.

1. The velocity decreases quadratically to zero, then becomes negative (reversing direction).

2. Position increases to a maximum, then starts to decrease as the boat reverses direction. The maximum position corresponds to the time when velocity is zero.

Check Your Understanding 3.8

A particle starts from rest and has an acceleration function \(a(t) = (5 - (\frac{10}{1 \text{ s}})t) \text{ m/s}^2\).

1. What is the velocity function?
2. What is the position function?
3. When is the velocity zero?

Apply the integration methods. Remember that “starts from rest” means \(v_0=0\), and you can assume \(x_0=0\) if not specified.

Key Takeaways

• Position is location relative to a frame of reference.
• Displacement is change in position (\(\Delta x = x_f - x_0\)), a vector.
• Distance traveled is the total path length, a scalar.
• Average Velocity is total displacement / elapsed time (\(\bar{v} = \Delta x / \Delta t\)), a vector.
• Instantaneous Velocity is the derivative of position (\(v(t) = dx/dt\)), a vector.
• Speed is the magnitude of velocity (\(|v(t)|\)), a scalar.
• Average Acceleration is change in velocity / elapsed time (\(\bar{a} = \Delta v / \Delta t\)), a vector.
• Instantaneous Acceleration is the derivative of velocity (\(a(t) = dv/dt\)), a vector.
• Constant Acceleration Kinematics: A set of five equations simplify motion analysis when acceleration is constant.
• Free Fall: A special case of constant acceleration where \(a = \pm g\) (acceleration due to gravity, \(9.8 \text{ m/s}^2\)).
• Calculus in Kinematics: Integration allows finding velocity from acceleration and position from velocity, especially when acceleration is not constant.

These points summarize the core concepts from this chapter. They represent the fundamental tools for analyzing motion along a straight line.

Key Equations

Equation	Description
\(\Delta x = x_f - x_0\)	Displacement
\(\Delta x_{Total} = \sum \Delta x_i\)	Total displacement
\(x_{Total} = \sum |\Delta x_i|\)	Total distance traveled
\(\bar{v} = \frac{\Delta x}{\Delta t}\)	Average velocity
\(v(t) = \frac{dx(t)}{dt}\)	Instantaneous velocity
\(\bar{s} = \frac{\text{Total distance}}{\text{Elapsed time}}\)	Average speed
Instantaneous speed = \(|v(t)|\)	Instantaneous speed
\(\bar{a} = \frac{\Delta v}{\Delta t}\)	Average acceleration
\(a(t) = \frac{dv(t)}{dt}\)	Instantaneous acceleration
\(\bar{v} = \frac{v_0 + v}{2}\)	Average velocity (constant acceleration)
\(v = v_0 + at\)	Final velocity (constant acceleration)
\(x = x_0 + v_0t + \frac{1}{2}at^2\)	Final position (constant acceleration)
\(v^2 = v_0^2 + 2a(x - x_0)\)	Final velocity from displacement (constant acceleration)
\(g = 9.8 \text{ m/s}^2\)	Acceleration due to gravity
\(v(t) = \int a(t)dt + C_1\)	Velocity from acceleration (integral)
\(x(t) = \int v(t)dt + C_2\)	Position from velocity (integral)

This table provides a concise reference for all the important equations introduced in Chapter 3.

Key Terms

Term	Definition
Acceleration	The rate at which an object’s velocity changes over time. A vector quantity.
Displacement	The change in an object’s position, a vector quantity representing the straight-line distance and direction from initial to final position.
Distance Traveled	The total length of the path an object has moved, a scalar quantity.
Frame of Reference	An arbitrary set of axes from which the position and motion of an object are described.
Free Fall	The motion of an object falling under the sole influence of gravity, neglecting air resistance and friction.
Gravity	The force of attraction between masses; on Earth, it causes objects to fall toward its center.
Instantaneous Acceleration	The acceleration of an object at a specific instant in time, the derivative of velocity with respect to time.
Instantaneous Velocity	The velocity of an object at a specific instant in time, the derivative of position with respect to time.
Position	The location of an object at any particular time, relative to a frame of reference.
Speed	The magnitude of velocity, a scalar quantity indicating how fast an object is moving.
Velocity	The rate at which an object’s position changes over time, a vector quantity (magnitude and direction).

Understanding these key terms is fundamental to comprehending the concepts of kinematics.